Task: Given an array of size N, find the inversion count of an array.
Example:
Input:
1
5
2 4 1 3 5
Output:
3
Here is the Code:
Program in C++:
- #include<iostream>
- #include<bits/stdc++.h>
- using namespace std;
- long long int c;
- void count(long long int *arr,int l,int m,int r)
- {
- int n1=m-l+1;
- int n2=r-m;
- long long int a[n1],b[n2];
- for(int i=0;i<n1;i++)
- {
- a[i]=arr[l+i];
- }
- for(int j=0;j<n2;j++)
- {
- b[j]=arr[m+1+j];
- }
- int i=0,j=0,k=l;
- while(i<n1&&j<n2)
- {
- if(a[i]<=b[j])
- {
- arr[k]=a[i];
- i++;
- }
- else
- {
- arr[k]=b[j];
- j++;
- c+=n1-i;
- }
- k++;
- }
- while(i<n1)
- {
- arr[k]=a[i];
- k++,i++;
- }
- while(j<n2)
- {
- arr[k]=b[j];
- k++,j++;
- }
- }
- void inversionOfArray(long long int *a,int i,int j)
- {
- if(i==j)
- {
- return ;
- }
- else
- {
- int m=floor((i+j)/2);
- inversionOfArray(a,i,m);
- inversionOfArray(a,m+1,j);
- count(a,i,m,j);
- }
- }
- int main()
- {
- int t;
- cin>>t;
- while(t--)
- {
- int n;
- cin>>n;
- long long int a[n];
- for(int i=0;i<n;i++)
- {
- cin>>a[i];
- }
- c=0;
- inversionOfArray(a,0,n-1);
- cout<<c<<endl;
- }
- return 0;
- }
Here is the Video:
Comments
Post a Comment