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 5 integers
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.)
Sample Input
1 2 3 4 5
Sample Output
10 14
Explanation
The numbers are 1, 2, 3, 4, and 5. Calculate the following sums using four of the five integers:
- Sum everything except 1, the sum is 2+3+4+5 .
- Sum everything except 2, the sum is 1+3+4+5.
- Sum everything except 3, the sum is 1+2+4+5.
- Sum everything except 4, the sum is 1+2+3+5.
- Sum everything except 5, the sum is 1+2+3+4.
Hints: Beware of integer overflow! Use 64-bit Integer.
Mini Max Sum HackerRank Solution in C
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { long long int arr[5],max=0,min,sum=0; for(int i=0;i<5;i++){ scanf("%lld",&arr[i]); if(arr[i]>max){ max=arr[i]; } sum+=arr[i]; } min =arr[0]; for(int i=0;i<5;i++) { if(arr[i]<min) min=arr[i]; } printf("%lld %lld",sum-max,sum-min);
Mini Max Sum HackerRank Solution in C++
#include <bits/stdc++.h> typedef long long LL; using namespace std; int main(){ LL s[5]; LL d = 0; for(int i = 0; i < 5; i++){ cin >> s[i]; d += s[i]; } sort(s,s+5); cout << d-s[4] << " " << d-s[0] << endl; }
Mini Max Sum HackerRank Solution in Java
import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); long sum = 0; long max = Long.MIN_VALUE; long min = Long.MAX_VALUE; for (int i = 0; i < 5; i++){ long n = in.nextLong(); sum += n; max = Math.max(max, n); min = Math.min(min, n); } System.out.println((sum - max) + " " + (sum - min)); } }
Mini Max Sum HackerRank Solution in Python
a = sorted(map(int,raw_input().split())) print sum(a[:4]),sum(a[1:])
Mini Max Sum HackerRank Solution in C#
using System; using System.Collections.Generic; using System.IO; using System.Linq; class Solution { static void Main(String[] args) { long[] A = Array.ConvertAll(Console.ReadLine().Split(' '), long.Parse); long max = A.Sum(), min = max; max -= A.Min(); min -= A.Max(); Console.WriteLine(min + " " + max); } }
Attempt – Mini Max Sum HackerRank Challenge
Link – https://www.hackerrank.com/challenges/mini-max-sum
Next Challenge – Birthday Cake Candles HackerRank Solution
Link – https://exploringbits.com/birthday-cake-candles-hackerrank-solution/
Aayush Kumar Gupta is the founder and creator of ExploringBits, a website dedicated to providing useful content for people passionate about Engineering and Technology. Aayush has completed his Bachelor of Technology (Computer Science & Engineering) from 2018-2022. From July 2022, Aayush has been working as a full-time Devops Engineer.
public static void miniMaxSum(List arr) {
Collections.sort(arr);
long min = 0;
long max = 0;
int count = 0;
int len = arr.size();
do{
min +=arr.get(count);
max += arr.get(len-1);
count++;
len–;
}while(count < 4);
System.out.println(min + " " + max);
}
For JavaScript
/*
* Complete the ‘miniMaxSum’ function below.
*
* The function accepts INTEGER_ARRAY arr as parameter.
*/
function miniMaxSum(arr) {
// Write your code here
let total = [];
let temp = 0 ;
for (let i=0; i< arr.length; i++){
temp = arr.reduce(function(a,b){return a+b},-arr[i]);
total.push(temp);
}
console.log(Math.min(…total)+" "+Math.max(…total));
}