Inversion Count for an array indicates – how far (or close) the array is from being sorted. If the array is already sorted, then the inversion count is 0, but if the array is sorted in the reverse order, the inversion count is the maximum.
Formally speaking, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j
Example:
Input: arr[] = {8, 4, 2, 1}
Output: 6
Explanation: Given array has six inversions:
(8, 4), (4, 2), (8, 2), (8, 1), (4, 1), (2, 1).
Input: arr[] = {3, 1, 2}
Output: 2
Explanation: Given array has two inversions:
(3, 1), (3, 2)
METHOD 1 (Simple)
- Approach: Traverse through the array, and for every index, find the number of smaller elements on its right side of the array. This can be done using a nested loop. Sum up the counts for all index in the array and print the sum.
- Algorithm:
- Traverse through the array from start to end
- For every element, find the count of elements smaller than the current number up to that index using another loop.
- Sum up the count of inversion for every index.
- Print the count of inversions.
- Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int getInvCount( int arr[], int n)
{
int inv_count = 0;
for ( int i = 0; i < n - 1; i++)
for ( int j = i + 1; j < n; j++)
if (arr[i] > arr[j])
inv_count++;
return inv_count;
}
int main()
{
int arr[] = { 1, 20, 6, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << " Number of inversions are "
<< getInvCount(arr, n);
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
int getInvCount( int arr[], int n)
{
int inv_count = 0;
for ( int i = 0; i < n - 1; i++)
for ( int j = i + 1; j < n; j++)
if (arr[i] > arr[j])
inv_count++;
return inv_count;
}
int main()
{
int arr[] = { 1, 20, 6, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
printf ( " Number of inversions are %d \n" ,
getInvCount(arr, n));
return 0;
}
|
Java
class Test {
static int arr[] = new int [] { 1 , 20 , 6 , 4 , 5 };
static int getInvCount( int n)
{
int inv_count = 0 ;
for ( int i = 0 ; i < n - 1 ; i++)
for ( int j = i + 1 ; j < n; j++)
if (arr[i] > arr[j])
inv_count++;
return inv_count;
}
public static void main(String[] args)
{
System.out.println( "Number of inversions are "
+ getInvCount(arr.length));
}
}
|
Python3
def getInvCount(arr, n):
inv_count = 0
for i in range (n):
for j in range (i + 1 , n):
if (arr[i] > arr[j]):
inv_count + = 1
return inv_count
arr = [ 1 , 20 , 6 , 4 , 5 ]
n = len (arr)
print ( "Number of inversions are" ,
getInvCount(arr, n))
|
C#
using System;
using System.Collections.Generic;
class GFG {
static int [] arr = new int [] { 1, 20, 6, 4, 5 };
static int getInvCount( int n)
{
int inv_count = 0;
for ( int i = 0; i < n - 1; i++)
for ( int j = i + 1; j < n; j++)
if (arr[i] > arr[j])
inv_count++;
return inv_count;
}
public static void Main()
{
Console.WriteLine( "Number of "
+ "inversions are "
+ getInvCount(arr.Length));
}
}
|
PHP
<?php
function getInvCount(& $arr , $n )
{
$inv_count = 0;
for ( $i = 0; $i < $n - 1; $i ++)
for ( $j = $i + 1; $j < $n ; $j ++)
if ( $arr [ $i ] > $arr [ $j ])
$inv_count ++;
return $inv_count ;
}
$arr = array (1, 20, 6, 4, 5 );
$n = sizeof( $arr );
echo "Number of inversions are " ,
getInvCount( $arr , $n );
?>
|
Javascript
<script>
arr = [1, 20, 6, 4, 5];
function getInvCount(arr){
let inv_count = 0;
for (let i=0; i<arr.length-1; i++){
for (let j=i+1; j<arr.length; j++){
if (arr[i] > arr[j]) inv_count++;
}
}
return inv_count;
}
document.write( "Number of inversions are " + getInvCount(arr));
</script>
|
Output Number of inversions are 5
- Complexity Analysis:
- Time Complexity: O(n^2), Two nested loops are needed to traverse the array from start to end, so the Time complexity is O(n^2)
- Space Complexity:O(1), No extra space is required.

METHOD 2(Enhance Merge Sort)
- Approach:
Suppose the number of inversions in the left half and right half of the array (let be inv1 and inv2); what kinds of inversions are not accounted for in Inv1 + Inv2? The answer is – the inversions that need to be counted during the merge step. Therefore, to get the total number of inversions that needs to be added are the number of inversions in the left subarray, right subarray, and merge().

- How to get the number of inversions in merge()?
