01 Matrix

Given an m x n binary matrix mat, return the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example 1:

Input: mat = [[0,0,0],[0,1,0],[0,0,0]]
Output: [[0,0,0],[0,1,0],[0,0,0]]

Example 2:

Input: mat = [[0,0,0],[0,1,0],[1,1,1]]
Output: [[0,0,0],[0,1,0],[1,2,1]]

Solution

We can do this with dfs or bfs. We copy over the matrix, and then replace all 1s with a distance of float('inf') since we don’t know their distance, and enqueue the squares with 0s to our queue.

With the 0s in our queue, we can flood fill out, terminating when we encounter a square where the distance is less than what we could offer.

class Solution:
    def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
        m = len(mat)
        n = len(mat[0])
        ans = mat.copy()
        dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        q = deque()

        for y in range(m):
            for x in range(n):
                if ans[y][x] == 1:
                    ans[y][x] = float('inf')
                else:
                    q.append((y, x))

        def inbounds(y, x):
            return 0 <= y < m and 0 <= x < n

        while q:
            y, x = q.popleft()
            for dy, dx in dirs:
                new_y, new_x = dy + y, dx + x
                if inbounds(new_y, new_x) and ans[new_y][new_x] > ans[y][x] + 1:
                    ans[new_y][new_x] = ans[y][x] + 1
                    q.append((new_y, new_x))

        return ans

Table of Contents

01 Matrix

Solution