0073-set-matrix-zeroes¶
Try it on leetcode
Description¶
Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's.
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:
- A straightforward solution using
O(mn)space is probably a bad idea. - A simple improvement uses
O(m + n)space, but still not the best solution. - Could you devise a constant space solution?
Solution(Python)¶
class Solution:
def setZeroes(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
# collect rows and columns to be filled with zero
# rows = []
# cols = []
# irtate the cells and if zero add row add cols
# fill rows and cols with zeroes
#
def hashset():
m,n = len(matrix), len(matrix[0])
rows,cols = [], []
for i in range(m):
for j in range(n):
if matrix[i][j] == 0:
rows.append(i)
cols.append(j)
for row in rows:
for j in range(n):
matrix[row][j] = 0
for col in cols:
for i in range(m):
matrix[i][col] = 0
m, n = len(matrix), len(matrix[0])
first_row_zero = False
first_col_zero = False
# Step 1: Check if first row or first column has any zeros
for j in range(n):
if matrix[0][j] == 0:
first_row_zero = True
break
for i in range(m):
if matrix[i][0] == 0:
first_col_zero = True
break
# Step 2: Use first row and column as markers
for i in range(1, m):
for j in range(1, n):
if matrix[i][j] == 0:
matrix[i][0] = 0
matrix[0][j] = 0
# Step 3: Zero out cells based on markers
for i in range(1, m):
for j in range(1, n):
if matrix[i][0] == 0 or matrix[0][j] == 0:
matrix[i][j] = 0
# Step 4: Zero out first row and column if needed
if first_row_zero:
for j in range(n):
matrix[0][j] = 0
if first_col_zero:
for i in range(m):
matrix[i][0] = 0