Hey Guys ! I am solving this problem
on Hackerearth which ultimately boils down to finding distinct integers in a range from L to R. I am having a hard time understanding how the bitset class is used with segment trees to find the solution. I am posting one of the accepted solutions. Can anyone please explain the logic.
include<bits/stdc++.h>
using namespace std; bitset<5001> seg_a[400001],seg_b[400001],rst; int a[100005],b[100005]; void built_a(int node,int start,int end){ if(start==end){ seg_a[node].set(a[start]); return; } int mid=(start+end)/2; built_a(2*node,start,mid); built_a(2*node+1,mid+1,end); seg_a[node]=(seg_a[2*node]|seg_a[2*node+1]); } void built_b(int node,int start,int end){ if(start==end){ seg_b[node].set(b[start]); return; } int mid=(start+end)/2; built_b(2*node,start,mid); built_b(2*node+1,mid+1,end); seg_b[node]=(seg_b[2*node]|seg_b[2*node+1]); } bitset<5001> query_a(int node,int start,int end,int L,int R){ if(R<start||end<L)return rst; if(start>=L && end<=R)return seg_a[node]; int mid=(start+end)/2; return query_a(2*node,start,mid,L,R)|query_a(2*node+1,mid+1,end,L,R); } bitset<5001> query_b(int node,int start,int end,int L,int R){ if(R<start||end<L)return rst; if(start>=L && end<=R)return seg_b[node]; int mid=(start+end)/2; return (query_b(2*node,start,mid,L,R)|query_b(2*node+1,mid+1,end,L,R)); } int main(){ int n;cin>>n; for(int i=0;i<n;i++)cin>>a[i]; for(int i=0;i<n;i++)cin>>b[i];
built_a(1,0,n-1);
built_b(1,0,n-1);
cout<<seg_a[1].
int q;cin>>q;
int L1,R1,L2,R2;
while(q--){
cin>>L1>>R1>>L2>>R2;
int ans=(query_a(1,0,n-1,L1-1,R1-1)|query_b(1,0,n-1,L2-1,R2-1)).count();
cout<<ans<<"\n";
}
return 0;
}