Hello Programmers, In this post, you will know how to solve the HackerRank Mini Max Sum Solution. This problem is a part of the HackerRank Algorithms Series.HackerRank Mini Max Sum SolutionOne more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.HackerRank Mini Max Sum SolutionProblemGiven 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.Examplearr = [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 prints16 24Function DescriptionComplete the miniMaxSum function in the editor below.miniMaxSum has the following parameter(s):arr: an array of 5 integersPrintPrint two space-separated integers on one line: the minimum sum and the maximum sum of 4 of 5 elements.Input FormatA single line of five space-separated integers.Constraints1 <= arr[i] <= 109Output FormatPrint 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 Input1 2 3 4 5Sample Output10 14ExplanationThe 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 = 14.Sum everything except 2, the sum is 1 + 3 + 4 + 5 = 13.Sum everything except 3, the sum is 1 + 2 + 4 + 5 = 12.Sum everything except 4, the sum is 1 + 2 + 3 + 5 = 11.Sum everything except 5, the sum is 1 + 2 +3 + 4 = 10.HackerRank Mini Max Sum SolutionMini Max Sum Solution in C#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { long long int i,j,k,l,m=0,n=0,t=2000000000; for(i=0;i<5;i++) { scanf("%lld",&k); m+=k; if(n<k) { n=k; } if(t>k) { t=k; } } printf("%lld %lld",m-n,m-t); return 0; }Mini Max Sum Solution in Cpp#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 Solution in Javaimport java.io.*; import java.math.BigInteger; import java.util.*; class Main { static int ans[]=new int[1000001]; public static void main(String[] args) throws IOException { InputReader in = new InputReader(System.in); PrintWriter pw = new PrintWriter(System.out); int prime[]=new int[1000001]; prime[0]=1; prime[1]=1; for(int i=2;i*i<1000001;i++) if(prime[i]==0) for(int j=i*i;j<1000001;j+=i) prime[j]=1; int n=5; int a[]=in.nextIntArray(n); Arrays.sort(a); long sum=(long)a[0]+(long)a[1]+(long)a[2]+(long)a[3]; long sum1=sum+(long)a[4]-(long)a[0]; System.out.println(sum+" "+sum1); } private static int firstOccurrenceBinarySearch(int[] source, int needle) { int low = 0; int high = source.length - 1; int firstOccurrence = Integer.MIN_VALUE; while (low <= high) { int middle = low + ((high - low) >>> 1); if (source[middle] == needle) { // key found and we want to search an earlier occurrence firstOccurrence = middle; high = middle - 1; } else if (source[middle] < needle) { low = middle + 1; } else { high = middle - 1; } } if (firstOccurrence != Integer.MIN_VALUE) { return firstOccurrence; } return -(low + 1); // key not found } public static int binarySearchLastOccurrence(int arr[], int low, int high, int data) { int mid; // A simple implementation of Binary Search if(high >= low) { mid = low + (high - low)/2; // To avoid overflow if((mid == high && arr[mid] == data) || (arr[mid] == data && arr[mid+1] > data)) return mid; // We need to give preference to right part of the array // since we are concerned with the last occurrence else if(arr[mid] <= data) return binarySearchLastOccurrence(arr, mid+1, high, data); else // We need to search in the left half return binarySearchLastOccurrence(arr, low, mid-1, data); } return -1; } public static long pow(long n,long p,long m) { long result = 1; if(p==0) return 1; if (p==1) return n; while(p!=0) { if(p%2==1) result *= n; if(result>=m) result%=m; p >>=1; n*=n; if(n>=m) n%=m; } return result; } public static int BS(int val,int a[]) { int low=0; int high=a.length-1; int tt=0; while(low<high) { int mid=(low+high)/2; if(a[mid]<val) { tt=low; low=mid+1; } else high=mid-1; } return low; } static class Pair implements Comparable<Pair>{ int r; int v; Pair(int mr,int er){ r=mr;v=er; } @Override public int compareTo(Pair o) { if(o.r>this.r) return -1; else if(o.r<this.r) return 1; else { if(o.v>this.v) return -1; else return 1; } } } static class TVF implements Comparable<TVF>{ int index,size; TVF(int i,int c){ index=i; size=c; } @Override public int compareTo(TVF o) { if(o.size>this.size) return -1; else if(o.size<this.size) return 1; else return 0; } } public static long gcd(long a, long b) { if (b == 0) return a; return gcd(b, a%b); } static class InputReader { private InputStream stream; private byte[] buf = new byte[8192]; private int curChar, snumChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int snext() { if (snumChars == -1) throw new InputMismatchException(); if (curChar >= snumChars) { curChar = 0; try { snumChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (snumChars <= 0) return -1; } return buf[curChar++]; } public int nextInt() { int c = snext(); while (isSpaceChar(c)) c = snext(); int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public long nextLong() { int c = snext(); while (isSpaceChar(c)) c = snext(); int sgn = 1; if (c == '-') { sgn = -1; c = snext(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = snext(); } while (!isSpaceChar(c)); return res * sgn; } public int[] nextIntArray(int n) { int a[] = new int[n]; for (int i = 0; i < n; i++) a[i] = nextInt(); return a; } public String readString() { int c = snext(); while (isSpaceChar(c)) c = snext(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = snext(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } } Mini Max Sum Solution in Python# Enter your code here. Read input from STDIN. Print output to STDOUT a = sorted(map(int,raw_input().split())) print sum(a[:4]),sum(a[1:])Mini Max Sum Solution using JavaScriptfunction processData(input) { var m = input.map(function(e) { return parseInt(e) }) m.sort(function(n1, n2){ return n1-n2; }) var min = 0, max = 0; for(var i=0; i<5; i++){ i!=4 && (min+=m[i]); i!=0 && (max+=m[i]); } 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.split(" ")); });Mini Max Sum Solution in Scalaobject Solution { def main(args: Array[String]) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */ val nums = io.StdIn.readLine.split(" ").map(_.toLong).sorted val mini = nums.take(4).sum; val maxi = nums.drop(1).sum; println(mini + " " + maxi) } }Mini Max Sum Solution in Pascal{$mode objfpc} program A; var ai: array [0..5] of Int64; S: Int64; i: Integer; Min, Max: Int64; begin S := 0; Min := MaxInt; Max := -1; for i := 1 to 5 do begin Read(ai[i]); S := S + ai[i]; if Max < ai[i] then Max := ai[i]; if ai[i] < Min then Min := ai[i]; end; WriteLn(S - Max, ' ', S - Min); end.Disclaimer: This problem (Mini Max Sum) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: HackerRank Birthday Cake Candles Solution Post navigationHackerRank Time Conversion Solution HackerRank Birthday Cake Candles Solution