LeetCode 01: Matrix challenge

Question

Recently, I've solved the "01 Matrix" LeetCode problem and the solution was accepted by the LeetCode OJ:

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example:
Input:
0 0 0
0 1 0
1 1 1

Output:
0 0 0
0 1 0
1 2 1
Note:

The number of elements of the given matrix will not exceed 10,000.

There are at least one 0 in the given matrix.

The cells are adjacent in only four directions: up, down, left and right.

The idea behind the solution above is to use Dynamic Programming - starting with 0 cells work outwards putting not processed cells on the queue:

from collections import deque


class Solution(object):
    def updateMatrix(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[List[int]]
        """
        if not matrix:
            return matrix

        row_length = len(matrix)
        col_length = len(matrix[0])

        queue = deque()

        # put all 0 cells on queue, set all other cells to a big number
        for row_index in range(row_length):
            for col_index in range(col_length):
                if matrix[row_index][col_index] == 0:
                    queue.append((row_index, col_index))
                else:
                    matrix[row_index][col_index] = 10000

        # work from the 0 cells outwards while the queue is not empty
        while queue:
            row_index, col_index = queue.popleft()
            for i, j in [(row_index - 1, col_index),
                         (row_index + 1, col_index),
                         (row_index, col_index - 1),
                         (row_index, col_index + 1)]:
                if 0 <= i < row_length and \
                   0 <= j < col_length and \
                   matrix[i][j] > matrix[row_index][col_index] + 1:
                        matrix[i][j] = matrix[row_index][col_index] + 1
                        queue.append((i, j))

        return matrix

Even though the code works, I am not happy with the readability, in particular:

setting the non-zero cells to "magical" 10000 does not look good
getting the cell neighbors and checking if they are not out-of-bounds seems overly complicated

What would you improve code style and organization or time and space complexity wise?

D Krueger · Accepted Answer · 2017-03-20 01:02:42Z

By storing the return value in a different variable and placing the distance in the queue, the magic number can be avoided:

class Solution(object):
    def updateMatrix(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[List[int]]
        """
        if not matrix:
            return matrix

        row_length = len(matrix)
        col_length = len(matrix[0])

        result = [[None for j in range(col_length)] for i in range(row_length)]

        queue = deque()

        # put all 0 cells in queue, set all other cells to a big number
        for row_index in range(row_length):
            for col_index in range(col_length):
                if 0 == matrix[row_index][col_index]:
                    queue.append((row_index, col_index, 0))

        # work from the 0 cells outwards while the queue is not empty
        while queue:
            i, j, dist = queue.popleft()
            if 0 <= i < row_length and 0 <= j < col_length and result[i][j] is None:
                result[i][j] = dist
                queue.append((i-1, j, dist+1))
                queue.append((i+1, j, dist+1))
                queue.append((i, j-1, dist+1))
                queue.append((i, j+1, dist+1))

        return result

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