Breaking an Integer to get Maximum Product

Given a number n, the task is to break n in such a way that multiplication of its parts is maximized. 

Input : n = 10
Output: 36
Explanation: 10 = 4 + 3 + 3 and 4 * 3 * 3 = 36 is the maximum possible product.

Input: n = 8
Output: 18
Explanation: 8 = 2 + 3 + 3 and 2 * 3 * 3 = 18 is the maximum possible product.

Mathematically, we are given n and we need to maximize a1 * a2 * a3 …. * aK such that n = a1 + a2 + a3 … + aK and a1, a2, ... ak > 0.
Note that we need to break given Integer in at least two parts in this problem for maximizing the product.

Method 1 - 

Now we know from maxima-minima concept that, If an integer need to break in two parts, then to maximize their product those part should be equal. Using this concept lets break n into (n/x) x's then their product will be x^(n/x), now if we take derivative of this product and make that equal to 0 for maxima, we will get to know that value of x should be e (base of the natural logarithm) for maximum product. As we know that 2 < e < 3, so we should break every Integer into 2 or 3 only for maximum product. 
Next thing is 6 = 3 + 3 = 2 + 2 + 2, but 3 * 3 > 2 * 2 * 2, that is every triplet of 2 can be replaced with tuple of 3 for maximum product, so we will keep breaking the number in terms of 3 only, until number remains as 4 or 2, which we will be broken into 2*2 (2*2 > 3*1) and 2 respectively and we will get our maximum product. 
In short, procedure to get maximum product is as follows – Try to break integer in power of 3 only and when integer remains small (<5) then use brute force. 
The complexity of below program is O(log N), because of repeated squaring power method. 

Follow the below steps to implement the above idea:

• Define a function breakInteger that takes an integer N as input and returns the maximum product that can be obtained by breaking N into a sum of positive integers.
  • Check for the two base cases:
    • If N is 2, return 1.
    • If N is 3, return 2.
  • Define a variable maxProduct to store the maximum product.
  • Determine the remainder of N when divided by 3:
               a. If the remainder is 0, the maximum product is 3 raised to the power of N/3.
               b. If the remainder is 1, the maximum product is 2 multiplied by 2 multiplied by 3 raised to the power of (N/3)-1.
               c. If the remainder is 2, the maximum product is 2 multiplied by 3 raised to the power of N/3.
  • Return the value of maxProduct.

Below is the implementation of the above approach: 

// C/C++ program to find maximum product by breaking
// the Integer
#include <bits/stdc++.h>
using namespace std;

// method return x^a in log(a) time
int power(int x, int a)
{
    int res = 1;
    while (a) {
        if (a & 1)
            res = res * x;
        x = x * x;
        a >>= 1;
    }
    return res;
}

// Method returns maximum product obtained by
// breaking N
int breakInteger(int N)
{
    //  base case 2 = 1 + 1
    if (N == 2)
        return 1;

    //  base case 3 = 2 + 1
    if (N == 3)
        return 2;

    int maxProduct;

    //  breaking based on mod with 3
    switch (N % 3) {
    // If divides evenly, then break into all 3
    case 0:
        maxProduct = power(3, N / 3);
        break;

    // If division gives mod as 1, then break as
    // 4 + power of 3 for remaining part
    case 1:
        maxProduct = 2 * 2 * power(3, (N / 3) - 1);
        break;

    // If division gives mod as 2, then break as
    // 2 + power of 3 for remaining part
    case 2:
        maxProduct = 2 * power(3, N / 3);
        break;
    }
    return maxProduct;
}

//  Driver code to test above methods
int main()
{
    int maxProduct = breakInteger(10);
    cout << maxProduct << endl;
    return 0;
}

// Java program to find maximum product by breaking
// the Integer

class GFG {
    // method return x^a in log(a) time
    static int power(int x, int a)
    {
        int res = 1;
        while (a > 0) {
            if ((a & 1) > 0)
                res = res * x;
            x = x * x;
            a >>= 1;
        }
        return res;
    }

    // Method returns maximum product obtained by
    // breaking N
    static int breakInteger(int N)
    {
        // base case 2 = 1 + 1
        if (N == 2)
            return 1;

        // base case 3 = 2 + 1
        if (N == 3)
            return 2;

        int maxProduct = -1;

        // breaking based on mod with 3
        switch (N % 3) {
        // If divides evenly, then break into all 3
        case 0:
            maxProduct = power(3, N / 3);
            break;

