leetcode 765 - Couples Holding Hands

https://leetcode.com/problems/couples-holding-hands/

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:

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

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
class Solution {
public:
int minSwapsCouples(vector<int>& row) {
int size = row.size();
vector<int> pos(size);

for (int i = 0; i < size; i++)
pos[row[i]] = i;

int min_swap = 0;

for (int i = 0; i < size; i += 2) {
int val;
if (row[i] % 2 == 0)
val = row[i] + 1;
else
val = row[i] - 1;

int idx = pos[val];

if (idx != i + 1) {
pos[row[i+1]] = idx;
swap(row[idx], row[i+1]);
min_swap++;
}
}

return min_swap;
}
};