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An N x N board contains only 0s and 1s. In each move, you can swap any 2 rows with each other, or any 2 columns with each other.

What is the minimum number of moves to transform the board into a "chessboard" - a board where no 0s and no 1s are 4-directionally adjacent? If the task is impossible, return -1.

Examples:
Input: board = [[0,1,1,0],[0,1,1,0],[1,0,0,1],[1,0,0,1]]
Output: 2
Explanation:
One potential sequence of moves is shown below, from left to right:

0110     1010     1010
0110 --> 1010 --> 0101
1001     0101     1010
1001     0101     0101

The first move swaps the first and second column.
The second move swaps the second and third row.

Input: board = [[0, 1], [1, 0]]
Output: 0
Explanation:
Also note that the board with 0 in the top left corner,
01
10

is also a valid chessboard.

Input: board = [[1, 0], [1, 0]]
Output: -1
Explanation:
No matter what sequence of moves you make, you cannot end with a valid chessboard.

Note:

• board will have the same number of rows and columns, a number in the range [2, 30].
• board[i][j] will be only 0s or 1s.

一个 N x N的 board 仅由 0 和 1 组成 。每次移动，你能任意交换两列或是两行的位置。

输出将这个矩阵变为 “棋盘” 所需的最小移动次数。“棋盘” 是指任意一格的上下左右四个方向的值均与本身不同的矩阵。如果不存在可行的变换，输出 -1。

示例:
输入: board = [[0,1,1,0],[0,1,1,0],[1,0,0,1],[1,0,0,1]]
输出: 2
解释:
一种可行的变换方式如下，从左到右：

0110     1010     1010
0110 --> 1010 --> 0101
1001     0101     1010
1001     0101     0101

第一次移动交换了第一列和第二列。
第二次移动交换了第二行和第三行。

输入: board = [[0, 1], [1, 0]]
输出: 0
解释:
注意左上角的格值为0时也是合法的棋盘，如：

01
10

也是合法的棋盘.

输入: board = [[1, 0], [1, 0]]
输出: -1
解释:
任意的变换都不能使这个输入变为合法的棋盘。 

提示：

• board 是方阵，且行列数的范围是[2, 30]。
• board[i][j] 将只包含 0或 1。

 1 class Solution {
 2     func movesToChessboard(_ board: [[Int]]) -> Int {
 3         var n:Int = board.count
 4         var rowSum:Int = 0
 5         var colSum:Int = 0
 6         var rowDiff:Int = 0
 7         var colDiff:Int = 0
 8         for i in 0..<n
 9         {
10             for j in 0..<n
11             {
12                 if board[0][0] ^ board[i][0] ^ board[0][j] ^ board[i][j] != 0
13                 {
14                     return -1
15                 }
16             }
17         }
18         for i in 0..<n
19         {
20             rowSum += board[0][i]
21             colSum += board[i][0]
22             rowDiff += (board[i][0] == i % 2 ? 1 : 0)
23             colDiff += (board[0][i] == i % 2 ? 1 : 0)
24         }
25         if n / 2 > rowSum || rowSum > (n + 1) / 2 {return -1}
26         if n / 2 > colSum || colSum > (n + 1) / 2 {return -1}
27         if n % 2 != 0
28         {
29             if rowDiff % 2 != 0 {rowDiff = n - rowDiff}
30             if colDiff % 2 != 0 {colDiff = n - colDiff}
31         }
32         else
33         {
34             rowDiff = min(n - rowDiff, rowDiff)
35             colDiff = min(n - colDiff, colDiff)
36         }
37         return (rowDiff + colDiff) / 2
38     }
39 }