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Number of swaps to sort when only adjacent swapping allowed
• Difficulty Level : Hard
• Last Updated : 13 Apr, 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;  ``/* i is index for right subarray*/``    ``int` `k = left; ``/* i 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;``    ` `    ``/* i is index for right subarray*/``    ``int` `j = mid;``    ` `    ``/* i 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 array``static` `int` `countSwaps(``int` `arr[], ``int` `n)``{``    ``int` `temp[] = ``new` `int``[n];``    ``return` `_mergeSort(arr, temp, ``0``, n - ``1``);``}` `// Driver Code``public` `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 ``#i is index for right subarray*/``    ``k ``=` `left ``#i 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;``    ` `    ``/* i is index for``    ``right subarray*/``    ``int` `j = mid;``    ` `    ``/* i 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 array``static` `int` `countSwaps(``int` `[]arr, ``int` `n)``{``    ``int` `[]temp = ``new` `int``[n];``    ``return` `_mergeSort(arr, temp, 0, n - 1);``}` `// Driver Code``public` `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
References :
http://stackoverflow.com/questions/20990127/sorting-a-sequence-by-swapping-adjacent-elements-using-minimum-swaps
This article is contributed by Shivam Gupta. 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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