HackerRank Ice Cream Parlor problem solution YASH PAL, 31 July 2024 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 {“mode”:”full”,”isActive”:false} 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++; } } } } } {“mode”:”full”,”isActive”:false} 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; } {“mode”:”full”,”isActive”:false} 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; } {“mode”:”full”,”isActive”:false} algorithm coding problems