Chef in Fantasy League - FFL (April Lunch Time CodeChef Question)

Hello Everyone , Welcome back to my blog .
Today we will be discussing the latest Codechef's question of April Lunch Time which concluded just now :-

Here is the Problem

Chef is going to start playing Fantasy Football League (FFL) this season. In FFL, each team consists of exactly 15$15$ players: 2$2$ goalkeepers, 5$5$ defenders, 5$5$ midfielders and 3$3$ forwards. Chef has already bought 13$13$ players; he is only missing one defender and one forward.

There are N$N$ available players (numbered 1$1$ through N$N$). For each valid i$i$, the i$i$-th player is either a defender or a forward and has a price Pi$P_i$. The sum of prices of all players in a team must not exceed 100$100$ dollars and the players Chef bought already cost him S$S$ dollars.

Can you help Chef determine if he can complete the team by buying one defender and one forward in such a way that he does not exceed the total price limit?

Input

• The first line of the input contains a single integer T$T$ denoting the number of test cases. The description of T$T$ test cases follows.
• The first line of each test case contains two space-separated integers N$N$ and S$S$.
• The second line contains N$N$ space-separated integers P1,P2,…,PN$P_1,P_2,\dots ,P_N$.
• The last line contains N$N$ space-separated integers. For each valid i$i$, the i$i$-th of these integers is 0$0$ if the i$i$-th player is a defender or 1$1$ if the i$i$-th player is a forward.

Output

For each test case, print a single line containing the string "yes" if it is possible to build a complete team or "no" otherwise (without quotes).

Constraints

• 1≤T≤100$1\le T\le 100$
• 1≤N≤100$1\le N\le 100$
• 13≤S≤100$13\le S\le 100$
• 1≤Pi≤100$1\le P_i\le 100$ for each valid i$i$

Subtasks

Subtask #1 (100 points): original constraints

Example Input

2
4 90
3 8 6 5
0 1 1 0
4 90
5 7 6 5
0 1 1 0

Example Output

yes
no

Explanation

Example case 1: If Chef buys the 1$1$-st and 3$3$-rd player, the total price of his team is 90+9=99$90+9=99$, which is perfectly fine. There is no other valid way to pick two players.

Example case 2: Chef cannot buy two players in such a way that all conditions are satisfied.

SOLUTION

I guess the answer is fairly simple , calculate the amount of money Chef can spend more which is equal to 100-s  and then just find the minimum price of a defender and of a forward.

If their sum is less than or equal to 100-s simply print yes otherwise no.

Here is the C++ Code :-

#include <iostream>

using namespace std;

int main() {

    int t;

    cin>>t;

    while(t--)

    {

        int n,s;

        cin>>n;

        cin>>s;

        s=100-s;

        int arr[n],arr1[n];

        int i=0;

        while(i!=n)

        {

            cin>>arr[i];

            i++;

        }

         i=0;

        while(i!=n)

        {

            cin>>arr1[i];

            i++;

        }

        int min=500,min1=500;

        i=0;

        while(i!=n)

        {

            if(arr1[i]==0)

            {

                if(arr[i]<min1)

                {

                    min1=arr[i];

                }

                

            }

            if(arr1[i]==1)

            {

                if(arr[i]<min)

                {

                    min=arr[i];

                }

            }

            i++;

        }

        if(min+min1<=s)

        {

            cout<<"yes"<<endl;

        }

        else

        {

            cout<<"no"<<endl;

        }

        

    }

// your code goes here

return 0;

}