# Minimum swaps to reach permuted array with at most 2 positions left swaps allowed

Given a permuted array of length N of first N natural numbers, we need to tell the minimum number of swaps required in the sorted array of first N natural number to reach given permuted array where a number can be swapped with at most 2 positions left to it. If it is not possible to reach permuted array by above swap condition then print not possible.
Examples:

```Input : arr = [1, 2, 5, 3, 4]
Output : 2
We can reach to above-permuted array
in total 2 swaps as shown below,
[1, 2, 3, 4, 5] -> [1, 2, 3, 5, 4] ->
[1, 2, 5, 3, 4]

Input : arr[] = [5, 1, 2, 3, 4]
Output : Not Possible
It is not possible to reach above array
just by swapping numbers 2 positions left
to it.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We can solve this problem using inversions. As we can see that if a number is at a position which is more than 2 places away from its actual position then it is not possible to reach there just by swapping with elements at 2 left positions and if all element satisfy this property (there are <=2 elements smaller than it on the right) then answer will simply be total number of inversions in the array because that many swaps will be needed to transform the array into permuted array.
We can find the number of inversions in N log N time using merge sort technique explained here so total time complexity of solution will be O(N log N) only.

```// C++ program to find minimum number of swaps
// to reach a permutation wiht at most 2 left
// swaps allowed for every element
#include <bits/stdc++.h>
using namespace std;

/* This funt merges two sorted arrays and returns inversion
count in the arrays.*/
int merge(int arr[], int temp[], int left, int mid, int right)
{
int inv_count = 0;

int i = left; /* i is index for left subarray*/
int j = mid;  /* j is index for right subarray*/
int k = left; /* k is index for resultant merged subarray*/
while ((i <= mid - 1) && (j <= right))
{
if (arr[i] <= arr[j])
temp[k++] = arr[i++];
else
{
temp[k++] = arr[j++];
inv_count = inv_count + (mid - i);
}
}

/* Copy the remaining elements of left subarray
(if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];

/* Copy the remaining elements of right subarray
(if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];

/*Copy back the merged elements to original array*/
for (i = left; i <= right; i++)
arr[i] = temp[i];

return inv_count;
}

/* An auxiliary recursive function that sorts the
input array and returns the number of inversions
in the array. */
int _mergeSort(int arr[], int temp[], int left, int right)
{
int mid, inv_count = 0;
if (right > left)
{
/* Divide the array into two parts and
call _mergeSortAndCountInv() for each
of the parts */
mid = (right + left)/2;

/* Inversion count will be sum of inversions
in left-part, right-part and number of inversions
in merging */
inv_count  = _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid+1, right);

/*Merge the two parts*/
inv_count += merge(arr, temp, left, mid+1, right);
}
return inv_count;
}

/* This function sorts the input array and returns the
number of inversions in the array */
int mergeSort(int arr[], int array_size)
{
int *temp = (int *)malloc(sizeof(int)*array_size);
return _mergeSort(arr, temp, 0, array_size - 1);
}

// method returns minimum number of swaps to reach
// permuted array 'arr'
int minSwapToReachArr(int arr[], int N)
{
//  loop over all elements to check Invalid
// permutation condition
for (int i = 0; i < N; i++)
{
/*  if an element is at distance more than 2
from its actual position then it is not
possible to reach permuted array just
by swapping with 2 positions left elements
so returning -1   */
if ((arr[i] - 1) - i > 2)
return -1;
}

/*  If permuted array is not Invalid, then number
of Inversion in array will be our final answer */
int numOfInversion = mergeSort(arr, N);
return numOfInversion;
}

//  Driver code to test above methods
int main()
{
//  change below example
int arr[] = {1, 2, 5, 3, 4};
int N = sizeof(arr) / sizeof(int);
int res = minSwapToReachArr(arr, N);
if (res == -1)
cout << "Not Possible\n";
else
cout << res << endl;
return 0;
}
```

Output:

```2
```

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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