README_EN.md

765. Couples Holding Hands

中文文档

Description

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 (row[1]) 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:

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

Solutions

Python3

class Solution:
    def minSwapsCouples(self, row: List[int]) -> int:
        n = len(row) >> 1
        p = list(range(n))

        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        for i in range(0, len(row), 2):
            a, b = row[i] >> 1, row[i + 1] >> 1
            p[find(a)] = find(b)

        cnt = 0
        for i in range(n):
            if i == find(i):
                cnt += 1
        return n - cnt

Java

class Solution {
    private int[] p;

    public int minSwapsCouples(int[] row) {
        int n = row.length >> 1;
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        for (int i = 0; i < row.length; i += 2) {
            int a = row[i] >> 1, b = row[i + 1] >> 1;
            p[find(a)] = find(b);
        }
        int cnt = 0;
        for (int i = 0; i < n; ++i) {
            if (i == find(i)) {
                ++cnt;
            }
        }
        return n - cnt;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

C++

class Solution {
public:
    vector<int> p;

    int minSwapsCouples(vector<int> &row) {
        int n = row.size() >> 1;
        p.resize(n);
        for (int i = 0; i < n; ++i)
        {
            p[i] = i;
        }
        for (int i = 0; i < row.size(); i += 2)
        {
            int a = row[i] >> 1, b = row[i + 1] >> 1;
            p[find(a)] = find(b);
        }
        int cnt = 0;
        for (int i = 0; i < n; ++i)
        {
            if (i == find(i))
                ++cnt;
        }
        return n - cnt;
    }

    int find(int x) {
        if (p[x] != x)
        {
            p[x] = find(p[x]);
        }
        return p[x];
    }
};

Go

var p []int

func minSwapsCouples(row []int) int {
	n := len(row) >> 1
	p = make([]int, n)
	for i := 0; i < n; i++ {
		p[i] = i
	}
	for i := 0; i < len(row); i += 2 {
		a, b := row[i]>>1, row[i+1]>>1
		p[find(a)] = find(b)
	}
	cnt := 0
	for i := 0; i < n; i++ {
		if i == find(i) {
			cnt++
		}
	}
	return n - cnt
}

func find(x int) int {
	if p[x] != x {
		p[x] = find(p[x])
	}
	return p[x]
}

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