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.



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++

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// 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;
}

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Java

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// Java program to find minimum 
// number of swaps to reach a 
// permutation wiht at most 2 left 
// swaps allowed for every element 
class GFG
{
  
    /* This funt merges two sorted
    arrays and returns inversion 
    count in the arrays.*/
    static 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. */
    static 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 */
    static int mergeSort(int arr[], int array_size)
    {
        int[] temp = new int[array_size];
        return _mergeSort(arr, temp, 0, array_size - 1);
    }
  
    // method returns minimum number of  
    // swaps to reach permuted array 'arr' 
    static 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 
    public static void main(String[] args) 
    {
          
        // change below example 
        int arr[] = {1, 2, 5, 3, 4};
        int N = arr.length;
        int res = minSwapToReachArr(arr, N);
        System.out.println(res == -1 ? "Not Possible\n" : res);
    }
}
  
// This code contributed by Rajput-Ji

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Python3

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# Python3 program to find minimum number of
# swaps to reach a permutation wiht at most
# 2 left swaps allowed for every element
  
# This funt merges two sorted arrays and
# returns inversion count in the arrays.
def merge(arr, temp, left, mid, right):
  
    inv_count = 0
  
    i = left # i is index for left subarray
    j = mid # j is index for right subarray
    k = left # k is index for resultant merged subarray
    while (i <= mid - 1) and (j <= right):
      
        if arr[i] <= arr[j]:
            temp[k] = arr[i]
            k, i = k + 1, i + 1
          
        else:
            temp[k] = arr[j]
            k, j = k + 1, j + 1
            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]
        k, i = k + 1, i + 1
  
    # Copy the remaining elements of right
    # subarray (if there are any) to temp
    while j <= right:
        temp[k] = arr[j]
        k, j = k + 1, j + 1
  
    # Copy back the merged elements to original array
    for i in range(left, right + 1):
        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.
def _mergeSort(arr, temp, left, right):
  
    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 
def mergeSort(arr, array_size):
  
    temp = [None] * array_size
    return _mergeSort(arr, temp, 0, array_size - 1)
  
# method returns minimum number of
# swaps to reach permuted array 'arr'
def minSwapToReachArr(arr, N):
  
    # loop over all elements to check
    # Invalid permutation condition
    for i in range(0, N):
      
        # 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
    numOfInversion = mergeSort(arr, N)
    return numOfInversion
  
# Driver code to test above methods
if __name__ == "__main__":
  
    # change below example
    arr = [1, 2, 5, 3, 4]
    N = len(arr)
    res = minSwapToReachArr(arr, N)
    if res == -1:
        print("Not Possible")
    else:
        print(res)
  
# This code is contributed by Rituraj Jain

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C#

// C# program to find minimum
// number of swaps to reach a
// permutation wiht at most 2 left
// swaps allowed for every element
using System;
class GFG
{

/* This funt merges two sorted
arrays and returns inversion
count in the arrays.*/
static 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. */ static 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 */
static int mergeSort(int []arr, int array_size)
{
int[] temp = new int[array_size];
return _mergeSort(arr, temp, 0, array_size – 1);
}

// method returns minimum number of
// swaps to reach permuted array ‘arr’
static 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
static void Main()
{

// change below example
int []arr = {1, 2, 5, 3, 4};
int N = arr.Length;
int res = minSwapToReachArr(arr, N);
if(res == -1)
Console.WriteLine(“Not Possible”);
else
Console.WriteLine(res);
}
}

// This code is contributed by mits

Output:

2

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