About Me
Posts tagged algorithm
I haven't really been writing much since I left grad school, but I figured it would be a shame not to capture this year, even if all I can muster is a brain dump and bulleted lists. It was my first year in NYC after all.
The biggest challenge for me has been finding my voice and asserting myself. The sheer number of people and small amount of space turns everything into a competition here. Upon moving here, I moused about always feeling like I was in everyone's way, but I am slowly learning to exercise my right to take up space. Everyone's attention is being pulled in different directions. Another challenge was having enough self-esteem to handled being constantly ghosted 👻 and not taking it personally.
Algorithms
I am ending the year with my Leetcode rating at 2164. Apparently that makes me top 1.25%. I've been aiming for 2300+, but I admittedly don't work too hard at this. I try my luck at a competition maybe once per week. Let me dump some new algorithms I wrote if I need to reuse them later.
Segment Tree
See 2407. Longest Increasing Subsequence II.
#include <algorithm>
#include <memory>
#include <utility>
#include <vector>
class Solution {
public:
int lengthOfLIS(const std::vector<int>& nums, int k);
};
namespace {
template <typename T>
struct Max {
constexpr T operator()(const T& a, const T& b) const {
return std::max(a, b);
}
};
template <typename K, typename V, typename Reducer = Max<V>>
class SegmentTree {
public:
// Disable default constructor.
SegmentTree() = delete;
// Disable copy (and move) semantics.
SegmentTree(const SegmentTree&) = delete;
SegmentTree<K, V, Reducer>& operator=(const SegmentTree&) = delete;
const std::pair<K, K>& segment() const { return segment_; }
const V& value() const { return value_; }
const SegmentTree<K, V, Reducer>& left() const { return *left_; }
const SegmentTree<K, V, Reducer>& right() const { return *right_; }
void Update(K key, V value) {
if (key < segment_.first || segment_.second < key) return;
if (segment_.first == segment_.second) {
value_ = value;
return;
}
left_->Update(key, value);
right_->Update(key, value);
value_ = reducer_(left_->value(), right_->value());
}
V Query(const std::pair<K, K>& segment, V out_of_range_value) {
if (segment.second < segment_.first || segment.first > segment_.second)
return out_of_range_value;
if (segment.first <= segment_.first && segment_.second <= segment.second) {
return value();
}
if (segment.second <= left_->segment().second)
return left_->Query(segment, out_of_range_value);
if (right_->segment().first <= segment.first)
return right_->Query(segment, out_of_range_value);
return reducer_(left_->Query(segment, out_of_range_value),
right_->Query(segment, out_of_range_value));
}
// Makes a segment tree from a range with `value` for leaf nodes.
template <typename It>
static std::unique_ptr<SegmentTree<K, V, Reducer>> Make(It begin, It end,
V value) {
std::pair<int, int> segment{*begin, *end};
if (begin == end)
return std::unique_ptr<SegmentTree<K, V, Reducer>>(
new SegmentTree<K, V, Reducer>(segment, value));
const auto mid = begin + (end - begin) / 2;
std::unique_ptr<SegmentTree<K, V, Reducer>> left =
SegmentTree<K, V, Reducer>::Make(begin, mid, value);
std::unique_ptr<SegmentTree<K, V, Reducer>> right =
SegmentTree<K, V, Reducer>::Make(std::next(mid), end, value);
return std::unique_ptr<SegmentTree<K, V, Reducer>>(
new SegmentTree<K, V, Reducer>(segment, std::move(left),
std::move(right)));
}
private:
// Leaf node.
SegmentTree(std::pair<K, K> segment, V value)
: segment_(segment), value_(value) {}
// Internal node.
SegmentTree(std::pair<K, K> segment,
std::unique_ptr<SegmentTree<K, V, Reducer>> left,
std::unique_ptr<SegmentTree<K, V, Reducer>> right)
: segment_(segment),
left_(std::move(left)),
right_(std::move(right)),
value_(reducer_(left_->value(), right_->value())) {}
const std::pair<K, K> segment_;
const std::unique_ptr<SegmentTree<K, V, Reducer>> left_;
const std::unique_ptr<SegmentTree<K, V, Reducer>> right_;
const Reducer reducer_;
V value_;
};
} // namespace
int Solution::lengthOfLIS(const std::vector<int>& nums, int k) {
std::vector<int> sorted_nums = nums;
std::sort(sorted_nums.begin(), sorted_nums.end());
sorted_nums.erase(std::unique(sorted_nums.begin(), sorted_nums.end()),
sorted_nums.end());
const std::unique_ptr<SegmentTree<int, int>> tree =
SegmentTree<int, int>::Make(sorted_nums.begin(), --sorted_nums.end(), 0);
for (int x : nums)
tree->Update(x, tree->Query(std::make_pair(x - k, x - 1), 0) + 1);
return tree->value();
}
Disjoint Set Forest
See 2421. Number of Good Paths.
#include <algorithm>
#include <cstddef>
#include <iostream>
#include <memory>
#include <numeric>
#include <unordered_map>
#include <vector>
class Solution {
public:
int numberOfGoodPaths(const std::vector<int>& vals,
const std::vector<std::vector<int>>& edges);
};
namespace phillypham {
class DisjointSetForest {
public:
explicit DisjointSetForest(size_t size) : parent_(size), rank_(size, 0) {
std::iota(parent_.begin(), parent_.end(), 0);
}
void Union(size_t x, size_t y) {
auto i = Find(x);
auto j = Find(y);
if (i == j) return;
if (rank_[i] > rank_[j]) {
parent_[j] = i;
} else {
parent_[i] = j;
if (rank_[i] == rank_[j]) rank_[j] = rank_[j] + 1;
}
}
size_t Find(size_t x) {
while (parent_[x] != x) {
auto parent = parent_[x];
parent_[x] = parent_[parent];
x = parent;
}
return x;
}
size_t size() const { return parent_.size(); }
private:
std::vector<size_t> parent_;
std::vector<size_t> rank_;
};
} // namespace phillypham
int Solution::numberOfGoodPaths(const std::vector<int>& vals,
const std::vector<std::vector<int>>& edges) {
const int n = vals.size();
std::vector<int> ordered_nodes(n);
std::iota(ordered_nodes.begin(), ordered_nodes.end(), 0);
std::sort(ordered_nodes.begin(), ordered_nodes.end(),
[&vals](int i, int j) -> bool { return vals[i] < vals[j]; });
std::vector<std::vector<int>> adjacency_list(n);
for (const auto& edge : edges)
if (vals[edge[0]] < vals[edge[1]])
adjacency_list[edge[1]].push_back(edge[0]);
else
adjacency_list[edge[0]].push_back(edge[1]);
phillypham::DisjointSetForest disjoint_set_forest(n);
int num_good_paths = 0;
for (int i = 0; i < n;) {
const int value = vals[ordered_nodes[i]];
int j = i;
for (; j < n && vals[ordered_nodes[j]] == value; ++j)
for (int k : adjacency_list[ordered_nodes[j]])
disjoint_set_forest.Union(ordered_nodes[j], k);
std::unordered_map<int, int> component_sizes;
for (int k = i; k < j; ++k)
++component_sizes[disjoint_set_forest.Find(ordered_nodes[k])];
for (const auto& [_, size] : component_sizes)
num_good_paths += size * (size + 1) / 2;
i = j;
}
return num_good_paths;
}
Books
I managed to read 12 books this year. Not too bad. Unfortunately, I didn't manage to capture what I took away in the moment, so I will try to dump it here with a few sentences for each book.
Sex and Vanity by Kevin Kwan
This was a fun novel in the Crazy Rich Asians universe. It's like a modern day society novel with social media. In classic Kevin Kwan style, it's filled with over-the-top descriptions of wealth. While mostly light-hearted and silly, the protagonist grapples with an identity crisis familiar to many Asian Americans.
Yolk by Mary H. K. Choi
This novel centers on two Korean-American sisters coming of age in NYC. They are foils: one went to Columbia and works at a hedge fund and the other studies art and dates losers. Both sisters are rather flawed and unlikeable and antiheroes in some sense. The description of the eating disorder is especially disturbing. Probably the most memorable quote for me was this gem:
Because I'm not built for this. I tried it. I did all that Tinder shit. Raya. Bumble. Whatever the fuck, Hinge. I thought maybe it was a good idea. I've had a girlfriend from the time I was fifteen. It's like in high school, Asian dudes were one thing, but a decade later it's like suddenly we're all hot. It was ridiculous. I felt like such a trope, like one of those tech bros who gets allcut up and gets Lasik and acts like a totally different person.
I've never been so accurately dragged in my life. Actually, who I am kidding, I was never hot enough to be having sex with strangers on dating apps.
House of Sticks by Ly Tran
As a child of Vietnamese war refugees, I thought this would be more relatable. It really wasn't at all, but it was still very good. Ly's family came over in different circumstances than mine and her parents never had the opportunity to get an education. I appreciated the different perspective of her life in Queens versus my Pennsylvania suburb. The complex relationship with her father who clearly suffers from PTSD and never really adjusts to life in America is most interesting. He wants to be a strong patriarchical figure for his family, but ultimately, he ends up alienating and hurting his daughter.
The Sympathizer by Viet Thanh Nguyen
I really enjoyed this piece of historical fiction. I have only really ever learned about the Vietnam war from the American perspective which has either been that it was necessary to stem communism or that it was a tragic mistake that killed millions of Vietnamese. But the narrator remarks:
Now that we are the powerful, we don’t need the French or the Americans to fuck us over. We can fuck ourselves just fine.
