Hard
You are given an n x n
binary grid board
. In each move, you can swap any two rows with each other, or any two columns with each other.
Return the minimum number of moves to transform the board into a chessboard board. If the task is impossible, return -1
.
A chessboard board is a board where no 0
’s and no 1
’s are 4-directionally adjacent.
Example 1:
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.
The first move swaps the first and second column.
The second move swaps the second and third row.
Example 2:
Input: board = [[0,1],[1,0]]
Output: 0
Explanation: Also note that the board with 0 in the top left corner, is also a valid chessboard.
Example 3:
Input: board = [[1,0],[1,0]]
Output: -1
Explanation: No matter what sequence of moves you make, you cannot end with a valid chessboard.
Constraints:
n == board.length
n == board[i].length
2 <= n <= 30
board[i][j]
is either 0
or 1
.class Solution {
fun movesToChessboard(board: Array<IntArray>): Int {
val n = board.size
var colToMove = 0
var rowToMove = 0
var rowOneCnt = 0
var colOneCnt = 0
for (ints in board) {
for (j in 0 until n) {
if (board[0][0] xor ints[0] xor (ints[j] xor board[0][j]) == 1) {
return -1
}
}
}
for (i in 0 until n) {
rowOneCnt += board[0][i]
colOneCnt += board[i][0]
if (board[i][0] == i % 2) {
rowToMove++
}
if (board[0][i] == i % 2) {
colToMove++
}
}
if (rowOneCnt < n / 2 || rowOneCnt > (n + 1) / 2) {
return -1
}
if (colOneCnt < n / 2 || colOneCnt > (n + 1) / 2) {
return -1
}
if (n % 2 == 1) {
// we cannot make it when ..ToMove is odd
if (colToMove % 2 == 1) {
colToMove = n - colToMove
}
if (rowToMove % 2 == 1) {
rowToMove = n - rowToMove
}
} else {
colToMove = colToMove.coerceAtMost(n - colToMove)
rowToMove = rowToMove.coerceAtMost(n - rowToMove)
}
return (colToMove + rowToMove) / 2
}
}