Help me in solving MOOCHEF problem

help
,
1

My issue

i cannot get this problem

My code

#include <iostream>

using namespace std;

int main()
{
    int t;
    cin >> t;

    for (int i = 0; i < t; i++)
    {
        int n, l, r, h = 0, s = 0;
        cin >> n >> l >> r;
        int arr[n];

        for (int i = 0; i < n; i++)
        {
            cin >> arr[i];
        }

        for (int i = 0; i < n; i++)
        {
            if (arr[i] >= l && r >= arr[i])
            {
                h++;
            }
            else s--;
        }

        cout << h << " " << s << endl;
    }

    return 0;
}

Problem Link: Moody Chef Practice Coding Problem - CodeChef

2

This problem was about to monitor current “satisfaction” of Chef, determinated by each number he was looking while explorting the array.

  • He starts with 0
  • If a number is within L and R, then satisfaction += 1
  • If a number is NOT within L and R, then satisfaction -= 1
  • We should be monitoring all the current satisfactions, and store the maximal, and the minimal

Basically that. Print the maximal and the minimal stored.

Imagine the sample test case:

4 1 3
4 3 2 1

L = 1, R = 3

You start with:
current_satisfaction = 0
maximal_satisfaction = 0
minimal_satistaction = 0

  • Is “4” within 1 and 3?
    No
    current_satistaction = -1
    maximal_satisfaction = 0
    minimal_satistaction = -1

  • Is “3” within 1 and 3?
    Yes
    current_satistaction = 0
    maximal_satisfaction = 0
    minimal_satistaction = -1

  • Is “2” within 1 and 3?
    Yes
    current_satistaction = 1
    maximal_satisfaction = 1
    minimal_satistaction = -1

  • Is “1” within 1 and 3?
    Yes
    current_satistaction = 2
    maximal_satisfaction = 1
    minimal_satistaction = -1

Then:
Maximal = 2
Minimal = -1

#include <bits/stdc++.h> 
#define ll long long 
using namespace std; 

void solve(){ 

    ll N,L,R;
    cin >> N >> L >> R;
    
    vector<ll> A(N);
    for(ll i=0; i<N; i++){
        cin >> A[i];
    }
    
    ll current_satisfaction = 0;
    ll max_satis = 0;
    ll min_satis = 0;
    
    for(ll i=0; i<N; i++){
        if ((L <= A[i]) && (A[i] <= R)){
            current_satisfaction ++;
        }
        else if ((A[i] <= L) || (R <= A[i])){
            current_satisfaction --;
        }
        
        max_satis = max(max_satis, current_satisfaction);
        min_satis = min(min_satis, current_satisfaction);
    }

    cout << max_satis << " " << min_satis << "\n";

} 

int main() { 
    ios_base::sync_with_stdio(false); 
    cin.tie(NULL); 
    cout.tie(NULL); 
    int T; 
 
    cin >> T; 
    while(T--){ 
        solve(); 
    } 
}