        // If division gives mod as 1, then break as
        // 4 + power of 3 for remaining part
        case 1:
            maxProduct = 2 * 2 * power(3, (N / 3) - 1);
            break;

        // If division gives mod as 2, then break as
        // 2 + power of 3 for remaining part
        case 2:
            maxProduct = 2 * power(3, N / 3);
            break;
        }
        return maxProduct;
    }

    // Driver code to test above methods
    public static void main(String[] args)
    {
        int maxProduct = breakInteger(10);
        System.out.println(maxProduct);
    }
}
// This code is contributed by mits

# Python3 program to find maximum product by breaking
# the Integer

# method return x^a in log(a) time


def power(x, a):

    res = 1
    while (a):
        if (a & 1):
            res = res * x
        x = x * x
        a >>= 1

    return res

# Method returns maximum product obtained by
# breaking N


def breakInteger(N):
    #  base case 2 = 1 + 1
    if (N == 2):
        return 1

    #  base case 3 = 2 + 1
    if (N == 3):
        return 2

    maxProduct = 0

    #  breaking based on mod with 3
    if(N % 3 == 0):
        # If divides evenly, then break into all 3
        maxProduct = power(3, int(N/3))
        return maxProduct
    elif(N % 3 == 1):
        # If division gives mod as 1, then break as
        # 4 + power of 3 for remaining part
        maxProduct = 2 * 2 * power(3, int(N/3) - 1)
        return maxProduct
    elif(N % 3 == 2):
        # If division gives mod as 2, then break as
        # 2 + power of 3 for remaining part
        maxProduct = 2 * power(3, int(N/3))
        return maxProduct


#  Driver code to test above methods


maxProduct = breakInteger(10)
print(maxProduct)

# This code is contributed by mits

// C# program to find maximum product by breaking
// the Integer

class GFG {
    // method return x^a in log(a) time
    static int power(int x, int a)
    {
        int res = 1;
        while (a > 0) {
            if ((a & 1) > 0)
                res = res * x;
            x = x * x;
            a >>= 1;
        }
        return res;
    }

    // Method returns maximum product obtained by
    // breaking N
    static int breakInteger(int N)
    {
        // base case 2 = 1 + 1
        if (N == 2)
            return 1;

        // base case 3 = 2 + 1
        if (N == 3)
            return 2;

        int maxProduct = -1;

        // breaking based on mod with 3
        switch (N % 3) {
        // If divides evenly, then break into all 3
        case 0:
            maxProduct = power(3, N / 3);
            break;

        // If division gives mod as 1, then break as
        // 4 + power of 3 for remaining part
        case 1:
            maxProduct = 2 * 2 * power(3, (N / 3) - 1);
            break;

        // If division gives mod as 2, then break as
        // 2 + power of 3 for remaining part
        case 2:
            maxProduct = 2 * power(3, N / 3);
            break;
        }
        return maxProduct;
    }

    // Driver code to test above methods
    public static void Main()
    {
        int maxProduct = breakInteger(10);
        System.Console.WriteLine(maxProduct);
    }
}
// This code is contributed by mits

<script>

// Javascript program to find maximum 
// product by breaking the Integer

// Method return x^a in log(a) time
function power(x, a)
{
    let res = 1;
    
    while (a > 0)
    {
        if ((a & 1) > 0)
            res = res * x;
            
        x = x * x;
        a >>= 1;
    }
    return res;
}

// Method returns maximum product obtained by
// breaking N
function breakInteger(N)
{
    
    // Base case 2 = 1 + 1
    if (N == 2)
        return 1;

    // Base case 3 = 2 + 1
    if (N == 3)
        return 2;

    let maxProduct;

    // Breaking based on mod with 3
    switch (N % 3)
    {
        
        // If divides evenly, then break into all 3
        case 0:
            maxProduct = power(3, N / 3);
            break;

        // If division gives mod as 1, then break as
        // 4 + power of 3 for remaining part
        case 1:
            maxProduct = 2 * 2 * power(3, (N / 3) - 1);
            break;

        // If division gives mod as 2, then break as
        // 2 + power of 3 for remaining part
        case 2:
            maxProduct = 2 * power(3, N / 3);
            break;
    }
    return maxProduct;
}

// Driver code 
let maxProduct = breakInteger(10);
document.write(maxProduct);

// This code is contributed by rameshtravel07  

</script>

<?php
// PHP program to find maximum product by breaking 
// the Integer 
  
// method return x^a in log(a) time 
function power($x, $a) 
{ 
    $res = 1; 
    while ($a) 
    { 
        if ($a & 1) 
            $res = $res * $x; 
        $x = $x * $x; 
        $a >>= 1; 
    } 
    return $res; 
} 
  
