LeetCode Solutions

			

class UnionFind {
 public:
  vector<int> id;

  UnionFind(int n) : id(n, -1) {}

  void union_(int u, int v) {
    id[u] = v;
  }

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

class Solution {
 public:
  vector<int> numIslands2(int m, int n, vector<vector<int>>& positions) {
    const vector<int> dirs{0, 1, 0, -1, 0};
    vector<int> ans;
    vector<vector<bool>> seen(m, vector<bool>(n));
    UnionFind uf(m * n);
    int count = 0;

    for (const vector<int>& p : positions) {
      const int i = p[0];
      const int j = p[1];
      if (seen[i][j]) {
        ans.push_back(count);
        continue;
      }
      seen[i][j] = true;
      const int id = getId(i, j, n);
      uf.id[id] = id;
      ++count;
      for (int k = 0; k < 4; ++k) {
        const int x = i + dirs[k];
        const int y = j + dirs[k + 1];
        if (x < 0 || x == m || y < 0 || y == n)
          continue;
        const int neighborId = getId(x, y, n);
        if (uf.id[neighborId] == -1)  // Water
          continue;
        const int currentParent = uf.find(id);
        const int neighborParent = uf.find(neighborId);
        if (currentParent != neighborParent) {
          uf.union_(currentParent, neighborParent);
          --count;
        }
      }
      ans.push_back(count);
    }

    return ans;
  }

 private:
  int getId(int i, int j, int n) {
    return i * n + j;
  }
};

			

class UnionFind {
  public int[] id;

  public UnionFind(int n) {
    id = new int[n];
    Arrays.fill(id, -1); // Water
  }

  public void union(int u, int v) {
    id[u] = v;
  }

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

class Solution {
  public List<Integer> numIslands2(int m, int n, int[][] positions) {
    final int[] dirs = {0, 1, 0, -1, 0};
    List<Integer> ans = new ArrayList<>();
    boolean[][] seen = new boolean[m][n];
    UnionFind uf = new UnionFind(m * n);
    int count = 0;

    for (int[] p : positions) {
      final int i = p[0];
      final int j = p[1];
      if (seen[i][j]) {
        ans.add(count);
        continue;
      }
      seen[i][j] = true;
      final int id = getId(i, j, n);
      uf.id[id] = id;
      ++count;
      for (int k = 0; k < 4; ++k) {
        final int x = i + dirs[k];
        final int y = j + dirs[k + 1];
        if (x < 0 || x == m || y < 0 || y == n)
          continue;
        final int neighborId = getId(x, y, n);
        if (uf.id[neighborId] == -1) // Water
          continue;
        final int currentParent = uf.find(id);
        final int neighborParent = uf.find(neighborId);
        if (currentParent != neighborParent) {
          uf.union(currentParent, neighborParent);
          --count;
        }
      }
      ans.add(count);
    }

    return ans;
  }

  private int getId(int i, int j, int n) {
    return i * n + j;
  }
}

			

class UnionFind:
  def __init__(self, n: int):
    self.id = [-1] * n

  def union(self, u: int, v: int) -> None:
    self.id[u] = v

  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 numIslands2(self, m: int, n: int, positions: List[List[int]]) -> List[int]:
    dirs = [0, 1, 0, -1, 0]
    ans = []
    seen = [[False] * n for _ in range(m)]
    uf = UnionFind(m * n)
    count = 0

    def getId(i: int, j: int, n: int) -> int:
      return i * n + j

    for i, j in positions:
      if seen[i][j]:
        ans.append(count)
        continue
      seen[i][j] = True
      id = getId(i, j, n)
      uf.id[id] = id
      count += 1
      for k in range(4):
        x = i + dirs[k]
        y = j + dirs[k + 1]
        if x < 0 or x == m or y < 0 or y == n:
          continue
        neighborId = getId(x, y, n)
        if uf.id[neighborId] == -1:  # Water
          continue
        currentParent = uf.find(id)
        neighborParent = uf.find(neighborId)
        if currentParent != neighborParent:
          uf.union(currentParent, neighborParent)
          count -= 1
      ans.append(count)

    return ans