HackerRank Mini Max Sum problem solution

In this HackerRank Mini Max Sum problem solution Given five positive integers, find the minimum and maximum values that can be calculated by summing exactly four of the five integers. Then print the respective minimum and maximum values as a single line of two space-separated long integers.

Example

arr = [1,3,5,7,9]

The minimum sum is 1+3+5+7 = 16 and the maximum sum is 3+5+7+9 = 24. The function prints

16 24

Function Description

Complete the miniMaxSum function in the editor below.

miniMaxSum has the following parameter(s):

arr: an array of integers

Print

Print two space-separated integers on one line: the minimum sum and the maximum sum of of elements.

Input Format

A single line of five space-separated integers.

Constraints

1 <= arr[i] <= 10^9

Output Format

Print two space-separated long integers denoting the respective minimum and maximum values that can be calculated by summing exactly four of the five integers. (The output can be greater than a 32 bit integer.)

HackerRank Mini max sum problem solution in Python.

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the miniMaxSum function below.
def miniMaxSum(arr):
    arr.sort()
    hold = [None]*int(len(arr)-3)
    for i in range(0,len(arr)-3):
        temp = 0
        for j in range(i,i+4):
            temp = temp + arr[j]
        hold[i] = temp
    
    print(hold[0],hold[-1])


if __name__ == '__main__':
    arr = list(map(int, input().rstrip().split()))

    miniMaxSum(arr)

Problem solution in Java Programming.

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

public class Solution {

    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        long[] m=new long[5];
        for(int i=0;i<5;i++){
            m[i]=in.nextLong();
        }
        Arrays.sort(m);
        long x=0;
        long y=0;
        for(int i=0;i<4;i++){
            x+=m[i];
        }
        for(int i=1;i<5;i++){
            y+=m[i];
        }
        System.out.println(x+" "+y);
    }
}

Problem solution in C++ programming.

#include <algorithm>
#include <string.h>
#include <vector>
#include <cstdio>
#include <climits>
#include <iostream>
using namespace std;
typedef long long lld;

lld arr[6], N = 5;

int main ()
{
    //freopen("input.txt", "r", stdin);

    lld allsum = 0;
    lld MN = LLONG_MAX, MX = LLONG_MIN;

    for (int i=1; i<=N; i++)
    {
        scanf("%lld", &arr[i]);
        allsum += arr[i];
    }

    for (int i=1; i<=N; i++)
    {
        lld cur = allsum - arr[i];
        MN = min(MN, cur);
        MX = max(MX, cur);
    }

    printf("%lld %lldn", MN, MX);
}

Problem solution in C programming.

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

int main() {
    
    int a[5];
    long sum=0;
    for(int i=0;i<5;i++){
        scanf("%d",a+i);
        sum+=a[i];
    }
    int min=a[0];
    int max=a[0];
    for(int i=1;i<5;i++){
        if(a[i]>max)
            max=a[i];
        if(a[i]<min)
            min=a[i];
    }
    printf("%ld %ld",sum-max,sum-min);
    return 0;
}

Problem solution in JavaScript programming.

function processData(input) {
    var v = input.split(' ');
    for (var i = 0; i < v.length; i++) {
        v[i] = parseInt(v[i])
    }
    var max = -Infinity;
    var min = Infinity;
    for (var i = 0; i < v.length; i++) {
        var sum = 0;
        for (var j = 0; j < v.length; j++) {
            if ( i != j ) {
                sum += v[j];
            }
        }
        if (sum < min) min = sum;
        if (sum > max) max = sum;
    }
    console.log(min, max)
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});

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