In merge process, let i is used for indexing left sub-array and j for right sub-array. At any step in merge(), if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j]


- Algorithm:
- The idea is similar to merge sort, divide the array into two equal or almost equal halves in each step until the base case is reached.
- Create a function merge that counts the number of inversions when two halves of the array are merged, create two indices i and j, i is the index for the first half, and j is an index of the second half. if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j].
- Create a recursive function to divide the array into halves and find the answer by summing the number of inversions is the first half, the number of inversion in the second half and the number of inversions by merging the two.
- The base case of recursion is when there is only one element in the given half.
- Print the answer
- Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int _mergeSort( int arr[], int temp[], int left, int right);
int merge( int arr[], int temp[], int left, int mid,
int right);
int mergeSort( int arr[], int array_size)
{
int temp[array_size];
return _mergeSort(arr, temp, 0, array_size - 1);
}
int _mergeSort( int arr[], int temp[], int left, int right)
{
int mid, inv_count = 0;
if (right > left) {
mid = (right + left) / 2;
inv_count += _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
int merge( int arr[], int temp[], int left, int mid,
int right)
{
int i, j, k;
int inv_count = 0;
i = left;
j = mid;
k = left;
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
inv_count = inv_count + (mid - i);
}
}
while (i <= mid - 1)
temp[k++] = arr[i++];
while (j <= right)
temp[k++] = arr[j++];
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
int main()
{
int arr[] = { 1, 20, 6, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
int ans = mergeSort(arr, n);
cout << " Number of inversions are " << ans;
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
int _mergeSort( int arr[], int temp[], int left, int right);
int merge( int arr[], int temp[], int left, int mid,
int right);
int mergeSort( int arr[], int array_size)
{
int * temp = ( int *) malloc ( sizeof ( int ) * array_size);
return _mergeSort(arr, temp, 0, array_size - 1);
}
int _mergeSort( int arr[], int temp[], int left, int right)
{
int mid, inv_count = 0;
if (right > left) {
mid = (right + left) / 2;
inv_count += _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
int merge( int arr[], int temp[], int left, int mid,
int right)
{
int i, j, k;
int inv_count = 0;
i = left;
j = mid;
k = left;
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
inv_count = inv_count + (mid - i);
}
}
while (i <= mid - 1)
temp[k++] = arr[i++];
while (j <= right)
temp[k++] = arr[j++];
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
int main( int argv, char ** args)
{
int arr[] = { 1, 20, 6, 4, 5 };
printf ( " Number of inversions are %d \n" ,
mergeSort(arr, 5));
getchar ();
return 0;
}
|
Java
import java.util.Arrays;
public class GFG {
private static int mergeAndCount( int [] arr, int l,
int m, int r)
{
int [] left = Arrays.copyOfRange(arr, l, m + 1 );
int [] right = Arrays.copyOfRange(arr, m + 1 , r + 1 );
int i = 0 , j = 0 , k = l, swaps = 0 ;
while (i < left.length && j < right.length) {
if (left[i] <= right[j])
arr[k++] = left[i++];
else {
arr[k++] = right[j++];
swaps += (m + 1 ) - (l + i);
}
}
while (i < left.length)
arr[k++] = left[i++];
while (j < right.length)
arr[k++] = right[j++];
return swaps;
}
private static int mergeSortAndCount( int [] arr, int l,
int r)
{
int count = 0 ;
if (l < r) {
int m = (l + r) / 2 ;
count += mergeSortAndCount(arr, l, m);
count += mergeSortAndCount(arr, m + 1 , r);
count += mergeAndCount(arr, l, m, r);
}
return count;
}
public static void main(String[] args)
{
int [] arr = { 1 , 20 , 6 , 4 , 5 };
System.out.println(
mergeSortAndCount(arr, 0 , arr.length - 1 ));
}
}
|
Python3
def mergeSort(arr, n):
temp_arr = [ 0 ] * n
return _mergeSort(arr, temp_arr, 0 , n - 1 )
def _mergeSort(arr, temp_arr, left, right):
inv_count = 0
if left < right:
mid = (left + right) / / 2
inv_count + = _mergeSort(arr, temp_arr,
left, mid)
inv_count + = _mergeSort(arr, temp_arr,
mid + 1 , right)
inv_count + = merge(arr, temp_arr, left, mid, right)
