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Solution 16: Topographical Connectivity (Number of Islands)

Approach Explanation

We iterate through each cell. When we find a '1', we increment our island count and use DFS (or BFS) to "sink" the entire island (set all connected '1's to '0').

Step-by-Step Logic

  1. Initialize count = 0.
  2. Iterate through r and c.
  3. If grid[r][c] == "1":
    • count += 1.
    • Call dfs(r, c) to sink the island.
  4. dfs(r, c):
    • If out of bounds or grid[r][c] == "0", return.
    • grid[r][c] = "0".
    • Recurse on 4 neighbors.
  5. Return count.

Complexity

  • Time Complexity: O(M * N).
  • Space Complexity: O(M * N) for recursion stack.

Code

def num_islands(grid):
    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] == "0"):
            return
            
        grid[r][c] = "0" # Sink it
        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