15. Number of Islands
mediumAsked at CheggCount connected components of '1's in a 2D grid — a graph traversal problem that mirrors Chegg's recommendation engine grouping related educational content.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given an m x n 2D binary grid which represents a map of '1's (land) and '0's (water), return the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically.
Constraints
m == grid.length, n == grid[i].length1 <= m, n <= 300grid[i][j] is '0' or '1'
Examples
Example 1
grid = [["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]]1Example 2
grid = [["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],["0","0","0","1","1"]]3Approaches
1. Brute force
Use a visited set and run BFS/DFS from every unvisited land cell.
- Time
- O(m*n)
- Space
- O(m*n)
function numIslands(grid) {
const visited = new Set();
let count = 0;
for (let i = 0; i < grid.length; i++)
for (let j = 0; j < grid[0].length; j++)
if (grid[i][j] === '1' && !visited.has(`${i},${j}`)) {
count++;
bfs(grid, i, j, visited);
}
return count;
}Tradeoff:
2. In-place DFS flood fill
Mark visited cells by mutating '1' to '0' during DFS — eliminates the visited set entirely and reduces space to the call stack depth.
- Time
- O(m*n)
- Space
- O(min(m,n)) call stack
function numIslands(grid) {
let count = 0;
function dfs(i, j) {
if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || 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 < grid.length; i++)
for (let j = 0; j < grid[0].length; j++)
if (grid[i][j] === '1') { count++; dfs(i, j); }
return count;
}Tradeoff:
Chegg-specific tips
Chegg often asks about graph traversal in the context of grouping related content clusters — mention how this maps to connected-component analysis in recommendation graphs.
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