// Method returns maximum product obtained by 
// breaking N 
function breakInteger($N) 
{ 
    //  base case 2 = 1 + 1 
    if ($N == 2) 
        return 1; 
  
    //  base case 3 = 2 + 1 
    if ($N == 3) 
        return 2; 
  
    $maxProduct=0; 
  
    //  breaking based on mod with 3 
    switch ($N % 3) 
    { 
        // If divides evenly, then break into all 3 
        case 0: 
            $maxProduct = power(3, $N/3); 
            break; 
  
        // If division gives mod as 1, then break as 
        // 4 + power of 3 for remaining part 
        case 1: 
            $maxProduct = 2 * 2 * power(3, ($N/3) - 1); 
            break; 
  
        // If division gives mod as 2, then break as 
        // 2 + power of 3 for remaining part 
        case 2: 
            $maxProduct = 2 * power(3, $N/3); 
            break; 
    } 
    return $maxProduct; 
} 
  
//  Driver code to test above methods 

 
    $maxProduct = breakInteger(10); 
    echo $maxProduct; 
    
// This code is contributed by mits
?>

36

Method 2 - 

If we see some examples of this problems, we can easily observe following pattern.
The maximum product can be obtained be repeatedly cutting parts of size 3 while size is greater than 4, keeping the last part as size of 2 or 3 or 4. For example, n = 10, the maximum product is obtained by 3, 3, 4. For n = 11, the maximum product is obtained by 3, 3, 3, 2. Following is the implementation of this approach.

#include <iostream> 
using namespace std; 
  
/* The main function that returns the max possible product */
int maxProd(int n) 
{ 
   // n equals to 2 or 3 must be handled explicitly 
   if (n == 2 || n == 3) return (n-1); 
  
   // Keep removing parts of size 3 while n is greater than 4 
   int res = 1; 
   while (n > 4) 
   { 
       n -= 3; 
       res *= 3; // Keep multiplying 3 to res 
   } 
   return (n * res); // The last part multiplied by previous parts 
} 
  
/* Driver program to test above functions */
int main() 
{ 
    cout << "Maximum Product is " << maxProd(45); 
    return 0; 
} 

public class GFG
{
  /* The main function that returns the max possible product */
  static int maxProd(int n) 
  { 
    
    // n equals to 2 or 3 must be handled explicitly 
    if (n == 2 || n == 3) return (n - 1); 

    // Keep removing parts of size 3 while n is greater than 4 
    int res = 1; 
    while (n > 4) 
    { 
      n -= 3; 
      res *= 3; // Keep multiplying 3 to res 
    } 
    return (n * res); // The last part multiplied by previous parts 
  } 

  // Driver code
  public static void main(String[] args) {
    System.out.println("Maximum Product is " + maxProd(45));
  }
}

// This code is contributed by divyeshrabadiya07

''' The main function that returns the max possible product '''


def maxProd(n):

    # n equals to 2 or 3 must be handled explicitly
    if (n == 2 or n == 3):
        return (n - 1)

    # Keep removing parts of size 3 while n is greater than 4
    res = 1
    while (n > 4):

        n -= 3
        res *= 3  # Keep multiplying 3 to res

    return (n * res)  # The last part multiplied by previous parts


''' Driver program to test above functions '''
if __name__ == '__main__':
    print("Maximum Product is", maxProd(45))

    # This code is contributed by rutvik_56.

using System;
class GFG {
    
  /* The main function that returns the max possible product */
  static int maxProd(int n) 
  { 
     
    // n equals to 2 or 3 must be handled explicitly 
    if (n == 2 || n == 3) return (n - 1); 
 
    // Keep removing parts of size 3 while n is greater than 4 
    int res = 1; 
    while (n > 4) 
    { 
      n -= 3; 
      res *= 3; // Keep multiplying 3 to res 
    } 
    return (n * res); // The last part multiplied by previous parts 
  } 
  
  // Driver code 
  static void Main()
  {
    Console.WriteLine("Maximum Product is " + maxProd(45));
  }
}

// This code is contributed by divyesh072019.

