Number of swaps to sort when only adjacent swapping allowed

• Difficulty Level : Hard
• Last Updated : 16 Jul, 2021

Given an array arr[] of non negative integers. We can perform a swap operation on any two adjacent elements in the array. Find the minimum number of swaps needed to sort the array in ascending order.

Examples :

Input  : arr[] = {3, 2, 1}
Output : 3
We need to do following swaps
(3, 2), (3, 1) and (1, 2)

Input  : arr[] = {1, 20, 6, 4, 5}
Output : 5

There is an interesting solution to this problem. It can be solved using the fact that number of swaps needed is equal to number of inversions. So we basically need to count inversions in array.
The fact can be established using below observations:
1) A sorted array has no inversions.
2) An adjacent swap can reduce one inversion. Doing x adjacent swaps can reduce x inversions in an array.

C++

 // C++ program to count number of swaps required// to sort an array when only swapping of adjacent// elements is allowed.#include  /* This function 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++];             /* this is tricky -- see above explanation/              diagram for merge()*/            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 countSwaps(int arr[], int n){    int temp[n];    return _mergeSort(arr, temp, 0, n - 1);} /* Driver progra to test above functions */int main(int argv, char** args){    int arr[] = {1, 20, 6, 4, 5};    int n = sizeof(arr)/sizeof(arr);    printf("Number of swaps is %d \n", countSwaps(arr, n));    return 0;}

Java

 // Java program to count number of// swaps required to sort an array// when only swapping of adjacent// elements is allowed.import java.io.*; class GFG {     // This function 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;     /* i is index for left subarray*/    int i = left;         /* j is index for right subarray*/    int j = mid;         /* k is index for resultant merged subarray*/    int k = left;         while ((i <= mid - 1) && (j <= right))    {        if (arr[i] <= arr[j])            temp[k++] = arr[i++];        else        {            temp[k++] = arr[j++];             /* this is tricky -- see above /             explanation diagram for merge()*/            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 arraystatic int countSwaps(int arr[], int n){    int temp[] = new int[n];    return _mergeSort(arr, temp, 0, n - 1);} // Driver Codepublic static void main (String[] args){     int arr[] = {1, 20, 6, 4, 5};    int n = arr.length;        System.out.println("Number of swaps is "                           + countSwaps(arr, n));    }} // This code is contributed by vt_m

Python3

 # python 3 program to count number of swaps required# to sort an array when only swapping of adjacent# elements is allowed.#include  #This function 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 += 1            i += 1        else:            temp[k] = arr[j]            k += 1            j += 1             #this is tricky -- see above explanation/            # diagram for merge()*/            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 += 1        i += 1     #Copy the remaining elements of right subarray    # (if there are any) to temp*/    while (j <= right):        temp[k] = arr[j]        k += 1        j += 1     # Copy back the merged elements to original array*/    for i in range(left,right+1,1):        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 = int((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 countSwaps(arr, n):    temp = [0 for i in range(n)]    return _mergeSort(arr, temp, 0, n - 1) # Driver progra to test above functions */if __name__ == '__main__':    arr = [1, 20, 6, 4, 5]    n = len(arr)    print("Number of swaps is",countSwaps(arr, n))      # This code is contributed by# Surendra_Gangwar

C#

 // C# program to count number of// swaps required to sort an array// when only swapping of adjacent// elements is allowed.using System; class GFG{     // This function 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;     /* i is index for    left subarray*/    int i = left;         /* j is index for    right subarray*/    int j = mid;         /* k is index for resultant    merged subarray*/    int k = left;         while ((i <= mid - 1) &&           (j <= right))    {        if (arr[i] <= arr[j])            temp[k++] = arr[i++];        else        {            temp[k++] = arr[j++];             /* this is tricky -- see above /            explanation diagram for merge()*/            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 arraystatic int countSwaps(int []arr, int n){    int []temp = new int[n];    return _mergeSort(arr, temp, 0, n - 1);} // Driver Codepublic static void Main (){     int []arr = {1, 20, 6, 4, 5};    int n = arr.Length;        Console.Write("Number of swaps is " +                         countSwaps(arr, n));    }} // This code is contributed by nitin mittal.

PHP

 \$left)    {        /* Divide the array into two parts and call          _mergeSortAndCountInv() for each of the parts */        \$mid = intval((\$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 */function countSwaps(&\$arr, \$n){    \$temp = array_fill(0,\$n,NULL);    return _mergeSort(\$arr, \$temp, 0, \$n - 1);}  /* Driver progra to test above functions */ \$arr = array(1, 20, 6, 4, 5);\$n = sizeof(\$arr)/sizeof(\$arr);echo "Number of swaps is ". countSwaps(\$arr, \$n);return 0;?>

Javascript



Output:

Number of swaps is 5

Time Complexity : O(n Log n)
Related Post :
Minimum number of swaps required to sort an array

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