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CountInversion.cpp
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CountInversion.cpp
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#include<bits/stdc++.h>
using namespace std;
//Algorithm used is Merge Sort
//Time Complexity is O(n log n)
long long int merge(int a[], int low, int mid, int high);
int mergeSort(int a[], int low, int high)
{
long long int count=0;
if(low<high)
{
int mid=(low+high)/2;
count+= mergeSort(a,low,mid); //counts inversions in left half
count+= mergeSort(a,mid+1,high); //counts inversions in right half
count+= merge(a,low,mid,high); //counts inversions when two halves are merged
}
return count;
}
long long int merge(int arr[], int low, int mid, int high)
{
long long int count=0;
int n1=mid-low+1; //size of left half
int n2=high-mid; //size of right half (high-(mid+1)+1)==high-mid
int a[n1],b[n2];
for(int i=0;i<n1;i++)
a[i]=arr[low+i];//storing sorted left half in a temporary array
for(int i=0;i<n2;i++)
b[i]=arr[mid+i+1];//storing sorted right half in a temporary array
int i=0,j=0,k=low;
while((i<n1)&&(j<n2))
{
if(a[i]<=b[j]) //No inversion occurs
arr[k++]=a[i++];
else {
arr[k++]=b[j++]; //Inversion occurs!
count+=(n1-i); //if a[i] is greater, a[i+1], a[i+2] etc will also be greater. So all inversions are counted here
}
}
while(i<n1)
arr[k++]=a[i++];
while(j<n2)
arr[k++]=b[j++];
return count;
}
int main()
{
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++)
cin>>a[i];
long long int ans=mergeSort(a,0,n-1);
cout<<ans<<endl;
return 0;
}