Lintcode1508 · Score After Flipping Matrix go

/**
1508 · Score After Flipping Matrix
Algorithms
Medium
Accepted Rate
89%

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Description
We have a two dimensional matrix A where each value is 0 or 1.

A move consists of choosing any row or column, and toggling each value in that row or column: changing all 0s to 1s, and all 1s to 0s.

After making any number of moves, every row of this matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.

Return the highest possible score.

1 \leq A.length \leq 201≤A.length≤20
1 \leq A[0].length \leq 201≤A[0].length≤20
A[i][j] is 0 or 1.
Example
Example 1:

Input: [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
Output: 39
Explanation:
Toggled to [[1,1,1,1],[1,0,0,1],[1,1,1,1]].
0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39
Example 2:

Input: [[1,1],[0,0],[1,0],[1,0],[1,1]]
Output: 13
Explanation:
Toggled to [[1,1],[1,1],[1,0],[1,0],[1,1]].
0b11 + 0b11 + 0b10 + ob10 + ob11 = 3 + 3 + 2 + 2 + 3= 13
Tags
Company
IIT Bombay

https://blog.youkuaiyun.com/m0_37889928/article/details/82946883

https://www.lintcode.com/problem/1508/
*/

/**
 * @param A: a matrix
 * @return: the score
 */
func matrixScore(A [][]int) int {
	// Write your code here.
	var row int = len(A)
	if row == 0 {
		return 0
	}
	var col int = len(A[0])
	var res int = row * (1 << uint32(col-1))
	for i := 0; i < row; i++ {
		if A[i][0] == 1 {
			continue
		}
		for j := 0; j < col; j++ {
			A[i][j] = (A[i][j] + 1) % 2
		}
	}
	for j := 1; j < col; j++ {
		var ct int = 0
		for i := 0; i < row; i++ {
			ct += A[i][j]
		}
		ct = Max(ct, row - ct)
		res += ct * (1 << uint32(col - j - 1))
	}
	return res
}

func Max(a, b int) int {
	if a > b {
		return a
	}
	return b
}