# Breaking an Integer to get Maximum Product

• Difficulty Level : Medium
• Last Updated : 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++

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

 `// 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

 `>= 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` `?>`

## Javascript

 ``

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 ` `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.`

## Javascript

 ``

Output

`Maximum Product is 14348907`

Time Complexity: O(n)

Auxiliary Space: O(1)

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