TODOLIST - Editorial

editorial
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1

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

Setter: Hrishikesh
Tester: Harris Leung
Editorialist: Trung Dang

DIFFICULTY:

580

PREREQUISITES:

None

PROBLEM:

Chef is a beginner and should ideally try and solve only problems with difficulty rating strictly less than 1000. Given a list of difficulty ratings for problems in the Chef’s to-do list, please help him identify how many of those problems Chef should remove from his to-do list, so that he is only left with problems of difficulty rating less than 1000.

EXPLANATION:

We loop through the list of difficulty ratings and count the number of problems that are at least 1000 in rating.

TIME COMPLEXITY:

Time complexity is O(N) per test case.

SOLUTION:

Preparer's Solution 
#include<iostream>

using namespace std;

int q,n,a;

int main()
{
	cin>>q;

	while(q--)
	{
		cin>>n;
		int cnt = 0;
		for(int i=1;i<=n;i++)
		{
			cin>>a;
			if(a >= 1000)
				cnt++;
		}
		cout<<cnt<<"\n";
	}
}
Tester's Solution 
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define fi first
#define se second
ll n;
void solve(int rr){
	cin >> n;
	int ans=0;
	for(int i=1; i<=n ;i++){
		int x;cin >> x;ans+=(x>=1000);
	}
	cout << ans << '\n';
}
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	int t;cin >> t;
	for(int i=1; i<=t ;i++) solve(i);
}
Editorialist's Solution 
#include <bits/stdc++.h>
using namespace std;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t; cin >> t;
    while (t--) {
        int n; cin >> n;
        vector<int> a(n);
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }
        cout << count_if(a.begin(), a.end(), [](int u) { return u >= 1000; }) << '\n';
    }
}