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# 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 <bits/stdc++.h>

/* 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);
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 <bits/stdc++.h>

#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

```
<?php
// PHP program to count number of swaps required
// to sort an array when only swapping of adjacent
// elements is allowed.

/* This function merges two sorted arrays and returns inversion
count in the arrays.*/
function 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) && (\$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. */
function _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 = 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);
echo "Number of swaps is ". countSwaps(\$arr, \$n);
return 0;
?>

```

## Javascript

```<script>
// Javascript program to count number of
// swaps required to sort an array
// when only swapping of adjacent
// elements is allowed.

// This function merges two sorted
// arrays and returns inversion
// count in the arrays.
function merge(arr,temp,left,mid,right)
{
let inv_count = 0;

/* i is index for left subarray*/
let i = left;

/* j is index for right subarray*/
let j = mid;

/* k is index for resultant merged subarray*/
let 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.
function _mergeSort(arr,temp,left,right)
{
let mid, inv_count = 0;
if (right > left)
{
// Divide the array into two parts and
// call _mergeSortAndCountInv() for
// each of the parts
mid = Math.floor((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)
{
let temp = new Array(n);
for(let i = 0; i < n; i++)
{
temp[i] = 0;
}
return _mergeSort(arr, temp, 0, n - 1);
}

// Driver Code
let arr=[1, 20, 6, 4, 5];
let n = arr.length;
document.write("Number of swaps is "
+ countSwaps(arr, n));

// This code is contributed by rag2127
</script>```
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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