Given a 2D grid of '1's (land) and '0's (water), count the number of islands (Number of Islands).

Optimal Breadth-First Search (BFS) / Depth-First Search (DFS) Approach:
- Iterate through the grid; when unvisited land ('1') is found, increment island count and sink the connected land mass using DFS/BFS.

def num_islands(grid: list[list[str]]) -> int:
    if not grid:
        return 0
    rows, cols = len(grid), len(grid[0])
    count = 0
    def dfs(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 visited by sinking
        dfs(r + 1, c)
        dfs(r - 1, c)
        dfs(r, c + 1)
        dfs(r, c - 1)
    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == '1':
                count += 1
                dfs(r, c)
    return count

Complexity & Key Logic:
- Time Complexity: O(M * N) where M is rows and N is columns; each cell visited a constant number of times.
- Space Complexity: O(M * N) worst-case recursion stack if entire grid is filled with land.
- Algorithm Details:
- Mutate grid in-place (grid[r][c] = '0') to eliminate extra visited set memory.
- Explore 4-directionally (up, down, left, right) until boundary or water reached.