# Number of swaps to sort when only adjacent swapping allowed

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

Recommended Problem

There is an interesting solution to this problem. It can be solved using the fact that a 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.

Implementation:

## 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 program to test above functions */` `int` `main(``int` `argv, ``char``** args)` `{` `    ``int` `arr[] = {1, 20, 6, 4, 5};` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);` `    ``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 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 ``#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 program 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 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 program to test above functions */`   `\$arr` `= ``array``(1, 20, 6, 4, 5);` `\$n` `= sizeof(``\$arr``)/sizeof(``\$arr``[0]);` `echo` `"Number of swaps is "``. countSwaps(``\$arr``, ``\$n``);` `return` `0;` `?>`

## Javascript

 ``

Output

`Number of swaps is 5 `

Time Complexity: O(n Log n)
Auxiliary Space: O(n)

Related Post :
Minimum number of swaps required to sort an array

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