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Solution 20: Minimal Obstacle Bypass (Shortest Path in Binary Matrix)

Approach Explanation

This is a shortest path problem in an unweighted grid, so BFS is the optimal approach. 8-directional movement is allowed.

Step-by-Step Logic

  1. If grid[0][0] == 1 or grid[n-1][n-1] == 1, return -1.
  2. Initialize queue = deque([(0, 0, 1)]) (row, col, distance).
  3. Mark grid[0][0] = 1 (visited).
  4. While queue is not empty:
    • Pop (r, c, d).
    • If (r, c) is destination, return d.
    • Check all 8 neighbors:
      • If in bounds and grid == 0:
        • Mark grid = 1 and append to queue with d + 1.
  5. Return -1.

Complexity

  • Time Complexity: O(N^2).
  • Space Complexity: O(N^2).

Code

from collections import deque

def shortest_path_binary_matrix(grid):
    n = len(grid)
    if grid[0][0] or grid[n-1][n-1]: return -1
    
    q = deque([(0, 0, 1)]) # row, col, dist
    grid[0][0] = 1 # Mark as visited
    
    while q:
        r, c, d = q.popleft()
        if r == n - 1 and c == n - 1:
            return d
            
        # 8 directions
        for dr in [-1, 0, 1]:
            for dc in [-1, 0, 1]:
                if dr == 0 and dc == 0: continue
                nr, nc = r + dr, c + dc
                if 0 <= nr < n and 0 <= nc < n and not grid[nr][nc]:
                    grid[nr][nc] = 1 # Mark visited
                    q.append((nr, nc, d + 1))
                    
    return -1