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3532. Path Existence Queries in a Graph I 👍

  • Time: $O(n\log^* n + q)$
  • Space: $O(n + q)$
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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:
  vector<bool> pathExistenceQueries(int n, vector<int>& nums, int maxDiff,
                                    vector<vector<int>>& queries) {
    vector<bool> ans;
    UnionFind uf(n);

    for (int i = 1; i < n; ++i)
      if (abs(nums[i] - nums[i - 1]) <= maxDiff)
        uf.unionByRank(i, i - 1);

    for (const vector<int>& query : queries) {
      const int u = query[0];
      const int v = query[1];
      ans.push_back(uf.find(u) == uf.find(v));
    }

    return ans;
  }
};

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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[] pathExistenceQueries(int n, int[] nums, int maxDiff, int[][] queries) {
    boolean[] ans = new boolean[queries.length];
    UnionFind uf = new UnionFind(n);

    for (int i = 1; i < n; ++i)
      if (Math.abs(nums[i] - nums[i - 1]) <= maxDiff)
        uf.unionByRank(i, i - 1);

    for (int i = 0; i < queries.length; ++i) {
      final int u = queries[i][0];
      final int v = queries[i][1];
      ans[i] = uf.find(u) == uf.find(v);
    }

    return ans;
  }
}

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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 pathExistenceQueries(
      self,
      n: int,
      nums: list[int],
      maxDiff: int,
      queries: list[list[int]]
  ) -> list[bool]:
    uf = UnionFind(n)

    for i in range(1, n):
      if abs(nums[i] - nums[i - 1]) <= maxDiff:
        uf.unionByRank(i, i - 1)

    return [uf.find(u) == uf.find(v)
            for u, v in queries]