His experience of moving to America as an adult is also interesting. I was born here and the primary emotions about Vietnamese for me were shame and embarassement. The narrator arrives as an adult and while he has moments of shame and embarassement, he lets himself feel righteous anger.
I felt the ending and the focus on the idea of nothing captures how powerful and yet empty the ideals of communism and the war are.
The Committed by Viet Thanh Nguyen
The sequel to The Sympathizer takes place in France. I don't know if this novel really touches on any new themes. There is the contrast between the Asians and Arabs? But it is very funny, thrilling, and suspenseful as the narrator has to deal with consequences of his lies. Who doesn't enjoy poking fun at the hypocrisy of the French, too?
A Gentleman in Moscow by Amor Towles
This story is seemingly mundane as it takes place almost entirely in the confines of a hotel. But the Count's humor and descriptions of the dining experiences make this novel great fun. The Count teams up with loveable friends like his lover Anna and the barber to undermine the Bishop and the Party. I enjoyed the history lesson of this tumultuous period in Russia, too.
Being somewhat a student of the Russian greats like Tolstoy and Dostoevsky, I appreciated that he would occasionally make references to that body of work.
...the Russian masters could not compute up with a better plot device than two central characters resolving a matter of conscience by means of pistols at thirty-two paces...
Rules of Civility by Amor Towles
I've always enjoyed fiction from this era via Fitzgerald, Hemingway, and Edith Wharton. The modern take on the American society novel is refreshing. Old New York City and the old courtship rituals never fail to capture my imagination. There seems to be a lot to say about a women's place in society when marrying rich was the only option. The men seem to be yearning for a higher purpose than preserving dynastic wealth, too.
The Lincoln Highway by Amor Towles
A really charming coming of age story, where the imagination of teenage boys is brought to its logical conclusion. There is some story about redemption for past mistakes here. The headstrong Sally who has disgressions about doing thing the hard way and her supposed duty to make herself fit for marriage might be my favorite character, and I wish more was done with her. Abacus Abernathe lectures on the magic of New York City (it's just the right size) and deems this adventure worthy of the legends of the old world like Achilles.
I couldn't help but be touched and feeling like I am in on the joke with references to Chelsea Market (where I work) and characters in his previous novels.
Dune by Frank Herbert
I read this one because of the movie, but I didn't really enjoy it. Science fiction has never been of favorite of mine. I didn't like the messiah narrative and found Paul to be arrogant and unlikeable. But maybe that's the reality of playing politics.
The Left Hand of Darkness by Ursula K. Le Guin
This one was an interesting thought experiment about what would the world be like if we didn't have a concept of sex and gender. It makes you think how we immediately classify everyone we meet into a gender. The protagonist struggles with his need to feel masculine in a population with no concept of masculinity.
Bluebeard by Kurt Vonnegut
Rabo is a cranky old man. I wasn't able to identify any larger themes, but it's funny and Rabo's journey to make a meaningful work of art with soul is interesting itself. Abstract art is just too easy an target with its ridiculousness.
Maybe one useful tidbit is that motivation to write can be found by having an audience.
Travel
Somehow, I traveled to countries with breakout World Cup teams.
Croatia
Honestly, I don't know if I could have found Croatia on the map before I went there. It was surprisingly awesome. A beautiful country with great food. The people were proud and enjoyed opening up and sharing their country with us. I'll be back here to climb sometime, I hope.
Congratulations to them for their World Cup performance and adopting the euro.
Morocco
I don't know if I really enjoyed this one. There didn't seem to be much to do beyond shopping at the markets and seeing the historical sights. I liked the food, but I wasn't crazy about it.
Washington
The Enchantments were just breathtaking. The hike took us over 14 hours. Late in the night, I did begin to wonder if were going to make it, though, but it was worth it.
I also did my first multipitch (3 pitches!), here. Thanks Kyle for leading!
Recently, a problem from the USACO training pages has been bothering me. I had solved it years ago in Java, but my friend Robert Won challenged me to do in Python. Since Python is many times slower, this means my code has to be much smarter.
Problem
An arithmetic progression is a sequence of the form $a$, $a+b$, $a+2b$, $\ldots$, $a+nb$ where $n=0, 1, 2, 3, \ldots$. For this problem, $a$ is a non-negative integer and $b$ is a positive integer.
Write a program that finds all arithmetic progressions of length $n$ in the set $S$ of bisquares. The set of bisquares is defined as the set of all integers of the form $p^2 + q^2$ (where $p$ and $q$ are non-negative integers).
TIME LIMIT: 5 secs
PROGRAM NAME: ariprog
INPUT FORMAT
- Line 1: $N$ ($3 \leq N \leq 25$), the length of progressions for which to search
- Line 2: $M$ ($1 \leq M \leq 250$), an upper bound to limit the search to the bisquares with $0 \leq p,q \leq M$.
SAMPLE INPUT (file ariprog.in)
5
7
OUTPUT FORMAT
If no sequence is found, a single line reading NONE. Otherwise, output one or more lines, each with two integers: the first element in a found sequence and the difference between consecutive elements in the same sequence. The lines should be ordered with smallest-difference sequences first and smallest starting number within those sequences first.
There will be no more than 10,000 sequences.
SAMPLE OUTPUT (file ariprog.out)
1 4
37 4
2 8
29 8
1 12
5 12
13 12
17 12
5 20
2 24
Dynamic Programming Solution
My initial solution that I translated from C++ to Python was not fast enough. I wrote a new solution that I thought was clever. We iterate over all possible deltas, and for each delta, we use dynamic programming to find the longest sequence with that delta.
def find_arithmetic_progressions(N, M):
is_bisquare = [False] * (M * M + M * M + 1)
bisquare_indices = [-1] * (M * M + M * M + 1)
bisquares = []
for p in range(0, M + 1):
for q in range(p, M + 1):
x = p * p + q * q
if is_bisquare[x]: continue
is_bisquare[x] = True
bisquares.append(x)
bisquares.sort()
for i, bisquare in enumerate(bisquares):
bisquare_indices[bisquare] = i
sequences, i = [], 0
for delta in range(1, bisquares[-1] // (N - 1) + 1):
sequence_lengths = [1] * len(bisquares)
while bisquares[i] < delta: i += 1
for x in bisquares[i:]:
previous_idx = bisquare_indices[x - delta]
if previous_idx == -1: continue
idx, sequence_length = bisquare_indices[x], sequence_lengths[previous_idx] + 1
sequence_lengths[idx] = sequence_length
if sequence_length >= N:
sequences.append((delta, x - (N - 1) * delta))
return sequences
Too slow!
Executing...
Test 1: TEST OK [0.011 secs, 9352 KB]
Test 2: TEST OK [0.011 secs, 9352 KB]
Test 3: TEST OK [0.011 secs, 9168 KB]
Test 4: TEST OK [0.011 secs, 9304 KB]
Test 5: TEST OK [0.031 secs, 9480 KB]
Test 6: TEST OK [0.215 secs, 9516 KB]
Test 7: TEST OK [2.382 secs, 9676 KB]
> Run 8: Execution error: Your program (`ariprog') used more than
the allotted runtime of 5 seconds (it ended or was stopped at
5.242 seconds) when presented with test case 8. It used 12948 KB
of memory.
------ Data for Run 8 [length=7 bytes] ------
22
250
----------------------------
Test 8: RUNTIME 5.242>5 (12948 KB)
I managed to put my mathematics background to good use here: $p^2 + q^2 \not\equiv 3 \pmod 4$ and $p^2 + q^2 \not\equiv 6 \pmod 8$. This means that a bisquare arithmetic progression with more than 3 elements must have delta divisible by 4. If $b \equiv 1 \pmod 4$ or $b \equiv 3 \pmod 4$, there would have to be a bisquare $p^2 + q^2 \equiv 3 \pmod 4$, which is impossible. If $b \equiv 2 \pmod 4$, there would be have to be $p^2 + q^2 \equiv 6 \pmod 8$, which is also impossible.
This optimization makes it fast, enough.
def find_arithmetic_progressions(N, M):
is_bisquare = [False] * (M * M + M * M + 1)
bisquare_indices = [-1] * (M * M + M * M + 1)
bisquares = []
for p in range(0, M + 1):
for q in range(p, M + 1):
x = p * p + q * q
if is_bisquare[x]: continue
is_bisquare[x] = True
bisquares.append(x)
bisquares.sort()
for i, bisquare in enumerate(bisquares):
bisquare_indices[bisquare] = i
sequences, i = [], 0
for delta in (range(1, bisquares[-1] // (N - 1) + 1) if N == 3 else
range(4, bisquares[-1] // (N - 1) + 1, 4)):
sequence_lengths = [1] * len(bisquares)
while bisquares[i] < delta: i += 1
for x in bisquares[i:]:
previous_idx = bisquare_indices[x - delta]
if previous_idx == -1: continue
idx, sequence_length = bisquare_indices[x], sequence_lengths[previous_idx] + 1
sequence_lengths[idx] = sequence_length
if sequence_length >= N:
sequences.append((delta, x - (N - 1) * delta))
return sequences
Executing...
