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Find minimum sum of factors of number

Given a number N, the task is to find minimum sum of its factors.

Examples: 



Input: 12
Output: 7
Explanation:
Following are different ways to factorize 12 andsum of factors in different ways.12 = 12 * 1 = 12 + 1 = 1312 = 2 * 6 = 2 + 6 = 812 = 3 * 4 = 3 + 4 = 712 = 2 * 2 * 3 = 2 + 2 + 3 = 7Therefore minimum sum is 7

Input: 105
Output: 15 



Recommended Practice

Approach:

To minimize sum, we must factorize factors as long as possible. With this process, we prime factors. So to find minimum sum of product of number, we find sum of prime factors of product. 




// CPP program to find minimum
// sum of product of number
#include <bits/stdc++.h>
using namespace std;
 
// To find minimum sum of
// product of number
int findMinSum(int num)
{
    int sum = 0;
 
    // Find factors of number
    // and add to the sum
    for (int i = 2; i * i <= num; i++) {
        while (num % i == 0) {
            sum += i;
            num /= i;
        }
    }
    sum += num;
 
    // Return sum of numbers
    // having minimum product
    return sum;
}
 
// Driver program to test above function
int main()
{
    int num = 12;
 
    cout << findMinSum(num);
 
    return 0;
}




// Java program to find minimum
// sum of product of number
 
public class Main {
 
    // To find minimum sum of
    // product of number
    static int findMinSum(int num)
    {
        int sum = 0;
 
        // Find factors of number
        // and add to the sum
        for (int i = 2; i * i <= num; i++) {
            while (num % i == 0) {
                sum += i;
                num /= i;
            }
        }
        sum += num;
 
        // Return sum of numbers
        // having minimum product
        return sum;
    }
 
    // Driver program to test above function
    public static void main(String[] args)
    {
        int num = 12;
        System.out.println(findMinSum(num));
    }
}




# Python program to find minimum
# sum of product of number
  
# To find minimum sum of
# product of number
def findMinSum(num):
    sum = 0
     
    # Find factors of number
    # and add to the sum
    i = 2
    while(i * i <= num):
        while(num % i == 0):
            sum += i
            num //= i
        i += 1
    sum += num
     
    # Return sum of numbers
    # having minimum product
    return sum
 
# Driver Code
num = 12
print (findMinSum(num))
 
# This code is contributed by Sachin Bisht




// C# program to find minimum
// sum of product of number
using System;
 
public class GFG {
 
    // To find minimum sum of
    // product of number
    static int findMinSum(int num)
    {
        int sum = 0;
 
        // Find factors of number
        // and add to the sum
        for (int i = 2; i * i <= num; i++) {
            while (num % i == 0) {
                sum += i;
                num /= i;
            }
        }
        sum += num;
 
        // Return sum of numbers
        // having minimum product
        return sum;
    }
 
    // Driver Code
    public static void Main()
    {
        int num = 12;
        Console.Write(findMinSum(num));
    }
     
}
 
// This Code is contributed by Nitin Mittal.




<script>
 
// Javascript program to find minimum
// sum of product of number
   
// To find minimum sum of
// product of number
function findMinSum(num)
{
    let sum = 0;
   
    // Find factors of number
    // and add to the sum
    for (let i = 2; i * i <= num; i++) 
    {
        while (num % i == 0) 
        {
            sum += i;
            num /= i;
        }
    }
    sum += num;
   
    // Return sum of numbers
    // having minimum product
    return sum;
}
   
// Driver Code
let num = 12;
   
document.write(findMinSum(num));
   
// This code is contributed by _saurabh_jaiswal.
 
</script>




<?php
// PHP program to find minimum
// sum of product of number
 
// To find minimum sum of
// product of number
function findMinSum($num)
{
    $sum = 0;
 
    // Find factors of number
    // and add to the sum
    for ($i = 2; $i * $i <= $num; $i++)
    {
        while ($num % $i == 0)
        {
            $sum += $i;
            $num /= $i;
        }
    }
    $sum += $num;
 
    // Return sum of numbers
    // having minimum product
    return $sum;
}
 
// Driver Code
$num = 12;
 
echo(findMinSum($num));
 
// This code is contributed by Ajit.
?>

Output: 

7

Time Complexity : O(n1/2 * log n)
Auxiliary Space: O(1)



 


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