Suppose we have a 2D matrix of numbers, now for each zero in the given matrix and replace all values in its row and column with zero, and return the final matrix.
So, if the input is like matrix, then the output will be matrix as the 0th, 2nd and 3rd rows contain 0 and the final matrix contains 0 in those rows. Similarly 0th, 1st and 2nd columns contain 0 and the final matrix contains 0 in those columns.
To solve this, we will follow these steps:
n := row count, m := column count res := make a matrix of size n x m and fill with 0 transpose := transpose given matrix for each row i, do if 0 not in matrix[i], then for each column j in matrix, do if 0 not in transpose[j], then res[i, j] := matrix[i, j] return res
Let us see the following implementation to get better understanding −
Example
class Solution: def solve(self, matrix): n, m = len(matrix), len(matrix[0]) res = [[0 for __ in range(m)] for _ in range(n)] transpose = [list(row) for row in zip(*matrix)] for i in range(n): if 0 not in matrix[i]: for j in range(m): if 0 not in transpose[j]: res[i][j] = matrix[i][j] return res ob = Solution() matrix = [ [6, 0, 0, 6, 9], [4, 9, 9, 4, 8], [0, 8, 3, 4, 2], [9, 0, 7, 8, 3], [5, 2, 9, 6, 8] ] print(ob.solve(matrix))
Input
matrix = [ [6, 0, 0, 6, 9], [4, 9, 9, 4, 8], [0, 8, 3, 4, 2], [9, 0, 7, 8, 3], [5, 2, 9, 6, 8] ]
Output
[[0, 0, 0, 0, 0], [0, 0, 0, 4, 8], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 6, 8]]