Test 1: TEST OK [0.010 secs, 9300 KB]
Test 2: TEST OK [0.011 secs, 9368 KB]
Test 3: TEST OK [0.015 secs, 9248 KB]
Test 4: TEST OK [0.014 secs, 9352 KB]
Test 5: TEST OK [0.045 secs, 9340 KB]
Test 6: TEST OK [0.078 secs, 9464 KB]
Test 7: TEST OK [0.662 secs, 9756 KB]
Test 8: TEST OK [1.473 secs, 9728 KB]
Test 9: TEST OK [1.313 secs, 9740 KB]
All tests OK.
Even Faster!
Not content to merely pass, I wanted to see if we could pass all test cases with less than 1 second (time limit was 5 seconds). Indeed, we can. The solution in the official analysis take advantage of the fact that the sequence length is short. The dynamic programming optimization is not that helpful. It's better to optimize for traversing the bisquares less. Instead, we take pairs of bisquares carefully: we break out when the delta is too big. The official solution has some inefficiencies like using a hash map. If we instead use indexed array lookups, we can be very fast.
def find_arithmetic_progressions(N, M):
is_bisquare = [False] * (M * M + M * M + 1)
bisquares = []
for p in range(0, M + 1):
for q in range(p, M + 1):
x = p * p + q * q
if is_bisquare[x]: continue
is_bisquare[x] = True
bisquares.append(x)
bisquares.sort()
sequences = []
for i in reversed(range(len(bisquares))):
x = bisquares[i]
max_delta = x // (N - 1)
for j in reversed(range(i)):
y = bisquares[j]
delta = x - y
if delta > max_delta: break
if N > 3 and delta % 4 != 0: continue
z = x - (N - 1) * delta
while y > z and is_bisquare[y - delta]: y -= delta
if z == y: sequences.append((delta, z))
sequences.sort()
return sequences
Executing...
Test 1: TEST OK [0.013 secs, 9280 KB]
Test 2: TEST OK [0.012 secs, 9284 KB]
Test 3: TEST OK [0.013 secs, 9288 KB]
Test 4: TEST OK [0.012 secs, 9208 KB]
Test 5: TEST OK [0.018 secs, 9460 KB]
Test 6: TEST OK [0.051 secs, 9292 KB]
Test 7: TEST OK [0.421 secs, 9552 KB]
Test 8: TEST OK [0.896 secs, 9588 KB]
Test 9: TEST OK [0.786 secs, 9484 KB]
All tests OK.
Yay!
I finally did it! The one-armed pull up. There's a bit of a kip, but I'll take it.
Despite my best years being behind me, it's still good to see progress. I have somewhat dispensed with the fake humility. As I find myself getting older, I appreciate clarity and directness more and being honest with myself. I am proud of this whether anyone else cares or not, so I am going to share it. Now, if I could only actually climb.
A lot of taking myself less seriously comes from the daily humiliations brought about by life as 30-something bachelor in NYC. Being stood up and ghosted is just par for the week now. But I am learning to understand that some people just need space to deal with their problems, and some people are just terrible, too.
Another area where I've grown in is how I write code. In Minimum Spanning Trees and the Least Common Ancestor Problem, I described a binary lifting algorithm to find the least common ancestor. I tried way too hard to impress and now I can't figure out what I wrote.
I recently needed this algorithm again in 2322. Minimum Score After Removals on a Tree . If I had a simple implementation that I could understand and reuse, I may have been able to solve it during the contest, so now, I rewrote it.
#include <algorithm>
#include <limits>
#include <vector>
class Solution {
public:
int minimumScore(const std::vector<int>& nums,
const std::vector<std::vector<int>>& edges);
};
namespace {
void RootTree(const std::vector<std::vector<int>>& adjacency_list, int root,
std::vector<int>* depths, std::vector<int>* parents,
std::vector<int>* xors) {
for (auto child : adjacency_list[root]) {
if (child == (*parents)[root]) continue;
(*depths)[child] = (*depths)[root] + 1;
(*parents)[child] = root;
RootTree(adjacency_list, child, depths, parents, xors);
(*xors)[root] ^= (*xors)[child];
}
}
void BuildAncestors(std::vector<std::vector<int>>* ancestors) {
bool built = true;
for (auto& as : *ancestors) {
const int a = as.back();
if (a == -1) continue;
const int j = as.size() - 1;
as.push_back(j < (*ancestors)[a].size() ? (*ancestors)[a][j] : -1);
built = false;
}
if (!built) BuildAncestors(ancestors);
}
int LeastCommonAncestor(const std::vector<std::vector<int>>& ancestors,
const std::vector<int>& depth, int u, int v) {
if (u == v) return u;
int delta = depth[v] - depth[u];
if (delta == 0) {
int j = 0;
while (ancestors[u][j + 1] != -1 &&
ancestors[u][j + 1] != ancestors[v][j + 1])
++j;
return LeastCommonAncestor(ancestors, depth, ancestors[u][j],
ancestors[v][j]);
}
if (delta < 0) {
std::swap(u, v);
delta = -delta;
}
int j = 0;
while (1 << (j + 1) <= delta) ++j;
return LeastCommonAncestor(ancestors, depth, u, ancestors[v][j]);
}
} // namespace
int Solution::minimumScore(const std::vector<int>& nums,
const std::vector<std::vector<int>>& edges) {
const int n = nums.size();
// Root the tree at 0 and compute properties based on that.
std::vector<std::vector<int>> adjacency_list(n);
for (const auto& edge : edges) {
adjacency_list[edge.front()].push_back(edge.back());
adjacency_list[edge.back()].push_back(edge.front());
}
std::vector<int> depths(n, 0);
std::vector<int> parents(n, -1);
std::vector<int> xors = nums;
RootTree(adjacency_list, 0, &depths, &parents, &xors);
// ancestors[i][j] is the 2^j ancestor of node i.
std::vector<std::vector<int>> ancestors;
ancestors.reserve(n);
for (int i = 0; i < n; ++i) ancestors.push_back({parents[i]});
BuildAncestors(&ancestors);
// Determine minimum score by looping over all pairs of edges.
int min_score = std::numeric_limits<int>::max();
for (int i = 1; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
const int u = depths[i] <= depths[j] ? i : j;
const int v = depths[i] <= depths[j] ? j : i;
const int ancestor = LeastCommonAncestor(ancestors, depths, u, v);
const int xor0 =
ancestor == u ? xors[0] ^ xors[u] : xors[0] ^ xors[u] ^ xors[v];
const int xor1 = ancestor == u ? xors[u] ^ xors[v] : xors[u];
const int xor2 = xors[v];
min_score = std::min(min_score, std::max(xor0, std::max(xor1, xor2)) -
std::min(xor0, std::min(xor1, xor2)));
}
}
return min_score;
}
The only question that really stumped me during my Google interviews was The Skyline Problem. I remember only being able to write some up a solution in pseudocode after being given many hints before my time was up.
It's been banned for some time now, so I thought I'd dump the solution here. Maybe, I will elaborate and clean up the code some other time. It's one of the cleverest uses of an ordered map (usually implemented as a tree map) that I've seen.
#include <algorithm>
#include <iostream>
#include <map>
#include <sstream>
#include <utility>
#include <vector>
using namespace std;
namespace {
struct Wall {
enum Type : int {
LEFT = 1,
RIGHT = 0
};
int position;
int height;
Type type;
Wall(int position, int height, Wall::Type type) :
position(position), height(height), type(type) {}
bool operator<(const Wall &other) {
return position < other.position;
}
};
ostream& operator<<(ostream& stream, const Wall &w) {
return stream << "Position: " << to_string(w.position) << ';'
<< " Height: " << to_string(w.height) << ';'
<< " Type: " << (w.type == Wall::Type::LEFT ? "Left" : "Right");
}
} // namespace
class Solution {
public:
vector<vector<int>> getSkyline(vector<vector<int>>& buildings) {
vector<Wall> walls;
for (const vector<int>& building : buildings) {
walls.emplace_back(building[0], building[2], Wall::Type::LEFT);
walls.emplace_back(building[1], building[2], Wall::Type::RIGHT);
}
sort(walls.begin(), walls.end());
vector<vector<int>> skyline;
map<int, int> heightCount;
for (vector<Wall>::const_iterator wallPtr = walls.cbegin(); wallPtr != walls.cend();) {
int currentPosition = wallPtr -> position;
do {
if (wallPtr -> type == Wall::Type::LEFT) {
++heightCount[wallPtr -> height];
} else if (wallPtr -> type == Wall::Type::RIGHT) {
if (--heightCount[wallPtr -> height] == 0) {
heightCount.erase(wallPtr -> height);
}
}
++wallPtr;
} while (wallPtr != walls.cend() && wallPtr -> position == currentPosition);
if (skyline.empty() || heightCount.empty() ||
heightCount.crbegin() -> first != skyline.back()[1]) {
skyline.emplace_back(vector<int>{
currentPosition, heightCount.empty() ? 0 : heightCount.crbegin() -> first});
}
}
return skyline;
}
};
The easiest way to detect cycles in a linked list is to put all the seen nodes into a set and check that you don't have a repeat as you traverse the list. This unfortunately can blow up in memory for large lists.
Floyd's Tortoise and Hare algorithm gets around this by using two points that iterate through the list at different speeds. It's not immediately obvious why this should work.
