LeetCode 542 01 Matrix - Implementation

Here's the solution:

function updateMatrix(mat):
    rows := number of rows in mat
    cols := number of columns in mat
    result := 2D array of size rows x cols, initialized to -1
    queue := empty queue
    directions := [(0,1), (1,0), (0,-1), (-1,0)]

    for r from 0 to rows - 1:
        for c from 0 to cols - 1:
            if mat[r][c] = 0 then
                result[r][c] := 0
                push (r, c) to queue

    while queue is not empty:
        (r, c) := pop from queue
        for each (dr, dc) in directions:
            nr := r + dr
            nc := c + dc
            if nr, nc is in bounds and result[nr][nc] = -1 then
                result[nr][nc] := result[r][c] + 1
                push (nr, nc) to queue

    return result

Time: $O(rows \times cols)$ because each cell is visited once. Space: $O(rows \times cols)$ for the result matrix and queue.