<script>

    /* The main function that returns the max possible product */
    function maxProd(n)
    {

      // n equals to 2 or 3 must be handled explicitly
      if (n == 2 || n == 3) return (n - 1);

      // Keep removing parts of size 3 while n is greater than 4
      let res = 1;
      while (n > 4)
      {
        n -= 3;
        res *= 3; // Keep multiplying 3 to res
      }
      return (n * res); // The last part multiplied by previous parts
    }
    
    document.write("Maximum Product is " + maxProd(45));
    
</script>

Maximum Product is 14348907

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

Method 3: (Using Recursion)

Intuition:
Basically in this problem, We have to maximize the product of some integers which sums up to the given integer. Let's take an example of n = 5 and try to solve it. So, we can break 5 to 4,1 or 3,1,1 or 2,1,1,1 or 1,1,1,1,1. We can also break it instead to 3,2 or 1,2,2 or 1,1,1,2 and so on.After taking all these possibilities, [3,2] gives the max product which is 6. Basically we see that our problem is getting divided into subproblems. Thus we can make use of recursion to solve this. Further, if we want to solve for n = 6, we can make use of the previous max product value we got for n = 5 that is 6 to check possibilites for 6, so we can easily do memoization in our recursive solution to reduce our time complexity.

Approach:
The idea is that at every value, we can loop from 1 to n-1 for the first value (a). The remain value (b) has two options:
Keep the same, which means the product will be a * b
Or broken down, which means the product is: a * max(integerBreak(b))

#include <iostream> 
using namespace std; 
  
/* The main function that returns the max possible product */
int helper(int n, int idx)
    {
         //base condition
       if(n == 0 or idx == 0) return 1;
       //recursive call for each step
       if(idx > n) return helper(n, idx - 1);
         //return the maximum result obtained from recursive calls
       return max((idx * helper(n - idx, idx)), helper(n , idx - 1));
    }
    //max product function
    int maxProd(int n)
    {
       return helper(n, n - 1);
    }
  
/* Driver program to test above functions */
int main() 
{ 
    cout << "Maximum Product is " << maxProd(45); 
    return 0; 
} 

public class GFG {

    /* The main function that returns the max possible product */
    static int helper(int n, int idx) {
        // base condition
        if (n == 0 || idx == 0)
            return 1;
        // recursive call for each step
        if (idx > n)
            return helper(n, idx - 1);
        // return the maximum result obtained from recursive calls
        return Math.max((idx * helper(n - idx, idx)), helper(n, idx - 1));
    }

    // max product function
    static int maxProd(int n) {
        return helper(n, n - 1);
    }

    /* Driver program to test above functions */
    public static void main(String[] args) {
        System.out.println("Maximum Product is " + maxProd(45));
    }
}

# Python code for the above approach

# The main function that returns the max possible product


def helper(n, idx):
    # Base condition
    if n == 0 or idx == 0:
        return 1
    # Recursive call for each step
    if idx > n:
        return helper(n, idx - 1)
    # Return the maximum result obtained from recursive calls
    return max((idx * helper(n - idx, idx)), helper(n, idx - 1))

# Max product function


def maxProd(n):
    return helper(n, n - 1)


# Driver program to test above functions
if __name__ == "__main__":
    print("Maximum Product is", maxProd(45))

# This code is contributed by Susobhan Akhuli

// C# code for the above approach
using System;

public class GFG {
    /* The main function that returns the max possible
     * product */
    static int Helper(int n, int idx)
    {
        // base condition
        if (n == 0 || idx == 0)
            return 1;
        // recursive call for each step
        if (idx > n)
            return Helper(n, idx - 1);
        // return the maximum result obtained from recursive
        // calls
        return Math.Max((idx * Helper(n - idx, idx)),
                        Helper(n, idx - 1));
    }

    // max product function
    static int MaxProd(int n) { return Helper(n, n - 1); }

    /* Driver program to test above functions */
    public static void Main(string[] args)
    {
        Console.WriteLine("Maximum Product is "
                          + MaxProd(45));
    }
}

// This code is contributed by Susobhan Akhuli

// function to calculate the max product 
function helper(n, idx) {
    
   //  If n or idx is 0, return 1
    if (n === 0 || idx === 0) return 1;

    // If idx is > n, recursive call with a smaller idx
    if (idx > n) return helper(n, idx - 1);

   //return the maximum result obtained from recursive calls
    return Math.max(idx * helper(n - idx, idx), helper(n, idx - 1));
}

  //max product function
function maxProd(n) {
   
    return helper(n, n - 1);
}

// Driver code
console.log("Maximum Product is " + maxProd(45));

Maximum Product is 14348907

Time Complexity: O(n),fo making the recursive calls.
Auxiliary Space: O(n), recursive stack space