/*
* For your reference:
*
* SinglyLinkedListNode {
* int data;
* SinglyLinkedListNode* next;
* };
*
*/
namespace {
template <typename Node>
bool has_cycle(const Node* const tortoise, const Node* const hare) {
if (tortoise == hare) return true;
if (hare->next == nullptr || hare->next->next == nullptr) return false;
return has_cycle(tortoise->next, hare->next->next);
}
} // namespace
bool has_cycle(SinglyLinkedListNode* head) {
if (head == nullptr ||
head->next == nullptr ||
head->next->next == nullptr) return false;
return has_cycle(head, head->next->next);
}
The above algorithm solves HackerRank's Cycle Detection.
To see why this work, consider a cycle that starts at index $\mu$ and has length $l$. If there is a cycle, we should have $x_i = x_j$ for some $i,j \geq \mu$ and $i \neq j$. This should occur when \begin{equation} i - \mu \equiv j - \mu \pmod{l}. \label{eqn:cond} \end{equation}
In the tortoise and hare algorithm, the tortoise moves with speed 1, and the hare moves with speed 2. Let $i$ be the location of the tortoise. Let $j$ be the location of the hare.
The cycle starts at $\mu$, so the earliest that we could see a cycle is when $i = \mu$. Then, $j = 2\mu$. Let $k$ be the number of steps we take after $i = \mu$. We'll satisfy Equation \ref{eqn:cond} when \begin{align*} i - \mu \equiv j - \mu \pmod{l} &\Leftrightarrow \left(\mu + k\right) - \mu \equiv \left(2\mu + 2k\right) - \mu \pmod{l} \\ &\Leftrightarrow k \equiv \mu + 2k \pmod{l} \\ &\Leftrightarrow 0 \equiv \mu + k \pmod{l}. \end{align*}
This will happen for some $k \leq l$, so the algorithm terminates within $\mu + k$ steps if there is a cycle. Otherwise, if there is no cycle the algorithm terminates when it reaches the end of the list.
Consider the problem Overrandomized. Intuitively, one can see something like Benford's law. Indeed, counting the leading digit works:
#include <algorithm>
#include <iostream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
string Decode() {
unordered_map<char, int> char_counts; unordered_set<char> chars;
for (int i = 0; i < 10000; ++i) {
long long Q; string R; cin >> Q >> R;
char_counts[R[0]]++;
for (char c : R) chars.insert(c);
}
vector<pair<int, char>> count_chars;
for (const pair<char, int>& char_count : char_counts) {
count_chars.emplace_back(char_count.second, char_count.first);
}
sort(count_chars.begin(), count_chars.end());
string code;
for (const pair<int, char>& count_char : count_chars) {
code += count_char.second;
chars.erase(count_char.second);
}
code += *chars.begin();
reverse(code.begin(), code.end());
return code;
}
int main(int argc, char *argv[]) {
ios::sync_with_stdio(false); cin.tie(NULL);
int T; cin >> T;
for (int t = 1; t <= T; ++t) {
int U; cin >> U;
cout << "Case #" << t << ": " << Decode() << '\n';
}
cout << flush;
return 0;
}
Take care to read Q as a long long because it can be large.
It occurred to me that there's no reason the logarithms of the randomly generated numbers should be uniformly distributed, so I decided to look into this probability distribution closer. Let $R$ be the random variable representing the return value of a query.
\begin{align*} P(R = r) &= \sum_{m = r}^{10^U - 1} P(M = m, R = r) \\ &= \sum_{m = r}^{10^U - 1} P(R = r \mid M = m)P(M = m) \\ &= \frac{1}{10^U - 1}\sum_{m = r}^{10^U - 1} \frac{1}{m}. \end{align*} since $P(M = m) = 1/(10^U - 1)$ for all $m$.
The probability that we get a $k$ digit number that starts with a digit $d$ is then \begin{align*} P(d \times 10^{k-1} \leq R < (d + 1) \times 10^{k-1}) &= \frac{1}{10^U - 1} \sum_{r = d \times 10^{k-1}}^{(d + 1) \times 10^{k-1} - 1} \sum_{m = r}^{10^U - 1} \frac{1}{m}. \end{align*}
Here, you can already see that for a fixed $k$, smaller $d$s will have more terms, so they should occur as leading digits with higher probability. It's interesting to try to figure out how much more frequently this should happen, though. To get rid of the summation, we can use integrals! This will make the computation tractable for large $k$ and $U$. Here, I start dropping the $-1$s in the approximations.
\begin{align*} P\left(d \times 10^{k-1} \leq R < (d + 1) \times 10^{k-1}\right) &= \frac{1}{10^U - 1} \sum_{r = d \times 10^{k-1}}^{(d + 1) \times 10^{k-1} - 1} \sum_{m = r}^{10^U - 1} \frac{1}{m} \\ &\approx \frac{1}{10^U} \sum_{r = d \times 10^{k-1}}^{(d + 1) \times 10^{k-1} - 1} \left[\log 10^U - \log r \right] \\ &=\frac{10^{k - 1}}{10^{U}}\left[ U\log 10 - \frac{1}{10^{k - 1}}\sum_{r = d \times 10^{k-1}}^{(d + 1) \times 10^{k-1} - 1} \log r \right]. \end{align*}
Again, we can apply integration. Using integration by parts, we have $\int_a^b x \log x \,dx = b\log b - b - \left(a\log a - a\right)$, so \begin{align*} \sum_{r = d \times 10^{k-1}}^{(d + 1) \times 10^{k-1} - 1} \log r &\approx 10^{k-1}\left[ (k - 1)\log 10 + (d + 1) \log (d + 1) - d \log d - 1 \right]. \end{align*}
Substituting, we end up with \begin{align*} P&\left(d \times 10^{k-1} \leq R < (d + 1) \times 10^{k-1}\right) \approx \\ &\frac{1}{10^{U - k + 1}}\left[ 1 + (U - k + 1)\log 10 - \left[(d + 1) \log(d+1) - d\log d\right] \right]. \end{align*}
We can make a few observations. Numbers with lots of digits are more likely to occur since for larger $k$, the denominator is much smaller. This makes sense: there are many more large numbers than small numbers. Independent of $k$, if $d$ is larger, the quantity inside the inner brackets is larger since $x \log x$ is convex, so the probability decreases with $d$. Thus, smaller digits occur more frequently. While the formula follows the spirit of Benford's law, the formula is not quite the same.
This was the first time I had to use integrals for a competitive programming problem!
One of my favorite things about personal coding projects is that you're free to over-engineer and prematurely optimize your code to your heart's content. Production code written in a shared code based needs to be maintained, and hence, should favor simplicity and readability. For personal projects, I optimize for fun, and what could be more fun than elaborate abstractions, unnecessary optimizations, and abusing recursion?
To that end, I present my solution to the Google Code Jam 2019 Round 1A problem, Alien Rhyme.
In this problem, we maximize the number of pairs of words that could possibly rhyme. I guess this problem has some element of realism as it's similar in spirit to using frequency analysis to decode or identify a language.
After reversing the strings, this problem reduces to greedily taking pairs of words with the longest common prefix. Each time we select a prefix, we update the sizes of the remaining prefixes. If where are $N$ words, this algorithm is $O\left(N^2\right)$ and can be implemented with a linked list in C++:
// Reverses and sorts suffixes to make finding common longest common suffix easier.
vector<string> NormalizeSuffixes(const vector<string>& words) {
vector<string> suffixes; suffixes.reserve(words.size());
for (const string& word : words) {
suffixes.push_back(word);
reverse(suffixes.back().begin(), suffixes.back().end());
}
sort(suffixes.begin(), suffixes.end());
return suffixes;
}
int CountPrefix(const string &a, const string &b) {
int size = 0;
for (int i = 0; i < min(a.length(), b.length()); ++i)
if (a[i] == b[i]) { ++size; } else { break; }
return size;
}
int MaximizePairs(const vector<string>& words) {
const vector<string> suffixes = NormalizeSuffixes(words);
// Pad with zeros: pretend there are empty strings at the beginning and end.
list<int> prefix_sizes{0};
for (int i = 1; i < suffixes.size(); ++i)
prefix_sizes.push_back(CountPrefix(suffixes[i - 1], suffixes[i]));
prefix_sizes.push_back(0);
// Count the pairs by continually finding the longest common prefix.
list<int>::iterator max_prefix_size;
while ((max_prefix_size = max_element(prefix_sizes.begin(), prefix_sizes.end())) !=
prefix_sizes.begin()) {
// Claim this prefix and shorten the other matches.
while (*next(max_prefix_size) == *max_prefix_size) {
--(*max_prefix_size);
++max_prefix_size;
}
// Use transitivity to update the common prefix size.
*next(max_prefix_size) = min(*prev(max_prefix_size), *next(max_prefix_size));
prefix_sizes.erase(prefix_sizes.erase(prev(max_prefix_size)));
}
return suffixes.size() - (prefix_sizes.size() - 1);
}
A single file example can be found on GitHub. Since $N \leq 1000$ in this problem, this solution is more than adequate.
Asymptotically Optimal Solution
We can use the fact that the number of characters in each word $W$ is at most 50 and obtain a $O\left(N\max\left(\log N, W\right)\right)$ solution.
Induction
Suppose we have a tree where each node is a prefix (sometimes called a trie). In the worst case, each prefix will have a single character. The title image shows such a tree for the words: PREFIX, PRELIM, PROF, SUFFER, SUFFIX, SUM, SWIFT, SWIFTER, SWOLE.
