Breaking an Integer to get Maximum Product

Given an number n, the task is to broken 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 = 36
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.

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 

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PHP

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<?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
?>

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Output:

36

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