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++



filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// 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 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# 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 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// 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 



















                        chevron_right







                        filter_none










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.


Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts: