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:
- A sorted array has no inversions.
- An adjacent swap can reduce one inversion. Doing x adjacent swaps can reduce x inversions in an array.
// 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));
}
}
Output :
Number of swaps is 5
Time Complexity : O(n Log n)