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Problem of the day

You are provided with a list of numbers from ‘0’ to (2 * ’N’ - 1). You have to find the minimum number of swaps needed to make every even number ‘E’ (present in the list) adjacent to (‘E’ + 1).

```
List = [3, 0, 2, 1]
We have to make ‘0’ adjacent to ‘1’ and ‘2’ to ‘3’. And, to achieve this we can swap ‘0’ with ‘2’.
New list = [3, 2, 0, 1].
Therefore, the answer (minimum number of swaps) is equal to 1.
```

```
There will be only distinct numbers present in the given list.
```

Detailed explanation

```
1 <= T <= 10
1 <= N <= 100
0 <= ARR[ i ] < 2 * N
Time limit: 1 sec
```

```
1
2
3 0 2 1
```

```
1
```

```
For the first test case, an explanation is given in the description.
```

```
2
1
1 0
3
1 0 2 3 5 4
```

```
0
0
```

```
In the first test case, the required pairing of all the even numbers is already done.
In the second test case, the required pairing of all the even numbers is already done.
```