Count Islands in a Grid

mediumcodingAlgorithmsGraphsData Structures~35 min

Asked at Google, Amazon, Meta

Overview

Given an m × n 2D grid where "1" represents land and "0" represents water, count the number of distinct islands. An island is surrounded by water and is formed by connecting adjacent land cells horizontally or vertically.

Implement numIslands(grid).

Constraints

grid is a 2D array of strings ("0" or "1").
1 ≤ m, n ≤ 300.
You may modify the grid in place (mark visited cells).
Aim for $O(m \times n)$ time.

Examples

numIslands([
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]);
// → 1

numIslands([
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]);
// → 3

Notes

Classic DFS/BFS flood-fill: iterate every cell. When you find a "1", increment the count and flood-fill (mark all connected land as visited).
Mark visited cells by setting them to "0" to avoid extra space for a visited set.
Alternatively, use Union-Find for a different approach.

Solution

Reveal solution

function numIslands(grid) {
  if (!grid.length) return 0;
  const rows = grid.length, cols = grid[0].length;
  let count = 0;

  function dfs(r, c) {
    if (r < 0 || r >= rows || c < 0 || c >= cols || grid[r][c] === "0") 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 < rows; r++) {
    for (let c = 0; c < cols; c++) {
      if (grid[r][c] === "1") {
        count++;
        dfs(r, c);
      }
    }
  }

  return count;
}

Resources

count-islands-in-grid.js

Count Islands in a Grid

mediumcodingAlgorithmsGraphs

Overview

Given an m × n 2D grid where "1" represents land and "0" represents water, count the number of distinct islands. An island is surrounded by water and is formed by connecting adjacent land cells horizontally or vertically.

Implement numIslands(grid).

Constraints

grid is a 2D array of strings ("0" or "1").
1 ≤ m, n ≤ 300.
You may modify the grid in place (mark visited cells).
Aim for $O(m \times n)$ time.

Examples

numIslands([
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]);
// → 1

numIslands([
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]);
// → 3

Notes

Classic DFS/BFS flood-fill: iterate every cell. When you find a "1", increment the count and flood-fill (mark all connected land as visited).
Mark visited cells by setting them to "0" to avoid extra space for a visited set.
Alternatively, use Union-Find for a different approach.

Solution

Reveal solution

function numIslands(grid) {
  if (!grid.length) return 0;
  const rows = grid.length, cols = grid[0].length;
  let count = 0;

  function dfs(r, c) {
    if (r < 0 || r >= rows || c < 0 || c >= cols || grid[r][c] === "0") 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 < rows; r++) {
    for (let c = 0; c < cols; c++) {
      if (grid[r][c] === "1") {
        count++;
        dfs(r, c);
      }
    }
  }

  return count;
}

Resources

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