You have N cup noodles. Each takes ci time to cook and satisfies hi hunger. You can cook a cup noodle by pouring hot water, you can only pour hot water in a noodle cup in one minute. You have M minutes, find the maximum possible hunger satisfaction.

constraints :

1 <= N <= 10^5

1 <= M <= 10^5

1 <= ci <= 10^5

1 <= hi <= 10^5

testcases -

n = 3 m = 5

c → 2 5 5

h → 10 5 10

Output - 20

explaination - We cook the third cup first and then the second cup to get maximum satisfaction of 20.

n = 3 m = 5

c → 5 5 3

h → 5 10 2

Output - 12

exp - Cook the second cup first and second one in the next minute. Both will be done by 5 minutes.

My thought process -

Afaik, the constraints rule out a DP approach to this problem. So this problem probably boils down to some sorting and greedy algorithm.

so I sort by hunger satisfaction and greedily try to cook noodles with large times first with maximum satisfaction.

```
#include<bits/stdc++.h>
using namespace std;
bool comp(pair<int,int>& a, pair<int,int>& b) {
if(a.second == b.second) return a.first > b.first;
return a.second > b.second;
}
int solve (int n, int m, vector<int> c, vector<int> h) {
// Write your code here
vector<pair<int,int>> a(n);
for(int i = 0; i < n; i++) {
a[i].first = c[i];
a[i].second = h[i];
}
sort(a.begin(), a.end(), comp);
int ans = 0, taken = 0;
for(int i = 0; i < n; i++) {
if(a[i].first <= m - taken) {
ans += a[i].second;
taken++;
}
}
return ans;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int T;
cin >> T;
for(int t_i = 0; t_i < T; t_i++)
{
int N;
cin >> N;
int M;
cin >> M;
vector<int> c(N);
for(int i_c = 0; i_c < N; i_c++)
{
cin >> c[i_c];
}
vector<int> h(N);
for(int i_h = 0; i_h < N; i_h++)
{
cin >> h[i_h];
}
int out_;
out_ = solve(N, M, c, h);
cout << out_;
cout << "\n";
}
}
```

I could pass 7/10 testcases in the problem. Any solutions to this problem will be greatly appreciated, thanks in advance.

P.S. If anyone can point out a few similar problems, would be great.