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LeetCode Solutions

547. Number of Provinces

Time: $O(n^2)$

Space: $O(n)$

			


class UnionFind {
 public:
  UnionFind(int n) : count(n), id(n) {
    iota(begin(id), end(id), 0);
  }

  void union_(int u, int v) {
    const int i = find(u);
    const int j = find(v);
    if (i == j)
      return;
    id[i] = j;
    --count;
  }

  int getCount() const {
    return count;
  }

 private:
  int count;
  vector<int> id;

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

class Solution {
 public:
  int findCircleNum(vector<vector<int>>& M) {
    const int n = M.size();
    UnionFind uf(n);

    for (int i = 0; i < n; ++i)
      for (int j = i; j < n; ++j)
        if (M[i][j] == 1)
          uf.union_(i, j);

    return uf.getCount();
  }
};




			


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

  public void union(int u, int v) {
    final int i = find(u);
    final int j = find(v);
    if (i == j)
      return;
    id[i] = j;
    --count;
  }

  public int getCount() {
    return count;
  }

  private int count;
  private int[] id;

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

class Solution {
  public int findCircleNum(int[][] M) {
    final int n = M.length;
    UnionFind uf = new UnionFind(n);

    for (int i = 0; i < n; ++i)
      for (int j = i; j < n; ++j)
        if (M[i][j] == 1)
          uf.union(i, j);

    return uf.getCount();
  }
}




			


class UnionFind:
  def __init__(self, n: int):
    self.count = n
    self.id = list(range(n))

  def union(self, u: int, v: int) -> None:
    i = self.find(u)
    j = self.find(v)
    if i == j:
      return
    self.id[i] = j
    self.count -= 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 findCircleNum(self, M: List[List[int]]) -> int:
    n = len(M)
    uf = UnionFind(n)

    for i in range(n):
      for j in range(i, n):
        if M[i][j] == 1:
          uf.union(i, j)

    return uf.count