In this HackerEarth Common Goods problem solution, There are N packets of goods each having some number of items in it. The number of items is in the form of an array A (A[i] items of ith type). There is M number of persons in total each having a share in the goods. They have shares in the form of L and R which means that they hold a share of goods [L....R]. Bob wants Q items. He reports the items one by one.
Each time he takes an item you are required to determine how many people have lost all the items from their share.


HackerEarth Common Goods problem solution


HackerEarth Common Goods problem solution.

#include<bits/stdc++.h>
using namespace std;
#define ll int
ll n,m,q,fir[100005],a[100005],ans[100005];
pair<ll,ll>p[100005];
ll tree[400005];
void buildtree(ll node,ll st,ll en)
{
    if(st==en)
    {
        tree[node]=fir[st];
    }
    else
    {
        ll mid=(st+en)/2;
        buildtree(2*node,st,mid);
        buildtree(2*node+1,mid+1,en);
        tree[node]=max(tree[2*node],tree[2*node+1]);
    }
}
ll query(ll node,ll st,ll en,ll lft,ll rght)
{
    if(rght<st||lft>en)
    return 0;
    else if(st>=lft&&en<=rght)
    return tree[node];
    else
    {
        ll mid=(st+en)/2;
        return max(query(2*node,st,mid,lft,rght),query(2*node+1,mid+1,en,lft,rght));
    }
}
int main()
{
    //ios::sync_with_stdio(0);
    ll i,j,k,l,r;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    cin>>m>>j;
    for(i=1;i<=m;i++)
    {
        cin>>p[i].first>>p[i].second;
    }
    cin>>q;
    for(i=1;i<=q;i++)
    {
        cin>>j;
        if(a[j]>0)
        {
            a[j]--;
            if(a[j]==0)
            fir[j]=i;
        }
    }
    for(i=1;i<=n;i++)
    {
        if(fir[i]==0)
        {
            fir[i]=INT_MAX;
        }
    }
    buildtree(1,1,n);
    for(i=1;i<=m;i++)
    {
        l=p[i].first;
        r=p[i].second;
        j=query(1,1,n,l,r);
        if(j!=INT_MAX)
        {
            ans[j]++;
        }
    }
    for(i=1;i<=q;i++)
    {
        ans[i]+=ans[i-1];
    }
    for(i=1;i<=q;i++)
    cout<<ans[i]<<" ";
    return 0;
}

Second solution

#include<bits/stdc++.h>
#define inf 1000000
using namespace std;
int n,a[100005],m,lft[100005],rgt[100005],ans[100005],ex[100005],st[400005];
int query(int ss,int se,int si,int l,int r)
{
    if(ss>se || se<l || ss>r)return 0;
    if(ss>=l && se<=r)return st[si];
    int mid=(ss+se)/2;
    return max(query(ss,mid,2*si,l,r),query(mid+1,se,2*si+1,l,r));
}
void build_segment(int ss,int se,int si)
{
    if(ss==se)
    {
        st[si]=ex[ss];
        return;
    }
    int mid=(ss+se)/2;
    build_segment(ss,mid,2*si);
    build_segment(mid+1,se,2*si+1);
    st[si]=max(st[2*si],st[2*si+1]);
}
int main()
{
    cin>>n;
    assert(n>=1 && n<=1e5);
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        assert(a[i]>=1 && a[i]<=1e5);
    }
    int two;
    cin>>m>>two;
    assert(m>=1 && m<=1e5);assert(two==2);
    for(int i=1;i<=m;i++)
    {
        cin>>lft[i]>>rgt[i];
        assert(lft[i]>=1 && lft[i]<=n);
        assert(rgt[i]>=lft[i] && rgt[i]<=n);
    }
    int q;
    cin>>q;
    assert(q>=1 && q<=1e5);
    for(int i=0;i<=100000;i++)ex[i]=inf;
    for(int i=1;i<=q;i++)
    {
        int temp;
        cin>>temp;
        assert(temp>=1 && temp<=n);
        a[temp]--;
        if(a[temp]==0)ex[temp]=i;
        //assert(a[temp]>=0);
    }
    build_segment(1,n,1);
    for(int i=1;i<=m;i++)
    {
        int num=query(1,n,1,lft[i],rgt[i]);
        if(num<=q)
            ans[num]++;
    }
    for(int i=1;i<=q;i++)
    {
        ans[i]+=ans[i-1];
        cout<<ans[i]<<" ";
    }
    return 0;
}