[leetcode] 1072. Flip Columns For Maximum Number of Equal Rows

[leetcode] 1072. Flip Columns For Maximum Number of Equal Rows

Description

You are given an m x n binary matrix matrix.

You can choose any number of columns in the matrix and flip every cell in that column (i.e., Change the value of the cell from 0 to 1 or vice versa).

Return the maximum number of rows that have all values equal after some number of flips.

Example 1:

Input: matrix = [[0,1],[1,1]]Output: 1Explanation: After flipping no values, 1 row has all values equal.

Example 2:

Input: matrix = [[0,1],[1,0]]Output: 2Explanation: After flipping values in the first column, both rows have equal values.

Example 3:

Input: matrix = [[0,0,0],[0,0,1],[1,1,0]]Output: 2Explanation: After flipping values in the first two columns, the last two rows have equal values.

Constraints:

m == matrix.lengthn == matrix[i].length1 <= m, n <= 300matrix[i][j] is either 0 or 1.

分析

题目的意思是：给定一个mxn的二进制矩阵，思路是把每一行变成全0和全1，并记录改变的位置，放在字典里面进行统计。

代码

class Solution: def maxEqualRowsAfterFlips(self, matrix: List[List[int]]) -> int: cnt=Counter() for row in matrix: cnt[tuple(i for i,x in enumerate(row) if x!=0)]+=1 cnt[tuple(i for i,x in enumerate(row) if x==0)]+=1 return max(cnt.values())

代码二

from collections import Counterclass Solution: def maxEqualRowsAfterFlips(self, matrix) -> int: cnt=Counter() for row in matrix: cnt[tuple(i for i,x in enumerate(row) if x!=0)]+=1 cnt[tuple(i for i,x in enumerate(row) if x==0)]+=1 print(cnt) return max(cnt.values())if __name__=="__main__": solution=Solution() # matrix = [[0,1],[1,1]] matrix = [[0,0,0],[0,0,1],[1,1,0]] res=solution.maxEqualRowsAfterFlips(matrix) print(res) # Counter({(2,): 2, (0, 1): 2, (): 1, (0, 1, 2): 1})

参考文献

​​Python use dictionary of row flip Ops​​

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