Find the value of N when F(N) = f(a)+f(b) where a+b is the minimum possible and a*b = N

Given an integer N, the task is to find the value of F(N) if:  

1. F(1) = 0
2. F(2) = 2
3. F(N) = 0, if N is odd prime.
4. F(N) = F(a) + F(b), where a and b are factors of N and (a + b) is minimum among all factors. Also, a * b = N

Examples:

Input: N = 5 
Output: 0 
Since 5 is an odd prime.
Input: N = 4 
Output: 4 
4 can be written as 2 * 2, hence f(2) + f(2) = 4
Input: N = 20 
Output: 4 
20 can be written as f(4) + f(5), and f(4) can be written as f(2) + f(2), which is 4.  

Approach: The following steps can be followed to solve the problem:  

• If N is 1 or 2, the answer is 0 or 2 respectively.
• On breaking the recurrence f(n) = f(a) + f(b), we get it is the number of times a number is divisible by 2.
• The answer for f(n) is 2 * (number of times a number is divisible by 2)

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the value of F(N)
int getValueOfF(int n)
{

    // Base cases
    if (n == 1)
        return 0;
    if (n == 2)
        return 1;

    int cnt = 0;

    // Count the number of times a number
    // if divisible by 2
    while (n % 2 == 0) {
        cnt += 1;
        n /= 2;
    }

    // Return the summation
    return 2 * cnt;
}

// Driver code
int main()
{
    int n = 20;
    cout << getValueOfF(n);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;

class GFG
{

// Function to return the value of F(N)
static int getValueOfF(int n)
{

    // Base cases
    if (n == 1)
        return 0;
    if (n == 2)
        return 1;

    int cnt = 0;

    // Count the number of times a number
    // if divisible by 2
    while (n % 2 == 0)
    {
        cnt += 1;
        n /= 2;
    }

    // Return the summation
    return 2 * cnt;
}

// Driver code
public static void main (String[] args)
{
    int n = 20;
    System.out.println (getValueOfF(n));
}
}

// This code is contributed by ajit.

Python3

# Python3 implementation of the approach

# Function to return the value of F(N)
def getValueOfF(n):

    # Base cases
    if (n == 1):
        return 0
    if (n == 2):
        return 1

    cnt = 0

    # Count the number of times a number
    # if divisible by 2
    while (n % 2 == 0):
        cnt += 1
        n /= 2

    # Return the summation
    return 2 * cnt

# Driver code
n = 20
print(getValueOfF(n))

# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the value of F(N)
static int getValueOfF(int n)
{

    // Base cases
    if (n == 1)
        return 0;
    if (n == 2)
        return 1;

    int cnt = 0;

    // Count the number of times a number
    // if divisible by 2
    while (n % 2 == 0)
    {
        cnt += 1;
        n /= 2;
    }

    // Return the summation
    return 2 * cnt;
}

// Driver code
static public void Main ()
{
    int n = 20;
    Console.WriteLine(getValueOfF(n));
}
}

// This code is contributed by akt_mit.

PHP

<?php
//PHP implementation of the approach
// Function to return the value of F(N)

function getValueOfF($n)
{

    // Base cases
    if ($n == 1)
        return 0;
    if ($n == 2)
        return 1;

    $cnt = 0;

    // Count the number of times a number
    // if divisible by 2
    while ($n % 2 == 0)
    {
        $cnt += 1;
        $n /= 2;
    }

    // Return the summation
    return 2 * $cnt;
}

    // Driver code
    $n = 20;
    echo getValueOfF($n);

// This code is contributed by Tushil..
?>

Javascript

<script>
    // Javascript implementation of the approach

    // Function to return the value of F(N)
    function getValueOfF(n)
    {

        // Base cases
        if (n == 1)
            return 0;
        if (n == 2)
            return 1;

        let cnt = 0;

        // Count the number of times a number
        // if divisible by 2
        while (n % 2 == 0)
        {
            cnt += 1;
            n = parseInt(n / 2, 10);
        }

        // Return the summation
        return 2 * cnt;
    }

    let n = 20;
    document.write(getValueOfF(n));

</script>

Output:

4

Time Complexity: O(log n), as we are using a loop to traverse and in each traversal we are decrementing n by floor division of 2 therefore the effective time will be 1+1/2+1/4+…..+1/2^n which is equivalent to log(n).

Auxiliary Space: O(1), as we are not using any extra space.