Problem: Smallest String With Swaps

You are given a string `s`, and an array of pairs of indices in the string
`pairs` where `pairs[i] = [a, b]` indicates 2 indices(0-indexed) of the
string.

You can swap the characters at any pair of indices in the given `pairs` any
number of times.

Return the lexicographically smallest string that `s` can be changed to after
using the swaps.


Python solution:
import math
import itertools
import bisect
import collections
import string
import heapq
import functools
import sortedcontainers
from typing import List, Dict, Tuple, Iterator

class UnionFind:
  def __init__(self, n: int):
    self.id = list(range(n))
    self.rank = [0] * n

  def unionByRank(self, u: int, v: int) -> None:
    i = self.find(u)
    j = self.find(v)
    if i == j:
      return
    if self.rank[i] < self.rank[j]:
      self.id[i] = j
    elif self.rank[i] > self.rank[j]:
      self.id[j] = i
    else:
      self.id[i] = j
      self.rank[j] += 1

  def find(self, u: int) -> int:
    if self.id[u] != u:
      self.id[u] = self.find(self.id[u])
    return self.id[u]


class Solution:
  def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
    ans = ''
    uf = UnionFind(len(s))
    map = collections.defaultdict(list)

    for a, b in pairs:
      uf.unionByRank(a, b)

    for i, c in enumerate(s):
      map[uf.find(i)].append(c)

    for key in map.keys():
      map[key].sort(reverse=True)

    for i in range(len(s)):
      ans += map[uf.find(i)].pop()

    return ans


Java solution:
class UnionFind {
  public UnionFind(int n) {
    id = new int[n];
    rank = new int[n];
    for (int i = 0; i < n; ++i)
      id[i] = i;
  }

  public void unionByRank(int u, int v) {
    final int i = find(u);
    final int j = find(v);
    if (i == j)
      return;
    if (rank[i] < rank[j]) {
      id[i] = j;
    } else if (rank[i] > rank[j]) {
      id[j] = i;
    } else {
      id[i] = j;
      ++rank[j];
    }
  }

  public int find(int u) {
    return id[u] == u ? u : (id[u] = find(id[u]));
  }

  private int[] id;
  private int[] rank;
}

class Solution {
  public String smallestStringWithSwaps(String s, List<List<Integer>> pairs) {
    StringBuilder sb = new StringBuilder();
    UnionFind uf = new UnionFind(s.length());
    Map<Integer, Queue<Character>> indexToLetters = new HashMap<>();

    for (List<Integer> pair : pairs) {
      final int a = pair.get(0);
      final int b = pair.get(1);
      uf.unionByRank(a, b);
    }

    for (int i = 0; i < s.length(); ++i)
      indexToLetters.computeIfAbsent(uf.find(i), k -> new PriorityQueue<>()).offer(s.charAt(i));

    for (int i = 0; i < s.length(); ++i)
      sb.append(indexToLetters.get(uf.find(i)).poll());

    return sb.toString();
  }
}


C++ solution:
class UnionFind {
 public:
  UnionFind(int n) : id(n), rank(n) {
    iota(id.begin(), id.end(), 0);
  }

  void unionByRank(int u, int v) {
    const int i = find(u);
    const int j = find(v);
    if (i == j)
      return;
    if (rank[i] < rank[j]) {
      id[i] = j;
    } else if (rank[i] > rank[j]) {
      id[j] = i;
    } else {
      id[i] = j;
      ++rank[j];
    }
  }

  int find(int u) {
    return id[u] == u ? u : id[u] = find(id[u]);
  }

 private:
  vector<int> id;
  vector<int> rank;
};

class Solution {
 public:
  string smallestStringWithSwaps(string s, vector<vector<int>>& pairs) {
    string ans;
    UnionFind uf(s.length());
    unordered_map<int, priority_queue<char, vector<char>, greater<>>>
        indexToLetters;

    for (const vector<int>& pair : pairs) {
      const int a = pair[0];
      const int b = pair[1];
      uf.unionByRank(a, b);
    }

    for (int i = 0; i < s.length(); ++i)
      indexToLetters[uf.find(i)].push(s[i]);

    for (int i = 0; i < s.length(); ++i)
      ans += indexToLetters[uf.find(i)].top(), indexToLetters[uf.find(i)].pop();

    return ans;
  }
};