Posts by Philip Pham tagged computer science

Photo URL is broken

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 {
  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);
      } while (wallPtr != walls.cend() && wallPtr -> position == currentPosition);
      if (skyline.empty() || heightCount.empty() ||
          heightCount.crbegin() -> first != skyline.back()[1]) {
            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;
    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;
  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!

Photo URL is broken

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) {
    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]));
  // 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) {
    // Use transitivity to update the common prefix size.
    *next(max_prefix_size) = min(*prev(max_prefix_size), *next(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.


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
    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 {
                  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.


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:

digraph {
  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"];
  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"];
  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"];
  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"];
  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"];

Photo URL is broken

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 {
    // 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))
        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.

Photo URL is broken

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


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!


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:

  1. Go right. Now all nodes in this subtree are greater.
  2. 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.


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) {

    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;
      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;

    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);
      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 ( -> left != nullptr) path.push( -> left);
          // swap with root
          node<T>* newRoot =; path.pop();
          newRoot -> left = root -> left;
          delete root;

          node<T>* currentNode = newRoot -> right;
          while (!path.empty()) {
   -> left = currentNode;
            currentNode =; path.pop();
            currentNode = rotate(currentNode);
          newRoot -> right = currentNode;
          return rotate(newRoot);
      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;
      while (idx != subtreeSize( -> left)) {
        if (idx < subtreeSize( -> left)) {
          path.push( -> left);
        } else {
          idx -= subtreeSize( -> left) + 1;
          path.push( -> 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 {
    cmp_fn comparator;
    avl::node<T>* root;
    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;
      cmp_fn comparator;
      stack<avl::node<T>*> path;
      avl::node<T>* beginNode;
      avl::node<T>* endNode;
      bool isEnded;

      const_iterator(avl::node<T>* root) {
        if (root != nullptr) {

      const_iterator(avl::node<T>* root, stack<avl::node<T>*>&& path): path(move(path)) {

      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;

      bool isBegin() const {
        return == beginNode;

      bool isEnd() const {
        return isEnded;

      const T& operator*() const {
        return -> key;

      const bool operator==(const const_iterator &other) const {
        if ( == {
          return != 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 ( == beginNode) return *this;
        if ( == endNode && isEnded) {
          isEnded = false;
          return *this;
        if ( -> left == nullptr) {
          T& key = -> key;
          do {
          } while (comparator(key, -> key));
        } else {
          path.push( -> left);
          while ( -> right != nullptr) {
            path.push( -> right);
        return *this;

      const_iterator& operator++() {
        if (path.empty()) return *this;
        if ( == endNode) {
          isEnded = true;
          return *this;
        if ( -> right == nullptr) {
          T& key = -> key;
          do {
          } while (comparator( -> key, key));
        } else {
          path.push( -> right);
          while ( -> left != nullptr) {
            path.push( -> 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() {

    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;


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':
      case 'D':
      case 'K':
        if (1 <= parameter && parameter <= orderStatisticTree.size()) {
          cout << *orderStatisticTree.find_by_order(parameter - 1) << '\n';
        } else {
          cout << "invalid\n";
      case 'C':
        cout << orderStatisticTree.order_of_key(parameter) << '\n';
  cout << flush;
  return 0;