Problem of the day
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.
The first line contains a single integer ‘T’ representing the number of test cases.
The first line of each test case will contain a single integer ‘N’ such that the size of the list is 2 * ‘N’.
The second line of each test case will contain 2 * ‘N’ integers separated by a single space that represents the elements/numbers present in the list initially.
For each test case, print the minimum number of swaps possible to make every even number ‘E’ adjacent to (‘E’ + 1).
Output for every test case will be printed in a separate line.
You don’t need to print anything; It has already been taken care of. Just implement the given function.
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.