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10. Number of Islands

mediumAsked at Redis

Count connected '1' regions in a 2D grid; Redis uses it to verify clean BFS/DFS hygiene before discussing cluster-partition reconnection.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Given an m x n 2D grid of '1' (land) and '0' (water), return the number of islands. An island is surrounded by water and formed by connecting adjacent land cells horizontally or vertically.

Constraints

  • 1 <= m, n <= 300
  • grid[i][j] is '0' or '1'

Examples

Example 1

Input
grid = [["1","1","0"],["1","0","0"],["0","0","1"]]
Output
2

Example 2

Input
grid = [["1","1","1"],["0","1","0"],["1","1","1"]]
Output
1

Approaches

1. Union-Find every land cell

Union each land cell with its right/down neighbor; count distinct roots.

Time
O(m*n * α)
Space
O(m*n)
// Build DSU sized m*n, union land-land neighbors, count roots.

Tradeoff:

2. DFS flood-fill

Scan the grid; on each unvisited '1' DFS-sink the whole island and increment a counter.

Time
O(m*n)
Space
O(m*n) worst case for recursion stack
function numIslands(grid) {
  let count = 0;
  const m = grid.length, n = grid[0].length;
  function dfs(r, c) {
    if (r < 0 || c < 0 || r >= m || c >= n || grid[r][c] !== '1') return;
    grid[r][c] = '0';
    dfs(r + 1, c); dfs(r - 1, c); dfs(r, c + 1); dfs(r, c - 1);
  }
  for (let r = 0; r < m; r++)
    for (let c = 0; c < n; c++)
      if (grid[r][c] === '1') { count++; dfs(r, c); }
  return count;
}

Tradeoff:

Redis-specific tips

Redis interviewers like a quick analogy to gossip protocols in Redis Cluster — mention how flood-fill mirrors how reachability info spreads across shard nodes.

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