765. Couples Holding Hands

LWC 67: 765. Couples Holding Hands

Problem:

N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so that every couple is sitting side by side. A swap consists of choosing any two people, then they stand up and switch seats. The people and seats are represented by an integer from 0 to 2N-1, the couples are numbered in order, the first couple being (0, 1), the second couple being (2, 3), and so on with the last couple being (2N-2, 2N-1). The couples’ initial seating is given by row[i] being the value of the person who is initially sitting in the i-th seat.

Example 1:

Input: row = [0, 2, 1, 3] Output: 1 Explanation: We only need to swap the second (row1) and third (row[2]) person.

Example 2:

Input: row = [3, 2, 0, 1] Output: 0 Explanation: All couples are already seated side by side.

Note:

• len(row) is even and in the range of [4, 60].
• row is guaranteed to be a permutation of 0…len(row)-1.

思路： 贪心，不过不好证明。从位置0开始，看位置1是否是它的partner，如果是，直接忽略这组couple，如果不是，从后面的数组中，找到partner直接交换，并且记录交换次数一次。遍历完整个数组即为最优解。

简单举例说明下：

首先，如果相邻两个位置已经是couple了，则直接忽略，那么原问题就变成了两种情况： 1. 交换位置后，能够使得两组couple得到匹配（最优） 2. 交换位置后，只能使得一组couple得到匹配（次优）

显然我们会进行最优的couple匹配，先把这部分给匹配了。针对上述贪心的做法，的确可以解决问题1.

问题2的关键在于不同的匹配顺序，是否会影响到子问题的最优解？答案是否定的。比如

[a, c, b, e, d, f]

可以得到三组block如下：
a --- c
b --- e
d --- f

考虑a --- c 的子状态：
1. 匹配a --- b,得到 c --- e  d --- f
2. 匹配c --- d,得到 b --- e  a --- f

两种可能的子状态实际是等价的，所以无关乎顺序，直接贪心选择a的partner或者c的partner即可。

Java版本：

    public int minSwapsCouples(int[] row) {
        int n = row.length;
        int ans = 0;
        for (int i = 0; i < n; i += 2) {
            int dest;
            if ((row[i] & 1) == 0) dest = row[i] + 1;
            else dest = row[i] - 1;
            if (row[i + 1] == dest) continue;
            ans ++;
            for (int j = i + 2; j < n; ++j) {
                if (row[j] == dest) {
                    int tmp = row[i + 1];
                    row[i + 1] = row[j];
                    row[j] = tmp;
                    break;
                }
            }
        }
        return ans;
    }

Python版本：

class Solution(object):
    def minSwapsCouples(self, row):
        """
        :type row: List[int]
        :rtype: int
        """
        n = len(row)
        ans = 0
        for i in range(0, n, 2):
            dest = row[i] + 1 if (row[i] & 1) == 0 else row[i] - 1
            if dest == row[i + 1]: continue
            ans += 1
            for j in range(i + 2, n):
                if row[j] == dest:
                    row[i + 1], row[j] = row[j], row[i + 1]
        return ans