Merge Sort: Counting Inversions


Problem Statement :


In an array, arr , the elements at indices i and j (where i < j ) form an inversion if arr[ i ] > arr[ j ]. In other words, inverted elements arr[ i ]  and arr[ j ]  are considered to be "out of order". To correct an inversion, we can swap adjacent elements.


Function Description

Complete the function countInversions in the editor below.

countInversions has the following parameter(s):

int arr[n]: an array of integers to sort
Returns

int: the number of inversions
Input Format

The first line contains an integer, d, the number of datasets.

Each of the next d pairs of lines is as follows:

1. The first line contains an integer, n , the number of elements in arr.
The second line contains n space-separated integers,  arr[ i ].

Constraints

1  <=  d   <=  15
1  <=   n  <=  10^5
1  <=  arr[ i ]  <=  10^7


Solution :



title-img 


                            Solution in C :

In C :




#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
long long int c=0;
void MergeIP(int a[],int l,int r){
	if(l<r){
		int k=(l+r)/2;
		MergeIP(a,l,k);
		MergeIP(a,k+1,r);
		MergeIPcount(a,l,r,k);
	}
}

void MergeIPcount(int a[],int l,int r,int k){
	int *b;    
	b=(int *)malloc(sizeof(int)*(r-l+1));
	int i=l,j=k+1,x=0;
	while(i<=k&&j<=r){
		if(a[i]<=a[j]){
            b[x++]=a[i++];            
        }			
		else if(a[i]>a[j]){
            b[x++]=a[j++];
            c=c+(k-i+1);
        }
        
            
	}	
	while(i<=k)
		b[x++]=a[i++];
	while(j<=r)
		b[x++]=a[j++];
	i=l;
	x=0;
	while(i<=r)
		a[i++]=b[x++];
	free(b);
}
int main(){
    int t; 
    scanf("%d",&t);
    for(int a0 = 0; a0 < t; a0++){
        int n; 
        scanf("%d",&n);
        int *arr = malloc(sizeof(int) * n);
        for(int arr_i = 0; arr_i < n; arr_i++){
           scanf("%d",&arr[arr_i]);
        }
        MergeIP(arr,0,n-1);
        printf("%lld\n",c);
        c=0;
    }
    return 0;
}
                        


                        Solution in C++ :

In  C++ :




#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;
long long contador=0;
typedef vector<int> vi;
vi va, vb, vc;
void combinar(int a, int b)
{
    //vi r;
    int i=a, mitad=(a+b)/2,j=mitad, w=a;
    while (i<mitad and j<b)
    {
        //cout<<"coi\n";
        if (va[i]<=va[j])
        {
            vb[w]=va[i];
            ++i;
        }
        else
        {
            contador += mitad-i;
            //cout<<"aqui\n";
            vb[w]=va[j];
            ++j;
        } 
            ++w;
    }
    while (i<mitad )
    {
        vb[w]=va[i];
        ++i;
        ++w;
    }
    while (j<b)
        {
            //contador += a.size()-i;
            vb[w]=va[j];
            ++j;
        ++w;
        } 
    for (int p=a; p<b; ++p)
        va[p]=vb[p];
    return ;
}

void mergesort(int a, int b)
{
   if (b-a<=1) return;
    //vi i(a.begin(), a.begin() + a.size()/2 );
    //vi d(a.begin() + a.size()/2, a.end() );
    mergesort(a, (a+b)/2);
    mergesort((a+b)/2,b);
    combinar(a, b);
}

long long count_inversions() {
  contador=0;
    vc=vb=va;
    mergesort(0, va.size());
    sort(vc.begin(), vc.end());
    if (vc!=va) 
        for (int co=-1; ;co--)
            vc[co];
    return contador;
}

int main(){
    int t;
    cin >> t;
    for(int a0 = 0; a0 < t; a0++){
        int n;
        cin >> n;
        vector<int> arr(n);
        for(int arr_i = 0;arr_i < n;arr_i++){
           cin >> arr[arr_i];
        }
        va= arr;
        //cout<<"popoku\n";
        cout << count_inversions() << endl;
    }
    return 0;
}
                    


                        Solution in Java :

In Java :



import java.util.*;

class Solution {
    public static long countInversions(int[] a){
        int n = a.length;
        
        // Base Case
        if(n <= 1) {
            return 0;
        }
        
        // Recursive Case
        int mid = n >> 1;
        int[] left = Arrays.copyOfRange(a, 0, mid);
        int[] right = Arrays.copyOfRange(a, mid, a.length);
        long inversions = countInversions(left) + countInversions(right);
        
        int range = n - mid;
        int iLeft = 0;
        int iRight = 0;
        for(int i = 0; i < n; i++) {
            if(
                iLeft < mid 
                && (
                    iRight >= range || left[iLeft] <= right[iRight]
                )
            ) {
                a[i] = left[iLeft++];
                inversions += iRight;
            }
            else if(iRight < range) {
                a[i] = right[iRight++];
            }
        }
        
        return inversions;
    }
  
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        
        for(int i = 0; i < t; i++){
            int n = scanner.nextInt();
            int[] a = new int[n];
            
            for(int j = 0; j < n; j++){
                a[j] = scanner.nextInt();
            }
            
            System.out.println(countInversions(a));
        }
        
        scanner.close();
    }
}
                    


                        Solution in Python : 
                            
In Python3 :





def merge(a, l, m, h):
    c = []
    i = l
    j = m + 1
    s = 0
    
    while i <= m and j <= h:
        if a[i] > a[j]:
            # there is an inversion
            s += (m - i + 1)
            c.append(a[j])
            j += 1
        else:
            c.append(a[i])
            i += 1
            
    # Adding remaning numbers
    while i <= m:
        c.append(a[i])
        i += 1
    while j <= h:
        c.append(a[j])
        j += 1
        
    
    a[l: h + 1] = c
    
    return s
            

def count(a, l, h):
    if l >= h:
        return 0
    #print(l, h)
    m = l + (h - l) // 2
    s = 0
    s += count(a, l, m)
    s += count(a, m + 1, h)
    
    s += merge(a, l, m, h)
    return s

def count_inversions(a):
    return count(a, 0, len(a) - 1)

t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    arr = list(map(int, input().strip().split(' ')))
    print(count_inversions(arr))


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