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Commit 4014528

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feat: add solutions to lc problem: No.0782
No.0782.Transform to Chessboard
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‎solution/0700-0799/0782.Transform to Chessboard/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:规律观察 + 状态压缩**
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在一个有效的棋盘中,有且仅有两种"行"。
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例如,如果棋盘中有一行为"01010011",那么任何其它行只能为"01010011"或者"10101100"。列也满足这种性质。
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另外,每一行和每一列都有一半 0ドル$ 和一半 1ドル$。假设棋盘为 $n \times n$:
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- 若 $n = 2 \times k,ドル则每一行和每一列都有 $k$ 个 1ドル$ 和 $k$ 个 0ドル$。
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- 若 $n = 2 \times k + 1,ドル则每一行都有 $k$ 个 1ドル$ 和 $k + 1$ 个 0ドル,ドル或者 $k + 1$ 个 1ドル$ 和 $k$ 个 0ドル$。
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基于以上的结论,我们可以判断一个棋盘是否有效。若有效,可以计算出最小的移动次数。
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若 $n$ 为偶数,最终的合法棋盘有两种可能,即第一行的元素为"010101...",或者"101010..."。我们计算出这两种可能所需要交换的次数的较小值作为答案。
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若 $n$ 为奇数,那么最终的合法棋盘只有一种可能。如果第一行中 0ドル$ 的数目大于 1ドル,ドル那么最终一盘的第一行只能是"01010...",否则就是"10101..."。同样算出次数作为答案。
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时间复杂度 $O(n^2)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def movesToChessboard(self, board: List[List[int]]) -> int:
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def f(mask, cnt):
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ones = mask.bit_count()
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if n & 1:
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if abs(n - 2 * ones) != 1 or abs(n - 2 * cnt) != 1:
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return -1
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if ones == n // 2:
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return n // 2 - (mask & 0xAAAAAAAA).bit_count()
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return (n + 1) // 2 - (mask & 0x55555555).bit_count()
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else:
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if ones != n // 2 or cnt != n // 2:
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return -1
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cnt0 = n // 2 - (mask & 0xAAAAAAAA).bit_count()
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cnt1 = n // 2 - (mask & 0x55555555).bit_count()
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return min(cnt0, cnt1)
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n = len(board)
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mask = (1 << n) - 1
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rowMask = colMask = 0
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for i in range(n):
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rowMask |= board[0][i] << i
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colMask |= board[i][0] << i
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revRowMask = mask ^ rowMask
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revColMask = mask ^ colMask
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sameRow = sameCol = 0
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for i in range(n):
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curRowMask = curColMask = 0
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for j in range(n):
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curRowMask |= board[i][j] << j
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curColMask |= board[j][i] << j
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if curRowMask not in (rowMask, revRowMask) or curColMask not in (colMask, revColMask):
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return -1
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sameRow += curRowMask == rowMask
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sameCol += curColMask == colMask
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t1 = f(rowMask, sameRow)
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t2 = f(colMask, sameCol)
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return -1 if t1 == -1 or t2 == -1 else t1 + t2
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private int n;
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public int movesToChessboard(int[][] board) {
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n = board.length;
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int mask = (1 << n) - 1;
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int rowMask = 0, colMask = 0;
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for (int i = 0; i < n; ++i) {
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rowMask |= board[0][i] << i;
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colMask |= board[i][0] << i;
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}
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int revRowMask = mask ^ rowMask;
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int revColMask = mask ^ colMask;
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int sameRow = 0, sameCol = 0;
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for (int i = 0; i < n; ++i) {
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int curRowMask = 0, curColMask = 0;
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for (int j = 0; j < n; ++j) {
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curRowMask |= board[i][j] << j;
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curColMask |= board[j][i] << j;
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}
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if (curRowMask != rowMask && curRowMask != revRowMask) {
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return -1;
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}
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if (curColMask != colMask && curColMask != revColMask) {
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return -1;
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}
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sameRow += curRowMask == rowMask ? 1 : 0;
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sameCol += curColMask == colMask ? 1 : 0;
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}
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int t1 = f(rowMask, sameRow);
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int t2 = f(colMask, sameCol);
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return t1 == -1 || t2 == -1 ? -1 : t1 + t2;
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}
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private int f(int mask, int cnt) {
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int ones = Integer.bitCount(mask);
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if (n % 2 == 1) {
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if (Math.abs(n - ones * 2) != 1 || Math.abs(n - cnt * 2) != 1) {
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return -1;
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}
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if (ones == n / 2) {
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return n / 2 - Integer.bitCount(mask & 0xAAAAAAAA);
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}
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return (n / 2 + 1) - Integer.bitCount(mask & 0x55555555);
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} else {
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if (ones != n / 2 || cnt != n / 2) {
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return -1;
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}
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int cnt0 = n / 2 - Integer.bitCount(mask & 0xAAAAAAAA);