Associated with each node is a count of how many words have that prefix as a maximal prefix. The depth of each node is the sum of the traversed prefix sizes.
The core observation is that at any given node, any words in the subtree can have a common prefix with length at least the depth of the node. Greedily selecting the longest common prefixes corresponds to pairing all possible prefixes in a subtree with length greater than the depth of the parent. The unused words can then be used higher up in the tree to make additional prefixes. Tree algorithms are best expressed recursively. Here's the Swift code.
func maximizeTreePairs<T: Collection>(
root: Node<T>, depth: Int, minPairWordCount: Int) -> (used: Int, unused: Int)
where T.Element: Hashable {
let (used, unused) = root.children.reduce(
(used: 0, unused: root.count),
{
(state: (used: Int, unused: Int), child) -> (used: Int, unused: Int) in
let childState = maximizeTreePairs(
root: child.value, depth: child.key.count + depth, minPairWordCount: depth)
return (state.used + childState.used, state.unused + childState.unused)
})
let shortPairUsed = min(2 * (depth - minPairWordCount), (unused / 2) * 2)
return (used + shortPairUsed, unused - shortPairUsed)
}
func maximizePairs(_ words: [String]) -> Int {
let suffixes = normalizeSuffixes(words)
let prefixTree = compress(makePrefixTree(suffixes))
return prefixTree.children.reduce(
0, { $0 + maximizeTreePairs(
root: $1.value, depth: $1.key.count, minPairWordCount: 0).used })
}
Since the tree has maximum depth $W$ and there are $N$ words, recursing through the tree is $O\left(NW\right)$.
Making the Prefix Tree
The simplest way to construct a prefix tree is to start at the root for each word and character-by-character descend into the tree, creating any nodes necessary. Update the count of the node when reaching the end of the word. This is $O\left(NW\right)$.
As far as I know, the wost case will always be $O\left(NW\right)$. In practice, though, if there are many words with lengthy shared common prefixes we can avoid retracing paths through the tree. In our example, consider SWIFT and SWIFTER. If we naively construct a tree, we will need to traverse through $5 + 7 = 12$ nodes. But if we insert our words in lexographic order, we don't need to retrace the first 5 characters and simply only need to traverse 7 nodes.
Swift has somewhat tricky value semantics. structs are always copied, so we need to construct this tree recursively.
func makePrefixTree<T: StringProtocol>(_ words: [T]) -> Node<T.Element> {
let prefixCounts = words.reduce(
into: (counts: [0], word: "" as T),
{
$0.counts.append(countPrefix($0.word, $1))
$0.word = $1
}).counts
let minimumPrefixCount = MinimumRange(prefixCounts)
let words = [""] + words
/// Inserts `words[i]` into a rooted tree.
///
/// - Parameters:
/// - root: The root node of the tree.
/// - state: The index of the word for the current path and depth of `root`.
/// - i: The index of the word to be inserted.
/// - Returns: The index of the next word to be inserted.
func insert(_ root: inout Node<T.Element>,
_ state: (node: Int, depth: Int),
_ i: Int) -> Int {
// Start inserting only for valid indices and at the right depth.
if i >= words.count { return i }
// Max number of nodes that can be reused for `words[i]`.
let prefixCount = state.node == i ?
prefixCounts[i] : minimumPrefixCount.query(from: state.node + 1, through: i)
// Either (a) inserting can be done more efficiently at a deeper node;
// or (b) we're too deep in the wrong state.
if prefixCount > state.depth || (prefixCount < state.depth && state.node != i) { return i }
// Start insertion process! If we're at the right depth, insert and move on.
if state.depth == words[i].count {
root.count += 1
return insert(&root, (i, state.depth), i + 1)
}
// Otherwise, possibly create a node and traverse deeper.
let key = words[i][words[i].index(words[i].startIndex, offsetBy: state.depth)]
if root.children[key] == nil {
root.children[key] = Node<T.Element>(children: [:], count: 0)
}
// After finishing traversal insert the next word.
return insert(
&root, state, insert(&root.children[key]!, (i, state.depth + 1), i))
}
var root = Node<T.Element>(children: [:], count: 0)
let _ = insert(&root, (0, 0), 1)
return root
}
While the naive implementation of constructing a trie would involve $48$ visits to a node (the sum over the lengths of each word), this algorithm does it in $28$ visits as seen in the title page. Each word insertion has its edges colored separately in the title image.
Now, for this algorithm to work efficiently, it's necessary to start inserting the next word at the right depth, which is the size of longest prefix that the words share.
Minimum Range Query
Computing the longest common prefix of any two words reduces to a minimum range query. If we order the words lexographically, we can compute the longest common prefix size between adjacent words. The longest common prefix size of two words $i$ and $j$, where $i < j$ is then:
\begin{equation} \textrm{LCP}(i, j) = \min\left\{\textrm{LCP}(i, i + 1), \textrm{LCP}(i + 1, i + 2), \ldots, \textrm{LCP}(j - 1, j)\right\}. \end{equation}
A nice dynamic programming $O\left(N\log N\right)$ algorithm exists to precompute such queries that makes each query $O\left(1\right)$.
We'll $0$-index to make the math easier to translate into code. Given an array $A$ of size $N$, let
\begin{equation} P_{i,j} = \min\left\{A_k : i \leq k < i + 2^{j} - 1\right\}. \end{equation}
Then, we can write $\mathrm{LCP}\left(i, j - 1\right) = \min\left(P_{i, l}, P_{j - 2^l, l}\right)$, where $l = \max\left\{l : l \in \mathbb{Z}, 2^l \leq j - i\right\}$ since $\left([i, i + 2^l) \cup [j - 2^l, j)\right) \cap \mathbb{Z} = \left\{i, i + 1, \ldots , j - 1\right\}$.
$P_{i,0}$ can be initialized $P_{i,0} = A_{i}$, and for $j > 0$, we can have
\begin{equation} P_{i,j} = \begin{cases} \min\left(P_{i, j - 1}, P_{i + 2^{j - 1}, j - 1}\right) & i + 2^{j -1} < N; \\ P_{i, j - 1} & \text{otherwise}. \\ \end{cases} \end{equation}
See a Swift implementation.
struct MinimumRange<T: Collection> where T.Element: Comparable {
private let memo: [[T.Element]]
private let reduce: (T.Element, T.Element) -> T.Element
init(_ collection: T,
reducer reduce: @escaping (T.Element, T.Element) -> T.Element = min) {
let k = collection.count
var memo: [[T.Element]] = Array(repeating: [], count: k)
for (i, element) in collection.enumerated() { memo[i].append(element) }
for j in 1..<(k.bitWidth - k.leadingZeroBitCount) {
let offset = 1 << (j - 1)
for i in 0..<memo.count {
memo[i].append(
i + offset < k ?
reduce(memo[i][j - 1], memo[i + offset][j - 1]) : memo[i][j - 1])
}
}
self.memo = memo
self.reduce = reduce
}
func query(from: Int, to: Int) -> T.Element {
let (from, to) = (max(from, 0), min(to, memo.count))
let rangeCount = to - from
let bitShift = rangeCount.bitWidth - rangeCount.leadingZeroBitCount - 1
let offset = 1 << bitShift
return self.reduce(self.memo[from][bitShift], self.memo[to - offset][bitShift])
}
func query(from: Int, through: Int) -> T.Element {
return query(from: from, to: through + 1)
}
}
Path Compression
Perhaps my most unnecessary optimization is path compression, especially since we only traverse the tree once in the induction step. If we were to traverse the tree multiple times, it might be worth it, however. This optimization collapses count $0$ nodes with only $1$ child into its parent.
/// Use path compression. Not necessary, but it's fun!
func compress(_ uncompressedRoot: Node<Character>) -> Node<String> {
var root = Node<String>(
children: [:], count: uncompressedRoot.count)
for (key, node) in uncompressedRoot.children {
let newChild = compress(node)
if newChild.children.count == 1, newChild.count == 0,
let (childKey, grandChild) = newChild.children.first {
root.children[String(key) + childKey] = grandChild
} else {
root.children[String(key)] = newChild
}
}
return root
}
Full Code Example
A full example with everything wired together can be found on GitHub.
