Breaking an Integer to get Maximum Product

23 Jun, 2022
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++
Java
Python3
C#
PHP
Javascript
// 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) timeint 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 Nint 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 methodsint 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    else if(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    else if(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
<?php// PHP program to find maximum product by breaking// the Integer   // method return x^a in log(a) timefunction power($x, $a){    $res = 1;    while ($a)    {        if ($a & 1)            $res = $res * $x;        $x = $x * $x;        $a >>= 1;    }    return $res;}   // Method returns maximum product obtained by// breaking Nfunction breakInteger($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?>
<script> // Javascript program to find maximum// product by breaking the Integer // Method return x^a in log(a) timefunction 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 Nfunction 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 codelet maxProduct = breakInteger(10);document.write(maxProduct); // This code is contributed by rameshtravel07  </script>
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++
Java
Python3
C#
Javascript
#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)
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