Skip to content
Programmingoneonone
Programmingoneonone
  • Engineering Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
    • 100+ C++ Programs
  • Solutions
    • HackerRank
      • Algorithms Solutions
      • C solutions
      • C++ solutions
      • Java solutions
      • Python solutions
    • Leetcode Solutions
    • HackerEarth Solutions
  • Work with US
Programmingoneonone
Programmingoneonone

HackerEarth Common Goods problem solution

YASH PAL, 31 July 202414 February 2026
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;
}
 
 
coding problems solutions HackerEarth HackerEarth

Post navigation

Previous post
Next post

Leave a Reply

Your email address will not be published. Required fields are marked *

Programmingoneonone

We at Programmingoneonone, also known as Programming101 is a learning hub of programming and other related stuff. We provide free learning tutorials/articles related to programming and other technical stuff to people who are eager to learn about it.

Pages

  • About US
  • Contact US
  • Privacy Policy

Practice

  • Java
  • C++
  • C

Follow US

  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2026 Programmingoneonone | WordPress Theme by SuperbThemes