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# Problems in your to-do list CodeChef Solution

## Problem – Problems in your to-do list CodeChef Solution

CodeChef recently revamped its practice page to make it easier for users to identify the next problems they should solve by introducing some new features:

• Recent Contest Problems – contains only problems from the last 2 contests
• Separate Un-Attempted, Attempted, and All tabs
• Problem Difficulty Rating – the Recommended dropdown menu has various difficulty ranges so that you can attempt the problems most suited to your experience
• Popular Topics and Tags

Like most users, Chef didn’t know that he could add problems to a personal to-do list by clicking on the magic ‘+’ symbol on the top-right of each problem page. But once he found out about it, he went crazy and added loads of problems to his to-do list without looking at their difficulty rating.

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.

### Input Format

• The first line of input will contain a single integer T, the number of test cases. Then the testcases follow.
• Each testcase consists of 2 lines of input.
• The first line of input of each test case contains a single integer, N, which is the total number of problems that the Chef has added to his to-do list.
• The second line of input of each test case contains N space-separated integers D1​,D2​,…,DN​, which are the difficulty ratings for each problem in the to-do list.

### Output Format

For each test case, output in a single line the number of problems that Chef will have to remove so that all remaining problems have a difficulty rating strictly less than 1000.

### Constraints

• 1≤T≤1000
• 1≤N≤1000
• 1≤Di​≤5000

• Subtask 1 (100 points):
• Original constraints

### Sample 1:

``````Input: 5
3
800 1200 900
4
999 1000 1001 1002
5
1 2 2 2 5000
5
1000 1000 1000 1000 1000
3
900 700 800
Output: 1
3
1
5
0``````

### Explanation:

Test case 1: Among the three difficulty ratings, Chef only needs to remove the problem with difficulty rating 1200, since it is ≥1000. So, the answer is 1.

Test case 2: Among the four difficulty ratings, Chef needs to remove the problems with difficulty ratings of 1000, 1001, and 1002, since they are ≥1000. So, the answer is 3.

Test case 3: Among the five difficulty ratings, Chef needs to remove the problem with a difficulty rating of 5000, since it is ≥1000. So, the answer is 1.

Test case 4: Chef needs to remove all the five problems, since they are all rated ≥1000. So, the answer is 5.

Test case 5: Chef does not need to remove any problem, since they are all rated <1000. So, the answer is 0.

## Problems in your to-do list CodeChef Solution in C++14

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

int main()
{
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int t; cin >> t;
while(t--)
{
int n; cin >> n;
int cnt = 0;
for(int i = 0; i < n; ++i)
{
int x; cin >> x;
cnt += x >= 1000;
}
cout<<cnt<<"\n";
}
}
``````

## Problems in your to-do list CodeChef Solution in Java

``````/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{

Scanner hello=new Scanner(System.in);
int T=hello.nextInt();
while(T>0){
T--;
int N=hello.nextInt();
int[] D=new int[N];
for(int i=0;i<N;i++){
D[i]=hello.nextInt();
}
int count=0;
for(int i=0;i<N;i++){
if(D[i]>=1000){
count++;
}
}
System.out.println(count);
}
}

}
``````

## Problems in your to-do list CodeChef Solution in Pyth 3

``````# cook your dish here
for x in range(int(input ())):
l=int(input ())
c=0
a=list(map(int, input ().split()))
for i in a:
if i>=1000:
c=c+1
print(c)``````
##### Problems in your to-do list CodeChef Solution Review:

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