HackerRank Ice Cream Parlor problem solution

In this HackerRank Ice Cream Parlor problem solution, we have given a list of prices for the flavors of ice cream, select the two that will cost all of the money they have.

Problem solution in Python.

for i in range(int(input())):
    target = int(input())
    loop = int(input())
    costs = input().split(" ")
    cur = 0
    done = False
    for j in range(loop - 1):
        for k in range(j+1, loop):
            if(int(costs[j]) + int(costs[k]) == target):
                print(str(j+1) + " " + str(k+1))
                done = True
                break;
        if(done): break

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Problem solution in Java.

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

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- != 0){
            int m = sc.nextInt();
            int n = sc.nextInt();
            int[] np = new int[n];
            Map<Integer, Integer> map = new HashMap<>();
            for(int i = 0; i < n ; i++){
                np[i] = sc.nextInt();
                if(map.containsKey(np[i]) == false)
                    map.put(np[i], i+1);
            }
            
            Arrays.sort(np);
            
            int s = 0;
            int e = n - 1;
            while(s < e){
                if(np[s] + np[e] > m){
                    e--;
                } else if (np[s] + np[e] < m){
                    s++;
                } else {
                    int i1 = map.get(np[s]);
                    int i2 = map.get(np[e]);
                    if(np[s] == np[e])
                        i2++;
                    
                    System.out.printf("%d %d%n", Math.min(i1, i2), Math.max(i1, i2));
                    e--;
                    s++;
                }
            }
        }
    }
}

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Problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    
    int tc, i, k, flag, x, y, sum, C, L, Li[10009];
    scanf("%d", &tc);
    while( tc-- )
    {
        flag=0;
        scanf("%d %d %d", &C, &L, &Li[0]);
        for(i=1; i<L; i++)
        {
            scanf("%d", &Li[i]);
            sum=Li[0]+Li[i];
            if(sum==C)
            {
                flag=1;
                y=i+1;
            }
        }
        if(flag) printf("1 %d\n", y);
        else
        {
            for(i=1; i<L; i++)
            {
                for(k=i+1; k<L; k++)
                {
                    sum=Li[i]+Li[k];
                    if(sum==C)
                    {
                        flag=1;
                        x=i+1;
                        y=k+1;
                        break;
                    }
                }
                if(flag) break;
            }
            printf("%d %d\n", x, y);
        }

    }
    return 0;
}

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Problem solution in C.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */  
    int n,m;
    scanf("%d",&n);
    int i=0;
    for(;i<n;i++)
    {
        int c ;
        scanf("%d",&c);
        scanf("%d",&m);
        int data[m];
        int j=0;
        for(;j<m;j++)
        {
            scanf("%d",&data[j]);
            //printf("%d ",data[j]);
        }
        int k=0;
        for(j=0;j<m;j++)
        {
            for(k=j+1;k<m;k++)
            {
                if(data[j]+data[k]==c)
                {
                    printf("%d %d\n",j+1,k+1);
                    goto flag;
                }
            }
        }
       flag:;
    }
    return 0;
}

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