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Breaking an Integer to get Maximum Product
  • Difficulty Level : Medium
  • Last Updated : 11 Feb, 2021

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
10 = 4 + 3 + 3 and 4 * 3 * 3 = 36
is maximum possible product.

Input : n = 8
Output : 18
8 = 2 + 3 + 3 and 2 * 3 * 3 = 18
is 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

Below is the implementation of above approach: 
 



C++




// 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




// 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




# 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#




// 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


PHP




<?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
?>


Output

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.

C++




#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;
}


Java




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


Python3




   
''' 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.


C#




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.


Output

Maximum Product is 14348907

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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