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Break a number such that sum of maximum divisors of all parts is minimum

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  • Difficulty Level : Hard
  • Last Updated : 23 Jun, 2022
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We need to split a number n such that sum of maximum divisors of all the parts is minimum. 

Examples : 

Input:  n = 27
Output: Minimum sum of maximum
        divisors of parts = 3
Explanation : We can split 27 as 
follows: 27 =  13 + 11 + 3,
Maximum divisor of 13 = 1,
                   11 = 1,
                   3 = 1.
Answer = 3 (1 + 1 + 1).

Input : n = 4
Output: Minimum sum of maximum
        divisors of parts = 2
Explanation : We can write 4 = 2 + 2
Maximum divisor of 2 = 1,
So answer is 1 + 1 = 2.

We need to minimize maximum divisors. It is obvious that if N is prime, maximum divisor = 1. If the number is not a prime, then the number should be atleast 2. 

According to Goldbach’s Conjecture, every even integer can be expressed as sum of two prime numbers. For our problem there will be two cases: 

1) When the number is even, it can be expressed as the sum of two prime numbers and our answer will be 2 because maximum divisor of a prime number is 1. 
2) When the number is odd, it can also be written as sum of prime numbers, n = 2 + (n-2); if (n-2) is a prime number(answer = 2), otherwise. Refer odd number as sum of primes for details. 
n = 3 + (n-3); (n-3) is an even number and it is sum of two primes(answer = 3). 

Below is the implementation of this approach.  

C++




// CPP program to break a number such
// that sum of maximum divisors of all
// parts is minimum
#include <iostream>
using namespace std;
 
// Function to check if a number is
// prime or not.
bool isPrime(int n)
{
    int i = 2;
    while (i * i <= n) {
        if (n % i == 0)
            return false;
        i++;
    }
    return true;
}
 
int minimumSum(int n)
{
    if (isPrime(n))
        return 1;
 
    // If n is an even number (we can
    // write it as sum of two primes)
    if (n % 2 == 0)
        return 2;
 
    // If n is odd and n-2 is prime.
    if (isPrime(n - 2))
        return 2;
 
    // If n is odd, n-3 must be even.
    return 3;
}
 
// Driver code
int main()
{
    int n = 27;
    cout << minimumSum(n);
    return 0;
}

Java




// JAVA Code for Break a number such that sum
// of maximum divisors of all parts is minimum
import java.util.*;
 
class GFG {
 
    // Function to check if a number is
    // prime or not.
    static boolean isPrime(int n)
    {
        int i = 2;
 
        while (i * i <= n) {
            if (n % i == 0)
                return false;
            i++;
        }
        return true;
    }
 
    static int minimumSum(int n)
    {
        if (isPrime(n))
            return 1;
 
        // If n is an even number (we can
        // write it as sum of two primes)
        if (n % 2 == 0)
            return 2;
 
        // If n is odd and n-2 is prime.
        if (isPrime(n - 2))
            return 2;
 
        // If n is odd, n-3 must be even.
        return 3;
    }
 
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 4;
        System.out.println(minimumSum(n));
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python3




# Python3 program to break a number
# such that sum of maximum divisors
# of all parts is minimum
 
# Function to check if a number is
# prime or not.
def isPrime( n):
    i = 2
     
    while (i * i <= n):
         
        if (n % i == 0):
            return False
        i+= 1
 
    return True
     
def minimumSum( n):
     
    if (isPrime(n)):
        return 1
     
    # If n is an even number (we can
    # write it as sum of two primes)
    if (n % 2 == 0):
        return 2
     
    # If n is odd and n-2 is prime.
    if (isPrime(n - 2)):
        return 2
         
    # If n is odd, n-3 must be even.
    return 3
 
# Driver code
n = 27
print(minimumSum(n))
 
# This code is contributed by "Abhishek Sharma 44"

C#




// C# Code for Break a number
// such that sum of maximum
// divisors of all parts is minimum
using System;
 
class GFG {
 
    // Function to check if a number is
    // prime or not.
    static bool isPrime(int n)
    {
        int i = 2;
 
        while (i * i <= n) {
            if (n % i == 0)
                return false;
            i++;
        }
        return true;
    }
 
    static int minimumSum(int n)
    {
        if (isPrime(n))
            return 1;
 
        // If n is an even number (we can
        // write it as sum of two primes)
        if (n % 2 == 0)
            return 2;
 
        // If n is odd and n-2 is prime.
        if (isPrime(n - 2))
            return 2;
 
        // If n is odd, n-3 must be even.
        return 3;
    }
 
    // Driver program
    public static void Main()
    {
        int n = 27;
        Console.WriteLine(minimumSum(n));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to break a number
// such that sum of maximum
// divisors of all parts is minimum
 
// Function to check if a
// number is prime or not.
function isPrime($n)
{
    $i = 2;
    while ($i * $i <= $n)
    {
        if ($n % $i == 0)
            return false;
        $i++;
    }
    return true;
}
 
function minimumSum($n)
{
    if (isPrime($n))    
        return 1;
 
    // If n is an even
    // number (we can
    // write it as sum
    // of two primes)
    if ($n % 2 == 0)
        return 2;
 
    // If n is odd and
    // n - 2 is prime.
    if (isPrime($n - 2))
        return 2;
 
    // If n is odd,
    // n - 3 must be even.
    return 3;
}
 
// Driver code
$n = 27;
echo minimumSum($n);
 
// This code is contributed by aj_36
?>

Javascript




<script>
 
// JavaScript program to break a number
// such that sum of maximum divisors of
// all parts is minimum
 
// Function to check if a number is
// prime or not.
function isPrime(n)
{
    let i = 2;
     
    while (i * i <= n)
    {
        if (n % i == 0)
            return false;
             
        i++;
    }
    return true;
}
 
function minimumSum(n)
{
    if (isPrime(n))
        return 1;
 
    // If n is an even number (we can
    // write it as sum of two primes)
    if (n % 2 == 0)
        return 2;
 
    // If n is odd and n-2 is prime.
    if (isPrime(n - 2))
        return 2;
 
    // If n is odd, n-3 must be even.
    return 3;
}
 
// Driver code
let n = 27;
 
document.write(minimumSum(n));
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output : 

    3

Time Complexity: O(√n) 
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above


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