Bit flip

861. Score After Flipping Matrix

Score After Flipping Matrix - LeetCode

“The idea is to flip all the rows, that has a 0 in the front.
Then for each column flip it, if the number of 0’s > number of 1's”

class Solution:
    def matrixScore(self, A: List[List[int]]) -> int:
        R, C = len(A), len(A[0])
        for i in range(R):
            if A[i][0]==1:
                new_a.append(A[i])
            else:
                new_a.append([int(not a) for a in A[i]])
        res = [['1']*R]
        for i, col in enumerate(zip(*new_a)):
            if i!=0:
                S = sum(col)
                if S<R-S:res.append([str(int(not c)) for c in col])
                else:
                    res.append([str(c) for c in col])
        return sum(int(''.join(col), 2) for col in zip(*res))

Or more tight

class Solution:
    def matrixScore(self, A: List[List[int]]) -> int:
        new_a = [row if row[0] else [int(not a) for a in row] for row in A]
        a_by_col = [[str(c) for c in col] if sum(col)>len(A)-sum(col) else
                    [str(int(not c)) for c in col]                    for col in zip(*new_a)]
        return sum(int(''.join(col), 2) for col in zip(*a_by_col))