Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's, and return the matrix.
You must do it in place.
Example 1:
Input: matrix = [[1,1,1],[1,0,1],[1,1,1]] Output: [[1,0,1],[0,0,0],[1,0,1]]
Example 2:
Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]] Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
Constraints:
m == matrix.lengthn == matrix[0].length1 <= m, n <= 200-231 <= matrix[i][j] <= 231 - 1
Follow up:
O(mn) space is probably a bad idea.O(m + n) space, but still not the best solution.For to solve it in O(mn), use a copy of given matrix and go through every element of original matrix and update the copy matrix's row and column to zero if a zero if found in original matrix.
O(m + n) Solution:
class Solution:
def setZeroes(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
ROWS, COLS = len(matrix), len(matrix[0])
rows = [0] * ROWS
cols = [0] * COLS
for r in range(ROWS):
for c in range(COLS):
if matrix[r][c] == 0:
rows[r] = 1
cols[c] = 1
for c in range(COLS):
for R in range(len(rows)):
if rows[R] == 1:
matrix[R][c] = 0
for r in range(ROWS):
for C in range(len(cols)):
if cols[C] == 1:
matrix[r][C] = 0
O(1) Solution:class Solution:
def setZeroes(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
ROWS, COLS = len(matrix), len(matrix[0])
rowZero = False
for r in range(ROWS):
for c in range(COLS):
if matrix[r][c] == 0:
matrix[0][c] = 0
if r > 0:
matrix[r][0] = 0
else:
rowZero = True
for r in range(1, ROWS):
for c in range(1, COLS):
if matrix[0][c] == 0 or matrix[r][0] == 0:
matrix[r][c] = 0
if matrix[0][0] == 0:
for r in range(ROWS):
matrix[r][0] = 0
if rowZero:
for c in range(COLS):
matrix[0][c] = 0
This article is contributed by Ashwini Verma. If you like dEexams.com and would like to contribute, you can write your article here or mail your article to admin@deexams.com . See your article appearing on the dEexams.com main page and help others to learn.