GraphViz
Also, if you're interested in the graph in the title image, I used GraphViz. It's pretty neat. About a year ago, I made a trivial commit to the project: https://gitlab.com/graphviz/graphviz/commit/1cc99f32bb1317995fb36f215fb1e69f96ce9fed.
digraph {
rankdir=LR;
// SUFFER
S1 [label="S"];
F3 [label="F"];
F4 [label="F"];
E2 [label="E"];
R2 [label="R"];
S1 -> U [label=12, color="#984ea3"];
U -> F3 [label=13, color="#984ea3"];
F3 -> F4 [label=14, color="#984ea3"];
F4 -> E2 [label=15, color="#984ea3"];
E2 -> R2 [label=16, color="#984ea3"];
// PREFIX
E1 [label="E"];
R1 [label="R"];
F1 [label="F"];
I1 [label="I"];
X1 [label="X"];
P -> R1 [label=1, color="#e41a1c"];
R1 -> E1 [label=2, color="#e41a1c"];
E1 -> F1 [label=3, color="#e41a1c"];
F1 -> I1 [label=4, color="#e41a1c"];
I1 -> X1 [label=5, color="#e41a1c"];
// PRELIM
L1 [label="L"];
I2 [label="I"];
M1 [label="M"];
E1 -> L1 [label=6, color="#377eb8"];
L1 -> I2 [label=7, color="#377eb8"];
I2 -> M1 [label=9, color="#377eb8"];
// PROF
O1 [label="O"];
F2 [label="F"];
R1 -> O1 [label=10, color="#4daf4a"];
O1 -> F2 [label=11, color="#4daf4a"];
// SUFFIX
I3 [label="I"];
X2 [label="X"];
F4 -> I3 [label=17, color="#ff7f00"];
I3 -> X2 [label=18, color="#ff7f00"];
// SUM
M2 [label="M"];
U -> M2 [label=19, color="#ffff33"];
// SWIFT
I4 [label="I"];
F5 [label="F"];
T1 [label="T"];
S1 -> W [label=20, color="#a65628"];
W -> I4 [label=21, color="#a65628"];
I4 -> F5 [label=22, color="#a65628"];
F5 -> T1 [label=23, color="#a65628"];
// SWIFTER
E3 [label="E"];
R3 [label="R"];
T1 -> E3 [label=24, color="#f781bf"];
E3 -> R3 [label=25, color="#f781bf"];
// SWOLE
O2 [label="O"];
L2 [label="L"];
E4 [label="E"];
W -> O2 [label=26, color="#999999"];
O2 -> L2 [label=27, color="#999999"];
L2 -> E4 [label=28, color="#999999"];
}
In general, the Hamiltonian path problem is NP-complete. But in some special cases, polynomial-time algorithms exists.
One such case is in Pylons, the Google Code Jam 2019 Round 1A problem. In this problem, we are presented with a grid graph. Each cell is a node, and a node is connected to every other node except those along its diagonals, in the same column, or in the same row. In the example, from the blue cell, we can move to any other cell except the red cells.
If there are $N$ cells, an $O\left(N^2\right)$ is to visit the next available cell with the most unavailable, unvisited cells. Why? We should visit those cells early because if we wait too long, we will become stuck at those cells when we inevitably need to visit them.
I've implemented the solution in findSequence (see full Swift solution on GitHub):
func findSequence(N: Int, M: Int) -> [(row: Int, column: Int)]? {
// Other cells to which we are not allowed to jump.
var badNeighbors: [Set<Int>] = Array(repeating: Set(), count: N * M)
for i in 0..<(N * M) {
let (ri, ci) = (i / M, i % M)
for j in 0..<(N * M) {
let (rj, cj) = (j / M, j % M)
if ri == rj || ci == cj || ri - ci == rj - cj || ri + ci == rj + cj {
badNeighbors[i].insert(j)
badNeighbors[j].insert(i)
}
}
}
// Greedily select the cell which has the most unallowable cells.
var sequence: [(row: Int, column: Int)] = []
var visited: Set<Int> = Set()
while sequence.count < N * M {
guard let i = (badNeighbors.enumerated().filter {
if visited.contains($0.offset) { return false }
guard let (rj, cj) = sequence.last else { return true }
let (ri, ci) = ($0.offset / M, $0.offset % M)
return rj != ri && cj != ci && rj + cj != ri + ci && rj - cj != ri - ci
}.reduce(nil) {
(state: (i: Int, count: Int)?, value) -> (i: Int, count: Int)? in
if let count = state?.count, count > value.element.count { return state }
return (i: value.offset, count: value.element.count)
}?.i) else { return nil }
sequence.append((row: i / M, column: i % M))
visited.insert(i)
for j in badNeighbors[i] { badNeighbors[j].remove(i) }
}
return sequence
}
The solution returns nil when no path exists.
Why this work hinges on the fact that no path is possible if there are $N_t$ remaining nodes and you visit a non-terminal node $x$ which has $N_t - 1$ unavailable neighbors. There must have been some node $y$ that was visited that put node $x$ in this bad state. This strategy guarantees that you visit $x$ before visiting $y$.
With that said, a path is not possible when there is an initial node that that has $N - 1$ unavailable neighbors. This situation describes the $2 \times 2$, $2 \times 3$, and $3 \times 3$ case. A path is also not possible if its impossible to swap the order of $y$ and $x$ because $x$ is in the unavailable set of node $z$ that must come before both. This describes the $2 \times 4$ case.
Unfortunately, I haven't quite figured out the conditions for this $O\left(N^2\right)$ algorithm to work in general. Certainly, it works in larger grids since we can apply the Bondy-Chvátal theorem in that case.
Never having taken any computer science courses beyond the freshman level, there are some glaring holes in my computer science knowledge. According to Customs and Border Protection, every self-respecting software engineer should be able to balance a binary search tree.
I used a library-provided red-black tree in Policy-Based Data Structures in C++, so I've been aware of the existence of these trees for quite some time. However, I've never rolled my own. I set out to fix this deficiency in my knowledge a few weeks ago.
I've created an augmented AVL tree to re-solve ORDERSET.
The Problem
The issue with vanilla binary search trees is that the many of the operations have worst-case $O(N)$ complexity. To see this, consider the case where your data is sorted, so you insert your data in ascending order. Then, the binary search tree degenerates into a linked list.
Definition of an AVL Tree
AVL trees solve this issue by maintaining the invariant that the height of the left and right subtrees differ by at most 1. The height of a tree is the maximum number of edges between the root and a leaf node. For any node $x$, we denote its height $h_x$. Now, consider a node $x$ with left child $l$ and right child $r$. We define the balance factor of $x$ to be $$b_x = h_l - h_r.$$ In an AVL tree, $b_x \in \{-1,0,1\}$ for all $x$.
Maintaining the Invariant
Insertion
As we insert new values into the tree, our tree may become unbalanced. Since we are only creating one node, if the tree becomes unbalanced some node has a balance factor of $\pm 2$. Without loss of generality, let us assume that it's $-2$ since the two cases are symmetric.
Now, the only nodes whose heights are affected are along the path between the root and the new node. So, only the parents of these nodes, who are also along this path, will have their balance factor altered. Thus, we should start at the deepest node with an incorrect balance factor. If we can find a a way to rebalance this subtree, we can recursively balance the whole tree by going up to the root and rebalancing on the way.
So, let us assume we have tree whose root has a balance factor of $-2$. Thus, the height of the right subtree exceeds the height of the left subtree by $2$. We have $2$ cases.
Right-left Rotation: The Right Child Has a Balance Factor of $1$
In these diagrams, circles denote nodes, and rectangles denote subtrees.
In this example, by making $5$ the new right child, we still have and unbalanced tree, where the root has balance factor $-2$, but now, the right child has a right subtree of greater height.
Right-right Rotation: The Right Child Has a Balance Factor of $-2$, $-1$, or $0$
To fix this situation, the right child becomes the root.
We see that after this rotation is finished, we have a balanced tree, so our invariant is restored!
Deletion
When we remove a node, we need to replace it with the next largest node in the tree. Here's how to find this node:
- Go right. Now all nodes in this subtree are greater.
- Find the smallest node in this subtree by going left until you can't.
We replace the deleted node, say $x$, with the smallest node we found in the right subtree, say $y$. Remember this path. The right subtree of $y$ becomes the new left subtree of the parent of $y$. Starting from the parent of $y$, the balance factor may have been altered, so we start here, go up to the root, and do any rotations necessary to correct the balance factors along the way.
Implementation
Here is the code for these rotations with templates for reusability and a partial implementation of the STL interface.
