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 reverse order, the inversion count is the maximum.
Given an array arr[]. The task is to find the inversion count of arr[]. Where two elements arr[i] and arr[j] form an inversion if a[i] > a[j] and i < j.
Examples:
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[] = {1, 20, 6, 4, 5}
Output: 5
Explanation: Given array has five inversions: (20, 6), (20, 4), (20, 5), (6, 4), (6, 5).
Naive 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 indices in the array and print the sum.
Follow the below steps to Implement the idea:
- 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.
Below is the Implementation of the above approach:
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
import java.io.*;
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));
}
}
|
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>
|
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);
?>
|
Output Number of inversions are 5
Time Complexity: O(N2), Two nested loops are needed to traverse the array from start to end.
Auxiliary Space: O(1), No extra space is required.
Count Inversions in an array using Merge Sort:
Below is the idea to solve the problem:
Use Merge sort with modification that every time an unsorted pair is found increment count by one and return count at the end.
Illustration:
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]
The complete picture:
Follow the below steps to Implement the idea:
- 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 in the first half, the number of inversions 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.
Below is the Implementation of the above approach:
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
Time Complexity: O(N * log N), The algorithm used is divide and conquer i.e. merge sort whose complexity is O(n log n).
Auxiliary Space: O(N), Temporary array.
Note: 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.
Count Inversions in an array using Heapsort and Bisection:
Follow the below steps to Implement the idea:
- Create a heap with new pair elements, (element, index).
- After sorting them, pop out each minimum sequentially 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.
Below is the idea to Implement the above approach:
C++
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
int getNumOfInversions(vector<int>& A) {
int N = A.size();
if (N <= 1) {
return 0;
}
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> sortList;
int result = 0;
for (int i = 0; i < N; i++) {
sortList.push(make_pair(A[i], i));
}
vector<int> x;
while (!sortList.empty()) {
int v = sortList.top().first;
int i = sortList.top().second;
sortList.pop();
int y = upper_bound(x.begin(), x.end(), i) - x.begin();
result += i - y;
x.insert(upper_bound(x.begin(), x.end(), i), i);
}
return result;
}
int main() {
vector<int> A = {1, 20, 6, 4, 5};
int result = getNumOfInversions(A);
cout << "Number of inversions are " << result << endl;
return 0;
}
|
Java
import java.util.ArrayList;
import java.util.List;
import java.util.PriorityQueue;
class Main {
public static int getNumOfInversions(List<Integer> A)
{
int N = A.size();
if (N <= 1) {
return 0;
}
PriorityQueue<int[]> sortList
= new PriorityQueue<>((a, b) -> a[0] - b[0]);
int result = 0;
for (int i = 0; i < N; i++) {
sortList.add(new int[] { A.get(i), i });
}
List<Integer> x = new ArrayList<>();
while (!sortList.isEmpty()) {
int[] v = sortList.poll();
int y = x.size()
- x.subList(0, x.size()).indexOf(v[1])
- 1;
int z = 0;
if (!x.isEmpty()) {
z = binarySearch(x, 0, x.size() - 1, v[1]);
if (z < 0) {
z = -(z + 1);
}
}
result += v[1] - z;
x.add(v[1]);
x.sort(null);
}
return result;
}
private static int binarySearch(List<Integer> list,
int start, int end,
int key)
{
while (start <= end) {
int mid = start + (end - start) / 2;
if (list.get(mid) == key) {
return mid;
}
else if (list.get(mid) > key) {
end = mid - 1;
}
else {
start = mid + 1;
}
}
return -(start + 1);
}
public static void main(String[] args)
{
List<Integer> A = List.of(1, 20, 6, 4, 5);
int result = getNumOfInversions(A);
System.out.println("Number of inversions are "
+ result);
}
}
|
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
if __name__ == '__main__':
A = [1, 20, 6, 4, 5]
result = getNumOfInversions(A)
print(f'Number of inversions are {result}')
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
namespace InversionCount
{
class MainClass
{
public static int GetNumOfInversions(List<int> A)
{
int N = A.Count;
if (N <= 1)
{
return 0;
}
var sortList = new List<int[]>();
int result = 0;
for (int i = 0; i < N; i++)
{
sortList.Add(new int[] { A[i], i });
}
List<int> x = new List<int>();
while (sortList.Any())
{
sortList = sortList.OrderBy(a => a[0]).ToList();
int[] v = sortList[0];
sortList.RemoveAt(0);
int y = x.Count
- x.GetRange(0, x.Count).IndexOf(v[1])
- 1;
int z = 0;
if (x.Any())
{
z = BinarySearch(x, 0, x.Count - 1, v[1]);
if (z < 0)
{
z = -(z + 1);
}
}
result += v[1] - z;
x.Add(v[1]);
x.Sort();
}
return result;
}
private static int BinarySearch(List<int> list,
int start, int end,
int key)
{
while (start <= end)
{
int mid = start + (end - start) / 2;
if (list[mid] == key)
{
return mid;
}
else if (list[mid] > key)
{
end = mid - 1;
}
else
{
start = mid + 1;
}
}
return -(start + 1);
}
public static void Main(string[] args)
{
List<int> A = new List<int> { 1, 20, 6, 4, 5 };
int result = GetNumOfInversions(A);
Console.WriteLine("Number of inversions are " + result);
}
}
}
|
Javascript
const GetNumOfInversions = (A) => {
const N = A.length;
if (N <= 1) {
return 0;
}
const sortList = [];
let result = 0;
for (let i = 0; i < N; i++) {
sortList.push([A[i], i]);
}
const x = [];
while (sortList.length) {
sortList.sort((a, b) => a[0] - b[0]);
const v = sortList[0];
sortList.shift();
const y = x.length - x.slice(0, x.length).indexOf(v[1]) - 1;
let z = 0;
if (x.length) {
z = BinarySearch(x, 0, x.length - 1, v[1]);
if (z < 0) {
z = -(z + 1);
}
}
result += v[1] - z;
x.push(v[1]);
x.sort();
}
return result;
}
const BinarySearch = (list, start, end, key) => {
while (start <= end) {
const mid = start + Math.floor((end - start) / 2);
if (list[mid] === key) {
return mid;
} else if (list[mid] > key) {
end = mid - 1;
} else {
start = mid + 1;
}
}
return -(start + 1);
}
const A = [1, 20, 6, 4, 5];
const result = GetNumOfInversions(A);
console.log(`Number of inversions are ${result}`);
|
OutputNumber of inversions are 5
Time Complexity: O(N * log N). Both heapsort and bisection can perform sorted insertion in (log n) in each element.
Auxiliary Space: O(N). A heap and a new list are the same length as the original array.
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Please write comments if you find any bug in the above program/algorithm or other ways to solve it.