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int cnt1 = n / 2 - Integer.bitCount(mask & 0x55555555);
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return Math.min(cnt0, cnt1);
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}
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int n;
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int movesToChessboard(vector<vector<int>>& board) {
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n = board.size();
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int mask = (1 << n) - 1;
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int rowMask = 0, colMask = 0;
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for (int i = 0; i < n; ++i) {
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rowMask |= board[0][i] << i;
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colMask |= board[i][0] << i;
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}
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int revRowMask = mask ^ rowMask;
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int revColMask = mask ^ colMask;
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int sameRow = 0, sameCol = 0;
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for (int i = 0; i < n; ++i) {
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int curRowMask = 0, curColMask = 0;
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for (int j = 0; j < n; ++j) {
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curRowMask |= board[i][j] << j;
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curColMask |= board[j][i] << j;
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}
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if (curRowMask != rowMask && curRowMask != revRowMask) return -1;
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if (curColMask != colMask && curColMask != revColMask) return -1;
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sameRow += curRowMask == rowMask;
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sameCol += curColMask == colMask;
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}
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int t1 = f(rowMask, sameRow);
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int t2 = f(colMask, sameCol);
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return t1 == -1 || t2 == -1 ? -1 : t1 + t2;
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}
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int f(int mask, int cnt) {
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int ones = __builtin_popcount(mask);
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if (n & 1) {
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if (abs(n - ones * 2) != 1 || abs(n - cnt * 2) != 1) return -1;
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if (ones == n / 2) return n / 2 - __builtin_popcount(mask & 0xAAAAAAAA);
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return (n + 1) / 2 - __builtin_popcount(mask & 0x55555555);
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} else {
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if (ones != n / 2 || cnt != n / 2) return -1;
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int cnt0 = (n / 2 - __builtin_popcount(mask & 0xAAAAAAAA));
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int cnt1 = (n / 2 - __builtin_popcount(mask & 0x55555555));
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return min(cnt0, cnt1);
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}
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}
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};
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```
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### **Go**
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```go
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func movesToChessboard(board [][]int) int {
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n := len(board)
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mask := (1 << n) - 1
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rowMask, colMask := 0, 0
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for i := 0; i < n; i++ {
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rowMask |= board[0][i] << i
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colMask |= board[i][0] << i
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}
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revRowMask := mask ^ rowMask
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revColMask := mask ^ colMask
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sameRow, sameCol := 0, 0
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for i := 0; i < n; i++ {
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curRowMask, curColMask := 0, 0
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for j := 0; j < n; j++ {
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curRowMask |= board[i][j] << j
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curColMask |= board[j][i] << j
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}
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if curRowMask != rowMask && curRowMask != revRowMask {
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return -1
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}
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if curColMask != colMask && curColMask != revColMask {
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return -1
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}
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if curRowMask == rowMask {
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sameRow++
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}
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if curColMask == colMask {
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sameCol++
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}
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}
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f := func(mask, cnt int) int {
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ones := bits.OnesCount(uint(mask))
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if n%2 == 1 {
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if abs(n-ones*2) != 1 || abs(n-cnt*2) != 1 {
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return -1
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}
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if ones == n/2 {
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return n/2 - bits.OnesCount(uint(mask&0xAAAAAAAA))
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}
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return (n+1)/2 - bits.OnesCount(uint(mask&0x55555555))
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} else {
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if ones != n/2 || cnt != n/2 {
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return -1
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}
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cnt0 := n/2 - bits.OnesCount(uint(mask&0xAAAAAAAA))
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cnt1 := n/2 - bits.OnesCount(uint(mask&0x55555555))
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return min(cnt0, cnt1)
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}
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}
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t1 := f(rowMask, sameRow)
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t2 := f(colMask, sameCol)
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if t1 == -1 || t2 == -1 {
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return -1
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}
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return t1 + t2
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

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