14. Number of Islands
mediumAsked at BaiduCount connected components of land cells in a 2D grid where '1' is land and '0' is water.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given an m x n grid of '1's (land) and '0's (water), return the number of islands. An island is a maximal group of land cells connected horizontally or vertically; assume the grid is bounded by water.
Constraints
1 <= m, n <= 300grid[i][j] is '0' or '1'
Examples
Example 1
grid = [['1','1','0'],['0','1','0'],['0','0','1']]2Example 2
grid = [['0','0'],['0','0']]0Approaches
1. Union find
Union each land cell with its right and down land neighbors; the final count of unique roots is the island count.
- Time
- O(mn * alpha(mn))
- Space
- O(mn)
// build parent[] of size m*n, union land-land neighbors; count distinct roots in land cells.
// Correct but more code than DFS for an on-screen whiteboard.Tradeoff:
2. DFS flood fill
Scan the grid; on each unvisited '1' increment the counter and DFS to mark its entire component as visited.
- Time
- O(mn)
- Space
- O(mn)
function numIslands(grid) {
const m = grid.length, n = grid[0].length;
let count = 0;
const dfs = (i, j) => {
if (i < 0 || j < 0 || i >= m || j >= n || grid[i][j] !== '1') return;
grid[i][j] = '0';
dfs(i+1,j); dfs(i-1,j); dfs(i,j+1); dfs(i,j-1);
};
for (let i = 0; i < m; i++)
for (let j = 0; j < n; j++)
if (grid[i][j] === '1') { count++; dfs(i, j); }
return count;
}Tradeoff:
Baidu-specific tips
Baidu uses flood-fill style traversals on heat-map grids for ad-ranking geographic CTR clusters, so they want the iterative-stack variant ready in case interviewer flags recursion depth on 300x300 grids.
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