#include <cassert>
#include <algorithm>
#include <iostream>
#include <functional>
#include <ostream>
#include <stack>
#include <string>
#include <utility>
using namespace std;
namespace phillypham {
namespace avl {
template <typename T>
struct node {
T key;
node<T>* left;
node<T>* right;
size_t subtreeSize;
size_t height;
};
template <typename T>
int height(node<T>* root) {
if (root == nullptr) return -1;
return root -> height;
}
template <typename T>
void recalculateHeight(node<T> *root) {
if (root == nullptr) return;
root -> height = max(height(root -> left), height(root -> right)) + 1;
}
template <typename T>
size_t subtreeSize(node<T>* root) {
if (root == nullptr) return 0;
return root -> subtreeSize;
}
template <typename T>
void recalculateSubtreeSize(node<T> *root) {
if (root == nullptr) return;
root -> subtreeSize = subtreeSize(root -> left) + subtreeSize(root -> right) + 1;
}
template <typename T>
void recalculate(node<T> *root) {
recalculateHeight(root);
recalculateSubtreeSize(root);
}
template <typename T>
int balanceFactor(node<T>* root) {
if (root == nullptr) return 0;
return height(root -> left) - height(root -> right);
}
template <typename T>
node<T>*& getLeftRef(node<T> *root) {
return root -> left;
}
template <typename T>
node<T>*& getRightRef(node<T> *root) {
return root -> right;
}
template <typename T>
node<T>* rotateSimple(node<T> *root,
node<T>*& (*newRootGetter)(node<T>*),
node<T>*& (*replacedChildGetter)(node<T>*)) {
node<T>* newRoot = newRootGetter(root);
newRootGetter(root) = replacedChildGetter(newRoot);
replacedChildGetter(newRoot) = root;
recalculate(replacedChildGetter(newRoot));
recalculate(newRoot);
return newRoot;
}
template <typename T>
void swapChildren(node<T> *root,
node<T>*& (*childGetter)(node<T>*),
node<T>*& (*grandChildGetter)(node<T>*)) {
node<T>* newChild = grandChildGetter(childGetter(root));
grandChildGetter(childGetter(root)) = childGetter(newChild);
childGetter(newChild) = childGetter(root);
childGetter(root) = newChild;
recalculate(childGetter(newChild));
recalculate(newChild);
}
template <typename T>
node<T>* rotateRightRight(node<T>* root) {
return rotateSimple(root, getRightRef, getLeftRef);
}
template <typename T>
node<T>* rotateLeftLeft(node<T>* root) {
return rotateSimple(root, getLeftRef, getRightRef);
}
template <typename T>
node<T>* rotate(node<T>* root) {
int bF = balanceFactor(root);
if (-1 <= bF && bF <= 1) return root;
if (bF < -1) { // right side is too heavy
assert(root -> right != nullptr);
if (balanceFactor(root -> right) != 1) {
return rotateRightRight(root);
} else { // right left case
swapChildren(root, getRightRef, getLeftRef);
return rotate(root);
}
} else { // left side is too heavy
assert(root -> left != nullptr);
// left left case
if (balanceFactor(root -> left) != -1) {
return rotateLeftLeft(root);
} else { // left right case
swapChildren(root, getLeftRef, getRightRef);
return rotate(root);
}
}
}
template <typename T, typename cmp_fn = less<T>>
node<T>* insert(node<T>* root, T key, const cmp_fn &comparator = cmp_fn()) {
if (root == nullptr) {
node<T>* newRoot = new node<T>();
newRoot -> key = key;
newRoot -> left = nullptr;
newRoot -> right = nullptr;
newRoot -> height = 0;
newRoot -> subtreeSize = 1;
return newRoot;
}
if (comparator(key, root -> key)) {
root -> left = insert(root -> left, key, comparator);
} else if (comparator(root -> key, key)) {
root -> right = insert(root -> right, key, comparator);
}
recalculate(root);
return rotate(root);
}
template <typename T, typename cmp_fn = less<T>>
node<T>* erase(node<T>* root, T key, const cmp_fn &comparator = cmp_fn()) {
if (root == nullptr) return root; // nothing to delete
if (comparator(key, root -> key)) {
root -> left = erase(root -> left, key, comparator);
} else if (comparator(root -> key, key)) {
root -> right = erase(root -> right, key, comparator);
} else { // actual work when key == root -> key
if (root -> right == nullptr) {
node<T>* newRoot = root -> left;
delete root;
return newRoot;
} else if (root -> left == nullptr) {
node<T>* newRoot = root -> right;
delete root;
return newRoot;
} else {
stack<node<T>*> path;
path.push(root -> right);
while (path.top() -> left != nullptr) path.push(path.top() -> left);
// swap with root
node<T>* newRoot = path.top(); path.pop();
newRoot -> left = root -> left;
delete root;
node<T>* currentNode = newRoot -> right;
while (!path.empty()) {
path.top() -> left = currentNode;
currentNode = path.top(); path.pop();
recalculate(currentNode);
currentNode = rotate(currentNode);
}
newRoot -> right = currentNode;
recalculate(newRoot);
return rotate(newRoot);
}
}
recalculate(root);
return rotate(root);
}
template <typename T>
stack<node<T>*> find_by_order(node<T>* root, size_t idx) {
assert(0 <= idx && idx < subtreeSize(root));
stack<node<T>*> path;
path.push(root);
while (idx != subtreeSize(path.top() -> left)) {
if (idx < subtreeSize(path.top() -> left)) {
path.push(path.top() -> left);
} else {
idx -= subtreeSize(path.top() -> left) + 1;
path.push(path.top() -> right);
}
}
return path;
}
template <typename T>
size_t order_of_key(node<T>* root, T key) {
if (root == nullptr) return 0ULL;
if (key == root -> key) return subtreeSize(root -> left);
if (key < root -> key) return order_of_key(root -> left, key);
return subtreeSize(root -> left) + 1ULL + order_of_key(root -> right, key);
}
template <typename T>
void delete_recursive(node<T>* root) {
if (root == nullptr) return;
delete_recursive(root -> left);
delete_recursive(root -> right);
delete root;
}
}
template <typename T, typename cmp_fn = less<T>>
class order_statistic_tree {
private:
cmp_fn comparator;
avl::node<T>* root;
public:
class const_iterator: public std::iterator<std::bidirectional_iterator_tag, T> {
friend const_iterator order_statistic_tree<T, cmp_fn>::cbegin() const;
friend const_iterator order_statistic_tree<T, cmp_fn>::cend() const;
friend const_iterator order_statistic_tree<T, cmp_fn>::find_by_order(size_t) const;
private:
cmp_fn comparator;
stack<avl::node<T>*> path;
avl::node<T>* beginNode;
avl::node<T>* endNode;
bool isEnded;
const_iterator(avl::node<T>* root) {
setBeginAndEnd(root);
if (root != nullptr) {
path.push(root);
}
}
const_iterator(avl::node<T>* root, stack<avl::node<T>*>&& path): path(move(path)) {
setBeginAndEnd(root);
}
void setBeginAndEnd(avl::node<T>* root) {
beginNode = root;
while (beginNode != nullptr && beginNode -> left != nullptr)
beginNode = beginNode -> left;
endNode = root;
while (endNode != nullptr && endNode -> right != nullptr)
endNode = endNode -> right;
if (root == nullptr) isEnded = true;
}
public:
bool isBegin() const {
return path.top() == beginNode;
}
bool isEnd() const {
return isEnded;
}
const T& operator*() const {
return path.top() -> key;
}
const bool operator==(const const_iterator &other) const {
if (path.top() == other.path.top()) {
return path.top() != endNode || isEnded == other.isEnded;
}
return false;
}
const bool operator!=(const const_iterator &other) const {
return !((*this) == other);
}
const_iterator& operator--() {
if (path.empty()) return *this;
if (path.top() == beginNode) return *this;
if (path.top() == endNode && isEnded) {
isEnded = false;
return *this;
}
if (path.top() -> left == nullptr) {
T& key = path.top() -> key;
do {
path.pop();
} while (comparator(key, path.top() -> key));
} else {
path.push(path.top() -> left);
while (path.top() -> right != nullptr) {
path.push(path.top() -> right);
}
}
return *this;
}
const_iterator& operator++() {
if (path.empty()) return *this;
if (path.top() == endNode) {
isEnded = true;
return *this;
}
if (path.top() -> right == nullptr) {
T& key = path.top() -> key;
do {
path.pop();
} while (comparator(path.top() -> key, key));
} else {
path.push(path.top() -> right);
while (path.top() -> left != nullptr) {
path.push(path.top() -> left);
}
}
return *this;
}
};
order_statistic_tree(): root(nullptr) {}
void insert(T newKey) {
root = avl::insert(root, newKey, comparator);
}
void erase(T key) {
root = avl::erase(root, key, comparator);
}
size_t size() const {
return subtreeSize(root);
}
// 0-based indexing
const_iterator find_by_order(size_t idx) const {
return const_iterator(root, move(avl::find_by_order(root, idx)));
}
// returns the number of keys strictly less than the given key
size_t order_of_key(T key) const {
return avl::order_of_key(root, key);
}
~order_statistic_tree() {
avl::delete_recursive(root);
}
const_iterator cbegin() const{
const_iterator it = const_iterator(root);
while (!it.isBegin()) --it;
return it;
}
const_iterator cend() const {
const_iterator it = const_iterator(root);
while (!it.isEnd()) ++it;
return it;
}
};
}
Solution
With this data structure, the solution is fairly short, and is in fact, identical to my previous solution in Policy-Based Data Structures in C++. Performance-wise, my implementation is very efficient. It differs by only a few hundredths of a second with the policy-based tree set according to SPOJ.
int main(int argc, char *argv[]) {
ios::sync_with_stdio(false); cin.tie(NULL);
phillypham::order_statistic_tree<int> orderStatisticTree;
int Q; cin >> Q; // number of queries
for (int q = 0; q < Q; ++q) {
char operation;
int parameter;
cin >> operation >> parameter;
switch (operation) {
case 'I':
orderStatisticTree.insert(parameter);
break;
case 'D':
orderStatisticTree.erase(parameter);
break;
case 'K':
if (1 <= parameter && parameter <= orderStatisticTree.size()) {
cout << *orderStatisticTree.find_by_order(parameter - 1) << '\n';
} else {
cout << "invalid\n";
}
break;
case 'C':
cout << orderStatisticTree.order_of_key(parameter) << '\n';
break;
}
}
cout << flush;
return 0;
}
After solving Kay and Snowflake, I wanted write this post just to serve as notes to my future self. There are many approaches to this problem. The solution contains three different approaches, and my approach differs from all of them, and has better complexity than all but the last solution. My solution contained a lot of code that I think is reusable in the future.
This problem involves finding the centroid of a tree, which is a node such that when removed, each of the new trees produced have at most half the number of nodes as the original tree. We need to be able to do this for any arbitrary subtree of the tree. Since there are as many as 300,000 subtrees we need to be a bit clever about how we find the centroid.
