# Couples Holding Hands Problem

## Description

LeetCode Problem 765.

There are n couples sitting in 2n seats arranged in a row and want to hold hands.

The people and seats are represented by an integer array row where row[i] is the ID of the person sitting in the i^th seat. 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).

Return 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.

Example 1:

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

Example 2:

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

Constraints:

• 2n == row.length
• 2 <= n <= 30
• n is even.
• 0 <= row[i] < 2n
• All the elements of row are unique.

## Sample C++ Code using Union Find

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
class UF {
public:
UF (int n) : parent(n) {}

void setParent(int x, int p) {
parent[x] = p;
}

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

bool union_find(int x, int y) {
int px = find(x);
int py = find(y);
if (px == py)
return false;
parent[px] = py;
return true;
}

private:
vector<int> parent;
};

class Solution {
public:
int minSwapsCouples(vector<int>& row) {
if (row.empty()) return 0;
UF uf(row.size());
for (int i = 0; i < row.size(); i += 2) {
uf.setParent(row[i], row[i]);
uf.setParent(row[i+1], row[i]);
}
int counter = 0;
for (int i = 0; i < row.size(); i += 2) {
if (uf.union_find(i, i + 1))
counter++;
}
return counter;
}
};
``````