HackerRank Mini-Max Sum Problem Solution
In this post, We are going to solve HackerRank Mini-Max Sum Problem. 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 24Function 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 4 of 5 elements.
Input Format
A single line of five space-separated integers.
Constraints
1 < arr [I] < 10 power 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
Mini-Max Sum C Solution
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
long long int a[5],temp;
int i, j;
scanf("%lld %lld %lld %lld %lld",&a[0],&a[1],&a[2],&a[3],&a[4]);
for (i=1;i<=4;i++)
{ j=i;
while (j>0 && (a[j-1] > a[j]))
{
temp= a[j];
a[j]= a[j-1];
a[j-1]=temp;
j=j-1;
}
}
long long int min=a[0]+a[1]+a[2]+a[3];
long long int max=a[1]+a[2]+a[3]+a[4];
printf("%lld %lld",min,max);
return 0;
}Mini-Max Sum C++ Solution
#include<bits/stdc++.h>
using namespace std;
int main()
{
long long a[5];
cin>>a[0]>>a[1]>>a[2]>>a[3]>>a[4];
sort(a,a+5);
cout<<a[0]+a[1]+a[2]+a[3]<<" "<<a[4]+a[1]+a[2]+a[3]<<endl;
return 0;
}Mini-Max Sum C Sharp Solution
using System;
using System.Collections.Generic;
using System.IO;
class Solution {
static void Main(String[] args) {
string[] arr = Console.ReadLine().Split(' ');
long sum = 0;
List<long> dizi = new List<long>();
for (int i = 0; i < arr.Length; i++)
{
long temp = Convert.ToInt64(arr[i]);
dizi.Add(temp);
}
for (int i = 0; i < dizi.Count; i++)
{
sum = sum + dizi[i];
}
for (int i = 0; i < dizi.Count - 1; i++)
{
for (int j = 1; j < dizi.Count - i; j++)
{
if (dizi[j] < dizi[j - 1])
{
long gecici = dizi[j - 1];
dizi[j - 1] = dizi[j];
dizi[j] = gecici;
}
}
}
long max = sum - dizi[0];
long min = sum - dizi[4];
Console.WriteLine("{0} {1}", min, max);
Console.ReadKey();
}
}Mini-Max Sum Java Solution
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
class Result {
/*
* Complete the 'miniMaxSum' function below.
*
256741038 623958417 467905213 714532089 938071625
2063136757 2744467344
* The function accepts INTEGER_ARRAY arr as parameter.
*/
public static void miniMaxSum(List<Integer> arr) {
int max = -1;
int min = Integer.MAX_VALUE; ;
long sumMax = 0;
long sumMin = 0;
long sum = 0;
int numberChecked =0;
for(int el : arr){
max = Math.max(el,max);
min = Math.min(el,min);
sum+=el;
}
if(min==max){
System.out.print(sum -max);
System.out.print(" ");
System.out.print(sum -max);
return;
}
for(int el : arr){
if(el!=max) sumMin+=el;
if(el!=min) sumMax+=el;
}
System.out.print(sumMin);
System.out.print(" ");
System.out.print(sumMax);
}
}
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
List<Integer> arr = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
.map(Integer::parseInt)
.collect(toList());
Result.miniMaxSum(arr);
bufferedReader.close();
}
}Mini-Max Sum JavaScript Solution
process.stdin.resume();
process.stdin.setEncoding('ascii');
var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;
process.stdin.on('data', function (data) {
input_stdin += data;
});
process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("\n");
main();
});
function readLine() {
return input_stdin_array[input_currentline++];
}
/////////////// ignore above this line ////////////////////
Array.prototype.min = function () {
var min = 777777777777;
for (var i = 0; i < this.length; i++) {
if (this[i] < min) min = this[i];
}
return min;
}
Array.prototype.max = function () {
var max = 0;
for (var i = 0; i < this.length; i++) {
if (this[i] > max) max = this[i];
}
return max;
}
Array.prototype.sum = function () {
return this.reduce((x, y) => { return x + y;});
}
function main() {
var vals = readLine().split(' ').map((i) => {return parseInt(i);});
//console.log(vals);
var lo = vals.sum() - vals.max();
var hi = vals.sum() - vals.min();
console.log(lo.toString() + ' ' + hi.toString())
}Mini-Max Sum Python Solution
def swap(arr,x,y):
mem = arr[x]
arr[x] = arr[y]
arr[y] = mem
return arr
def qsort(ar,start,end):
# base case
if end-start <= 0:
return ar
# pivot
pivot = ar[end]
# init index for pivot
ind = start
# loop less final value (pivot)
for i in range(start,end):
if ar[i] <= pivot:
ar = swap(ar,i,ind)
ind += 1
else:
pass
swap(ar,ind,end)
# left side
ar = qsort(ar,start,ind-1)
# right side
ar = qsort(ar,ind+1,end)
return ar
ar=[int(a) for a in input().strip().split(" ")]
ar=qsort(ar,0,len(ar)-1)
print(sum(ar[:4]), sum(ar[1:]))
Other Solutions
