Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.

Example 1:

Input: 
[
  [1,1,1],
  [1,0,1],
  [1,1,1]
]
Output: 
[
  [1,0,1],
  [0,0,0],
  [1,0,1]
]

Example 2:

Input: 
[
  [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]
]

Follow up:

  • A straight forward 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?

Solutions

1. Space O(m+n)

  注意理解位操作,以及最后是要成行或成列都要变成 0.

# Time Complexity: O(mn)
# Space Complexity: O(m+n)
class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        if not matrix:
            return matrix
        lr, ll = len(matrix), len(matrix[0])
        row = [0 for i in range(lr)]
        col = [0 for i in range(ll)]
        for i in range(lr):
            for j in range(ll):
                row[i] = row[i] | (matrix[i][j] == 0)
                col[j] = col[j] | (matrix[i][j] == 0)
        
        for i in range(lr):
            for j in range(ll):
                if row[i] or col[j]:
                    matrix[i][j] = 0

        return matrix
# Runtime: 160 ms, faster than 35.48% of Python3 online submissions for Set Matrix Zeroes.
# Memory Usage: 14.5 MB, less than 5.13% of Python3 online submissions for Set Matrix Zeroes.

2. Space O(1)

# Time Complexity: O(mn)
# Space Complexity: O(1)
class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        if not matrix:
            return matrix
        lr, ll = len(matrix), len(matrix[0])
        r0 = False
        c0 = False
        # if there is 0 in edge
        for i in range(lr):
            r0 = r0 | (matrix[i][0] == 0)
        for j in range(ll):
            c0 = c0 | (matrix[0][j] == 0)
        
        for i in range(1, lr):
            for j in range(1, ll):
                if matrix[i][j] == 0:
                    matrix[i][0] = 0
                    matrix[0][j] = 0
        
        for i in range(1, lr):
            for j in range(1, ll):
                if matrix[i][0] == 0 or matrix[0][j] == 0:
                    matrix[i][j] = 0
        
        if r0:
            for i in range(lr):
                matrix[i][0] = 0
        if c0:
            for j in range(ll):
                matrix[0][j] = 0

        return matrix
# Runtime: 148 ms, faster than 93.53% of Python3 online submissions for Set Matrix Zeroes.
# Memory Usage: 14.4 MB, less than 5.13% of Python3 online submissions for Set Matrix Zeroes.

References

  1. 73. Set Matrix Zeroes
  2. 花花酱