Problem: Find if Path Exists in Graph

There is a bi-directional graph with `n` vertices, where each vertex is
labeled from `0` to `n - 1` (inclusive). The edges in the graph are
represented as a 2D integer array `edges`, where each `edges[i] = [ui, vi]`
denotes a bi-directional edge between vertex `ui` and vertex `vi`. Every
vertex pair is connected by at most one edge, and no vertex has an edge to
itself.

You want to determine if there is a valid path that exists from vertex
`source` to vertex `destination`.

Given `edges` and the integers `n`, `source`, and `destination`, return `true`
if there is a valid path from `source` to `destination`, or `false` otherwise.


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 validPath(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
    uf = UnionFind(n)

    for u, v in edges:
      uf.unionByRank(u, v)

    return uf.find(source) == uf.find(destination)


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 boolean validPath(int n, int[][] edges, int source, int destination) {
    UnionFind uf = new UnionFind(n);

    for (int[] edge : edges) {
      final int u = edge[0];
      final int v = edge[1];
      uf.unionByRank(u, v);
    }

    return uf.find(source) == uf.find(destination);
  }
}


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:
  bool validPath(int n, vector<vector<int>>& edges, int source,
                 int destination) {
    UnionFind uf(n);

    for (const vector<int>& edge : edges) {
      const int u = edge[0];
      const int v = edge[1];
      uf.unionByRank(u, v);
    }

    return uf.find(source) == uf.find(destination);
  }
};