In this HackerEarth Inversions Revisited problem solution, Most of us know how to count the number of inversions in an array. An inversion in an array is a pair of indices(i,j) such that a[i]>a[j] and i < j.

In this problem you are given 2 arrays A and B and you have to return number of such pairs such that a[i]>b[j] and i < j.


HackerEarth Inversions Revisited problem solution


HackerEarth Inversions Revisited problem solution.

#include <bits/stdc++.h>
#define _ ios_base::sync_with_stdio(false);cin.tie(0);
using namespace std;
#define pb push_back
#define pob pop_back
#define pf push_front
#define pof pop_front
#define mp make_pair
#define all(a) a.begin(),a.end()
#define bitcnt(x) __builtin_popcountll(x)
#define MOD 1000000007
#define total 5000005
#define M 1000000000001
#define NIL 0
#define MAXN 100001
#define EPS 1e-5
#define INF (1<<28)
typedef unsigned long long int uint64;
typedef long long int int64;

int BIT[1000006];
int a[1000006];
int b[1000006];
void upd(int idx,int val){
for(int i=idx;i<=1000000;i+=(i&(-i)))
BIT[i]+=val;
}
int query(int val){
int ret=0;
while(val){
ret+=BIT[val];
val-=(val&(-val));
}
return ret;
}
int main(){
int n,i;
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
cin>>n;
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(i=1;i<=n;i++)
scanf("%d",&b[i]);
for(i=1;i<=n;i++){
upd(b[i],1);
}
int64 ans=0;
for(i=1;i<=n;i++){
upd(b[i],-1);
ans+=(query(a[i]-1));
}
cout<<ans;
fclose(stdout);
}

Second solution

#include <bits/stdc++.h>

using namespace std;

long long tree[200005];
int A[200005];
int B[200005];

long long query(int idx)
{
long long res = 0;
while ( idx > 0 ) {
res += tree[idx];
idx -= (idx & (-idx));
}
return res;
}

void update(int idx)
{
while ( idx <= 100001 ) {
tree[idx] += 1;
idx += (idx &(-idx));
}
return;
}
int main()
{
int t=1,n;
while ( t-- ){
cin >> n;
for ( int i = 0; i < n; i++ ) cin >> A[i];
for ( int i = 0; i < n; i++ ) cin >> B[i];
memset(tree, 0, sizeof(tree));
long long ans = 0;
for ( int i = n-1; i >= 0; i-- ) {
ans += query(A[i]-1);
update(B[i]);
}
cout << ans << endl;
}

return 0;
}