Help me in solving PROBLEMS problem

help
,
1

My issue

Can someone explain this question to me I am a bit confused , can someone explain the sample test case ??

My code

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

int main() {
	// your code goes here

}

Problem Link: Problem Sort Practice Coding Problem - CodeChef

2

The problem is really ad-hoc, but makes sense if you read it reeeeeally carefully.

You are given an P amount of problems, and a S amount of subtask.
For each subtask you have 2 data:

  • Score
  • Right attempts

You are told to sort the problems according to the difficulty, which is calculated by this:

  • Subtasks should be sorted by score
  • Each subtask has an amount of right attempts
  • If a bigger score has less right attempts, then you add n += 1 to the difficulty (which makes sense, if a score is bigger then it should be more difficult, so it should have less right attempts)

Let’s take these 3 problems with 3 subtasks of the example:
16 24 60
498 861 589
14 24 62
72 557 819
16 15 69
435 779 232

You first have to join each subtask with their right attempts like this:
P1 = [(16,498), (24,861), (60,589)]
P2 = [(14,72), (24,557), (62,819)]
P3 = [(16,435), (15,779), (69,232)]

Problem ask you to sort each problem by its score, so:
P1 = [(16,498), (24,861), (60,589)]
P2 = [(14,72), (24,557), (62,819)]
P3 = [(15,779), (16,435), (69,232)]

Now, a problem is harder if the number of right attempts is decreasing, calculated by

n += 1
if
P[i][1] > P[i+1][1]

where:
P are the problems
i is the index of each subtask.
1 is the number of right attempts.

Therefore, the “n” number of each problem is:
P1 = 1 (861 is bigger than 589)
P2 = 0
P3 = 2 (779 is bigger than 435, and 435 is bigger than 232)

Then, you are asked to sort them increasingly according to their n value, so:
P2 = 0
P1 = 1
P3 = 2

Finally, you are asked to print THE PROBLEM index increasingly, accordingly to its n value, do not print the n value, so:
2
1
3

If a problem has the same n, sort them by index.

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

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

    vector<pair<ll,ll>> probs;
    
    ll P, S;
    cin >> P >> S;
    
    for(ll i=0; i<P; i++){
        vector<ll> score(S);
        vector<ll> right(S);
        
        for(ll j=0; j<S; j++){
            cin >> score[j];
        }
        for(ll j=0; j<S; j++){
            cin >> right[j];
        }
        
        vector<pair<ll,ll>> subs;
        for(ll j=0; j<S; j++){
            subs.push_back(make_pair(score[j], right[j]));
        }
        
        sort(subs.begin(), subs.end(), [](pair<ll,ll> a, pair<ll,ll> b){
            if (a.first < b.first)
                return true;
            else if (a.first > b.first)
                return false;
            else{
                return a.second < b.second;
            }
        });
        
        ll n = 0;
        for(ll j=0; j<S-1; j++){
            if (subs[j].second > subs[j+1].second)
                n++;
        }
        
        probs.push_back(make_pair(n,i+1));
    }
    
    sort(probs.begin(), probs.end(), [](pair<ll,ll> a, pair<ll,ll> b){
        if (a.first < b.first)
            return true;
        else if (a.first > b.first)
            return false;
        else{
            return a.second < b.second;
        }
    });
    
    for(ll i=0; i<P; i++){
        cout << probs[i].second << "\n";
    }

}
1 Like
3

hey man that was a great reply , I Know it must have taken a lot of effort for creating this .
Really appreciate it man !

gracias hermano!

1 Like