Campus Monk by Rachit Rastogi

🎯 CODING CHALLENGE #11 (BONUS) - Number of Islands Difficulty: Medium Given a 2D grid of '1' (land) and '0' (water), count the number of islands (connected groups of land, horizontally/vertically).

Input:
11000
11000
00100
00011

Output: 3

💡 Hint: This is graph traversal on an implicit grid graph - each land cell is a node, adjacent land cells are connected edges. DFS or BFS, "sinking" each island as you find it so you don't count it twice. Solution:

python
def num_islands(grid):
    if not grid:
        return 0

    rows, cols = len(grid), len(grid[0])
    count = 0

    def sink(r, c):
        if r < 0 or r >= rows or c < 0 or c >= cols or grid[r][c] != '1':
            return
        grid[r][c] = '0'  # mark as visited by sinking it
        sink(r+1, c)
        sink(r-1, c)
        sink(r, c+1)
        sink(r, c-1)

    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == '1':
                count += 1
                sink(r, c)

    return count

Complexity: O(rows × cols) time - every cell is visited a constant number of times. Space is O(rows × cols) worst case for the recursion stack, if the entire grid is one giant island.

Common mistake: Modifying the grid in place without realizing that mutates the input the caller passed in - perfectly fine for most interview settings, but worth mentioning out loud: "I'm mutating the grid directly to track visited cells - if we need to preserve the original input, I'd use a separate visited set instead."

This exact pattern - grid + DFS/BFS + "sinking"/marking visited - solves a huge family of "connected regions" problems. Worth having memorized cold.

Would you use DFS or BFS here, and does it actually matter for this particular problem? 👇

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