Number of Islands - II
Online Queries using DSU (Disjoint Set Union)

class DSU {
  constructor(n) {
    this.parent = new Array(n).fill(0).map((_, i) => i);
    this.rank = new Array(n).fill(0);
  }

  find(x) {
    if (this.parent[x] !== x) {
      this.parent[x] = this.find(this.parent[x]);
    }
    return this.parent[x];
  }

  union(x, y) {
    let rootX = this.find(x);
    let rootY = this.find(y);
    if (rootX === rootY) return false;
    if (this.rank[rootX] < this.rank[rootY]) {
      this.parent[rootX] = rootY;
    } else if (this.rank[rootX] > this.rank[rootY]) {
      this.parent[rootY] = rootX;
    } else {
      this.parent[rootY] = rootX;
      this.rank[rootX]++;
    }
    return true;
  }
}

function numIslands2(n, m, operators) {
  const dsu = new DSU(n * m);
  const grid = Array.from({ length: n }, () => Array(m).fill(0));
  const dirs = [[-1,0], [1,0], [0,-1], [0,1]];
  const result = [];
  let count = 0;

  for (const [r, c] of operators) {
    if (grid[r][c] === 1) {
      result.push(count);
      continue;
    }
    grid[r][c] = 1;
    count++;
    const id = r * m + c;

    for (const [dr, dc] of dirs) {
      const nr = r + dr, nc = c + dc;
      if (nr >= 0 && nr < n && nc >= 0 && nc < m && grid[nr][nc] === 1) {
        const nid = nr * m + nc;
        if (dsu.union(id, nid)) {
          count--;
        }
      }
    }
    result.push(count);
  }

  return result;
}

console.log(numIslands2(3, 3, [[0,0], [0,1], [1,2], [2,1]])); // Output: [1, 1, 2, 3]