First, we can prove that the centroid of any tree is a node with smallest subtree such that the subtree has size at least half of the original tree. That is, if $T(i)$ is the subtree of node $i$, and $S(i)$ is its size, then the centroid of $T(i)$ is \begin{equation} C(i) = \underset{k \in \{j \in T(i) : 2S(j) - T(i) \geq 0\}}{\arg\min} \left(S(k) - \frac{T(i)}{2}\right). \end{equation}
To see this, first note the ancestors and descendents of $k = C(i)$ will be in separate subtrees. Since $S(k) \geq T(i)/2 \Rightarrow T(i) - S(k) \leq T(i)/2$, so all the ancestors will be part of new trees that are sufficiently small. Now, suppose for a contradiction that some subset of the descendents of $k$ belong to a tree that is greater than $T(i)/2$. This tree is also a subtree of $T(i)$, so $C(i)$ should be in this subtree, so this is ridiculous.
Now, that proof wasn't so hard, but actually finding $C(i)$ for each $i$ is tricky. The naive way would be to do an exhaustive search in $T(i)$, but that would give us an $O(NQ)$ solution.
The first observation that we can make to simplify our search is that if we start at $i$ and always follow the node with the largest subtree, the centroid is on that path. Why? Well, there can only be one path that has at least $T(i)/2$ nodes. Otherwise, the tree would have too many nodes.
So to find the centroid, we can start at the leaf of this path and go up until we hit the centroid. That still doesn't sound great though since the path might be very long, and it is still $O(N)$. The key thing to note is that the subtree size is strictly increasing, so we can do a binary search, reducing this step to $O(\log N)$. With a technique called binary lifting described in Least Common Ancestor, we can further reduce this to $O\left(\log\left(\log N\right)\right)$, which is probably not actually necessary, but I did it anyway.
Now, we need to know several statistics for each node. We need the leaf of path of largest subtrees, and we need the subtree size. Subtree size can be calculate recursively with depth-first search. Since subtree size of $i$ is the size of all the child subtrees plus 1 for $i$ itself. Thus, we do a post-order traversal to calculate subtree size and the special leaf. We also need depth of a node to precompute the ancestors for binary lifting. The depth is computed with a pre-order traversal. In the title picture, the numbers indicate the order of an in-order traversal. The letters correspond to a pre-order traversal, and the roman numerals show the order of a post-order traversal. Since recursion of large trees leads to stack overflow errors and usually you can't tell the online judge to increase the stack size, it's always better to use an explicit stack. I quite like my method of doing both pre-order and post-order traversal with one stack.
/**
* Does a depth-first search to calculate various statistics.
* We are given as input a rooted tree in the form each node's
* children and parent. The statistics returned:
* ancestors: Each node's ancestors that are a power of 2
* edges away.
* maxPathLeaf: If we continually follow the child that has the
* maximum subtree, we'll end up at this leaf.
* subtreeSize: The size of the node's subtree including itself.
*/
void calculateStatistics(const vector<vector<int>> &children,
const vector<int> &parent,
vector<vector<int>> *ancestorsPtr,
vector<int> *maxPathLeafPtr,
vector<int> *subtreeSizePtr) {
int N = parent.size();
if (N == 0) return;
// depth also serves to keep track of whether we've visited, yet.
vector<int> depth(N, -1); // -1 indicates that we have not visited.
vector<int> height(N, 0);
vector<int> &maxPathLeaf = *maxPathLeafPtr;
vector<int> &subtreeSize = *subtreeSizePtr;
stack<int> s; s.push(0); // DFS
while (!s.empty()) {
int i = s.top(); s.pop();
if (depth[i] == -1) { // pre-order
s.push(i); // Put it back in the stack so we visit again.
depth[i] = parent[i] == -1 ? 0: depth[parent[i]] + 1;
for (int j : children[i]) s.push(j);
} else { // post-order
int maxSubtreeSize = INT_MIN;
int maxSubtreeRoot = -1;
for (int j : children[i]) {
height[i] = max(height[j] + 1, height[i]);
subtreeSize[i] += subtreeSize[j];
if (maxSubtreeSize < subtreeSize[j]) {
maxSubtreeSize = subtreeSize[j];
maxSubtreeRoot = j;
}
}
maxPathLeaf[i] = maxSubtreeRoot == -1 ?
i : maxPathLeaf[maxSubtreeRoot];
}
}
// Use binary lifting to calculate a subset of ancestors.
vector<vector<int>> &ancestors = *ancestorsPtr;
for (int i = 0; i < N; ++i)
if (parent[i] != -1) ancestors[i].push_back(parent[i]);
for (int k = 1; (1 << k) <= height.front(); ++k) {
for (int i = 0; i < N; ++i) {
int j = ancestors[i].size();
if ((1 << j) <= depth[i]) {
--j;
ancestors[i].push_back(ancestors[ancestors[i][j]][j]);
}
}
}
}
With these statistics calculated, the rest our code to answer queries is quite small if we use C++'s built-in implementation of binary search. With upper_bound:
#include <algorithm>
#include <cassert>
#include <climits>
#include <iostream>
#include <iterator>
#include <stack>
#include <vector>
using namespace std;
/**
* Prints out vector for any type T that supports the << operator.
*/
template<typename T>
void operator<<(ostream &out, const vector<T> &v) {
copy(v.begin(), v.end(), ostream_iterator<T>(out, "\n"));
}
int findCentroid(int i,
const vector<vector<int>> &ancestors,
const vector<int> &maxPathLeaf,
const vector<int> &subtreeSize) {
int centroidCandidate = maxPathLeaf[i];
int maxComponentSize = (subtreeSize[i] + 1)/2;
while (subtreeSize[centroidCandidate] < maxComponentSize) {
// Alias the candidate's ancestors.
const vector<int> &cAncenstors = ancestors[centroidCandidate];
assert(!cAncenstors.empty());
// Check the immediate parent first. If this is an exact match, and we're done.
if (subtreeSize[cAncenstors.front()] >= maxComponentSize) {
centroidCandidate = cAncenstors.front();
} else {
// Otherwise, we can approximate the next candidate by searching ancestors that
// are a power of 2 edges away.
// Find the index of the first ancestor who has a subtree of size
// greater than maxComponentSize.
int centroidIdx =
upper_bound(cAncenstors.cbegin() + 1, cAncenstors.cend(), maxComponentSize,
// If this function evaluates to true, j is an upper bound.
[&subtreeSize](int maxComponentSize, int j) -> bool {
return maxComponentSize < subtreeSize[j];
}) - ancestors[centroidCandidate].cbegin();
centroidCandidate = cAncenstors[centroidIdx - 1];
}
}
return centroidCandidate;
}
int main(int argc, char *argv[]) {
ios::sync_with_stdio(false); cin.tie(NULL);
int N, Q;
cin >> N >> Q;
vector<int> parent; parent.reserve(N);
parent.push_back(-1); // the root has no parent
vector<vector<int>> children(N);
for (int i = 1; i < N; ++i) {
int p; cin >> p; --p; // Convert to 0-indexing.
children[p].push_back(i);
parent.push_back(p);
}
// ancestors[i][j] will be the (2^j)th ancestor of node i.
vector<vector<int>> ancestors(N);
vector<int> maxPathLeaf(N, -1);
vector<int> subtreeSize(N, 1);
calculateStatistics(children, parent,
&ancestors, &maxPathLeaf, &subtreeSize);
vector<int> centroids; centroids.reserve(Q);
for (int q = 0; q < Q; ++q) {
int v; cin >> v; --v; // Convert to 0-indexing.
// Convert back to 1-indexing.
centroids.push_back(findCentroid(v, ancestors, maxPathLeaf, subtreeSize) + 1);
}
cout << centroids;
cout << flush;
return 0;
}
Of course, we can use lower_bound in findCentroid, too. It's quite baffling to me that the function argument takes a different order of parameters in lower_bound, though. I always forget how to use these functions, which is why I've decided to write this post.
int findCentroid(int i,
const vector<vector<int>> &ancestors,
const vector<int> &maxPathLeaf,
const vector<int> &subtreeSize) {
int centroidCandidate = maxPathLeaf[i];
int maxComponentSize = (subtreeSize[i] + 1)/2;
while (subtreeSize[centroidCandidate] < maxComponentSize) {
// Alias the candidate's ancestors.
const vector<int> &cAncenstors = ancestors[centroidCandidate];
assert(!cAncenstors.empty());
// Check the immediate parent first. If this is an exact match, and we're done.
if (subtreeSize[cAncenstors.front()] >= maxComponentSize) {
centroidCandidate = cAncenstors.front();
} else {
// Otherwise, we can approximate the next candidate by searching ancestors that
// are a power of 2 edges away.
// Find the index of the first ancestor who has a subtree of size
// at least maxComponentSize.
int centroidIdx =
lower_bound(cAncenstors.cbegin() + 1, cAncenstors.cend(), maxComponentSize,
// If this function evaluates to true, j + 1 is a lower bound.
[&subtreeSize](int j, int maxComponentSize) -> bool {
return subtreeSize[j] < maxComponentSize;
}) - ancestors[centroidCandidate].cbegin();
if (centroidIdx < cAncenstors.size() &&
subtreeSize[cAncenstors[centroidIdx]] == maxComponentSize) {
centroidCandidate = cAncenstors[centroidIdx];
} else { // We don't have an exact match.
centroidCandidate = cAncenstors[centroidIdx - 1];
}
centroidCandidate = cAncenstors[centroidIdx - 1];
}
}
return centroidCandidate;
}
All in all, we need $O(N\log N)$ time to compute ancestors, and $O\left(Q\log\left(\log N\right)\right)$ time to answer queries, so total complexity is $O\left(N\log N + Q\log\left(\log N\right)\right)$.