HackerRank Ice Cream Parlor problem solution

In this HackerRank Ice Cream Parlor problem solution, Two friends like to pool their money and go to the ice cream parlor. They always choose two distinct flavors and they spend all of their money. Given a list of prices for the flavors of ice cream, select the two that will cost all of the money they have.

Function Description

Complete the icecreamParlor function in the editor below.

icecreamParlor has the following parameter(s):

• int m: the amount of money they have to spend
• int cost[n]: the cost of each flavor of ice cream

Returns

• int[2]: the indices of the prices of the two flavors they buy, sorted ascending

HackerRank Ice Cream Parlor 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

Ice Cream Parlor 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++;
                }
            }
        }
    }
}

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 %dn", 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 %dn", x, y);
        }

    }
    return 0;
}

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 %dn",j+1,k+1);
                    goto flag;
                }
            }
        }
       flag:;
    }
    return 0;
}