HackerEarth Customer satisfaction problem solution

In this HackerEarth Customer satisfaction problem solution Alice is the biggest seller in the town who sells notebooks. She has N customers to satisfy. Each customer has three parameters Li, Ri, and Zi denoting the arrival time, departure time, and quantity of notebooks required.

Each customer has to be supplied a total of Zi notebooks between Li and Ri inclusive. What is the minimum rate of W notebooks per unit time by which if Alice produces so that it satisfies the demand of each customer?

Note that Alice does not need to supply Zi to customer i for per unit time but the total of Zi between Li and Ri and distribution does not need to be uniform.

HackerEarth Customer satisfaction problem solution.

#include<bits/stdc++.h>
using namespace std;
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mod 1000000007
#define endl "\n"
#define test ll t; cin>>t; while(t--)
typedef long long int ll;
bool check(vector<pair<ll,ll>>&vp,ll n,ll quan){
    ll ok=true;
    ll ans=vp[0].first*quan;
    for(ll i=0;i<n;i++){
        if(i>0){
            ans+=(vp[i].first-vp[i-1].first)*quan;
        }
        if(ans<vp[i].second){
            ok=false;
        }
        ans-=vp[i].second;
    }
    return ok;
}
int main() {
    FIO;
    test
    {
      ll n,l,r,w; cin>>n;
      vector<pair<ll,ll>>vp;
      for(int i=0;i<n;i++){
          cin>>l>>r>>w;
          vp.push_back({r,w});
      }
      sort(vp.begin(),vp.end());
      for(auto it:vp){
          cout<<it.first<<" "<<it.second<<endl;
      }
      ll st=0,dr=1e18,ans,mid;
      while(st<=dr){
          mid=(st+dr)/2;
          if(check(vp,n,mid)){
              ans=mid;
              dr=mid-1;
          }
          else{
              st=mid+1;
          }
      }
      cout<<ans<<endl;
    }
    return 0;
}

Second solution

MAX = 10000
t = int(input())
while t > 0:
    t -= 1
    n = int(input())
    need = {}
    for i in range(n):
        [l, r, z] = map(int, input().split())
        if r not in need:
            need[r] = 0
        need[r] += z
    ans = 0
    li = list(need.items())
    li.sort()
    cur_need = 0
    for key, val in li:
        cur_need += val
        ans = max(ans, (cur_need + key - 1) // key)
    print(ans)