Coding-Problems-Solutions/Date-Shopping-Techgig.java at main · tejad245/Coding-Problems-Solutions · GitHub

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/*
Date Shopping (100 Marks)
Varun has got a date after a long time and wants to look his best for his partner.
He decides to go shopping to buy a new t-shirt and a pair of jeans but since it is the month’s end, he has a budget of only B rupees.
He wants to spend the maximum money on jeans and t-shirt combo within the given budget. The problem is that he is good with his cloth choices but not with the calculations and asks for your help.

Varun has selected T t-shirts and J jeans. He provides you the prices of each of them and you have to determine the maximum money he can spend shopping.
If it is not possible to buy the jeans and t-shirt with the given budget, you can tell him -1.


Example: 1
Budget, B = 20
Jeans, Ji = [ 10, 5, 8 ]
T-shirts, Ti = [ 11, 7, 4 ]
The maximum money he can spend is 19 by buying the J3 jeans along with the T1 t-shirt.

Example: 2
Budget, B = 10
Jeans, Ji = [ 8 , 9 , 10 ]
T-shirt, Ti = [ 4, 6, 3 ]

With the given budget, it is not possible for Varun to buy the jeans and t-shirt both. Thus, the maximum money he can spend is -1.


Can you help him shop and get ready for his date?

Input Format
The first line of input consists of the Budget, B
The second line of input consists of the number of jeans (J) and number of t-shirts (T) space-separately.
The third line of input consists of the J space-separated jeans prices, Ji
The fourth line of input consists of the T space-separated t-shirt prices, Ti

Constraints
1<= B <=10^6
1<= J, T <=1000
1<= Ji, Ti =10^6

Output Format
Print the maximum money Varun can spend shopping.

Sample TestCase 1
Input
10
3 3
5 7 9
6 2 7
Output
9
Explanation

The maximum money can be spent by selecting J2 jeans and T2 t-shirt.
Maximum money  = 7 + 2 = 9
*/


import java.io.*;
import java.util.*;
public class CandidateCode {
    public static void main(String args[] ) throws Exception {
        Scanner sc=new Scanner(System.in);
        int b=sc.nextInt();
        int a=sc.nextInt();
        int bl=sc.nextInt();
        int arr[]=new int[a];
        int brr[]=new int[b];
        for(int i=0;i<a;i++)
        {
            arr[i]=sc.nextInt();
        }
        for(int i=0;i<bl;i++)
        {
            brr[i]=sc.nextInt();
        }
        int k=0;
        int mn=0;
        int pre=-1;
        int flag=0;
        for(int i=0;i<a;i++)
        {
            for(int j=0;j<bl;j++)
            {
                int s=arr[i]+brr[j];
                if(s<=b && flag==0)
                {
                    pre=s;
                   // System.out.println(f[k]+"fis");
                    flag=1;
                }
                else if(s<=b && flag!=0)
                {
                     mn=s;
                     if(mn>pre)
                     {
                         pre=mn;
                     }
                     // System.out.println(f[k]+"sec");
                }
                else{
                }
                k=k+1;
            }
        }
            System.out.println(pre);
    }
}