return inv_count
def merge(arr, temp_arr, left, mid, right):
i = left
j = mid + 1
k = left
inv_count = 0
while i < = mid and j < = right:
if arr[i] < = arr[j]:
temp_arr[k] = arr[i]
k + = 1
i + = 1
else :
temp_arr[k] = arr[j]
inv_count + = (mid - i + 1 )
k + = 1
j + = 1
while i < = mid:
temp_arr[k] = arr[i]
k + = 1
i + = 1
while j < = right:
temp_arr[k] = arr[j]
k + = 1
j + = 1
for loop_var in range (left, right + 1 ):
arr[loop_var] = temp_arr[loop_var]
return inv_count
arr = [ 1 , 20 , 6 , 4 , 5 ]
n = len (arr)
result = mergeSort(arr, n)
print ( "Number of inversions are" , result)
|
C#
using System;
public class Test {
static int mergeSort( int [] arr, int array_size)
{
int [] temp = new int [array_size];
return _mergeSort(arr, temp, 0, array_size - 1);
}
static int _mergeSort( int [] arr, int [] temp, int left,
int right)
{
int mid, inv_count = 0;
if (right > left) {
mid = (right + left) / 2;
inv_count += _mergeSort(arr, temp, left, mid);
inv_count
+= _mergeSort(arr, temp, mid + 1, right);
inv_count
+= merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
static int merge( int [] arr, int [] temp, int left,
int mid, int right)
{
int i, j, k;
int inv_count = 0;
i = left;
j = mid;
k = left;
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
inv_count = inv_count + (mid - i);
}
}
while (i <= mid - 1)
temp[k++] = arr[i++];
while (j <= right)
temp[k++] = arr[j++];
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
public static void Main()
{
int [] arr = new int [] { 1, 20, 6, 4, 5 };
Console.Write( "Number of inversions are "
+ mergeSort(arr, 5));
}
}
|
Javascript
<script>
function mergeAndCount(arr,l,m,r)
{
let left = [];
for (let i = l; i < m + 1; i++)
{
left.push(arr[i]);
}
let right = [];
for (let i = m + 1; i < r + 1; i++)
{
right.push(arr[i]);
}
let i = 0, j = 0, k = l, swaps = 0;
while (i < left.length && j < right.length)
{
if (left[i] <= right[j])
{
arr[k++] = left[i++];
}
else
{
arr[k++] = right[j++];
swaps += (m + 1) - (l + i);
}
}
while (i < left.length)
{
arr[k++] = left[i++];
}
while (j < right.length)
{
arr[k++] = right[j++];
}
return swaps;
}
function mergeSortAndCount(arr, l, r)
{
let count = 0;
if (l < r)
{
let m = Math.floor((l + r) / 2);
count += mergeSortAndCount(arr, l, m);
count += mergeSortAndCount(arr, m + 1, r);
count += mergeAndCount(arr, l, m, r);
}
return count;
}
let arr= new Array(1, 20, 6, 4, 5 );
document.write(mergeSortAndCount(arr, 0, arr.length - 1));
</script>
|
Output:
Number of inversions are 5
Complexity Analysis:
- Time Complexity: O(n log n), The algorithm used is divide and conquer, So in each level, one full array traversal is needed, and there are log n levels, so the time complexity is O(n log n).
- Space Complexity: O(n), Temporary array.
Note that the above code modifies (or sorts) the input array. If we want to count only inversions, we need to create a copy of the original array and call mergeSort() on the copy to preserve the original array’s order.
METHOD 3(Heapsort + Bisection)
- Algorithm:
- Create a heap with new pair elements, (element, index).
- After sorting them, pop out each minimum sequently and create a new sorted list with the indexes.
- Calculate the difference between the original index and the index of bisection of the new sorted list.
- Sum up the difference.
Python3
from heapq import heappush, heappop
from bisect import bisect, insort
def getNumOfInversions(A):
N = len (A)
if N < = 1 :
return 0
sortList = []
result = 0
for i, v in enumerate (A):
heappush(sortList, (v, i))
x = []
while sortList:
v, i = heappop(sortList)
y = bisect(x, i)
result + = i - y
insort(x, i)
return result
A = [ - 1 , 6 , 3 , 4 , 7 , 4 ]
result = getNumOfInversions(A)
print (f 'Number of inversions are {result}' )
|
Complexity Analysis:
- Time Complexity: O(n log n). Both heapsort and bisection can perform sorted insertion in log n in each element, so the time complexity is O(n log n).
- Space Complexity: O(n). A heap and a new list are the same length as the original array.
You may like to see:
Count inversions in an array | Set 2 (Using Self-Balancing BST)
Counting Inversions using Set in C++ STL
Count inversions in an array | Set 3 (Using BIT)
Please write comments if you find any bug in the above program/algorithm or other ways to solve it.