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.
One 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 Solution
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 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 4 of 5 elements.
Input Format
A single line of five space-separated integers.
Constraints
- 1 <= arr[i] <= 109
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 = 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 Solution
Mini 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 Java
import 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 JavaScript
function 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 Scala
object 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.