Break a number such that sum of maximum divisors of all parts is minimum
Last Updated :
23 Jun, 2022
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++
#include <iostream>
using namespace std;
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 % 2 == 0)
return 2;
if (isPrime(n - 2))
return 2;
return 3;
}
int main()
{
int n = 27;
cout << minimumSum(n);
return 0;
}
|
Java
import java.util.*;
class GFG {
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 % 2 == 0 )
return 2 ;
if (isPrime(n - 2 ))
return 2 ;
return 3 ;
}
public static void main(String[] args)
{
int n = 4 ;
System.out.println(minimumSum(n));
}
}
|
Python3
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 % 2 = = 0 ):
return 2
if (isPrime(n - 2 )):
return 2
return 3
n = 27
print (minimumSum(n))
|
C#
using System;
class GFG {
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 % 2 == 0)
return 2;
if (isPrime(n - 2))
return 2;
return 3;
}
public static void Main()
{
int n = 27;
Console.WriteLine(minimumSum(n));
}
}
|
PHP
<?php
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 % 2 == 0)
return 2;
if (isPrime( $n - 2))
return 2;
return 3;
}
$n = 27;
echo minimumSum( $n );
?>
|
Javascript
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.
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