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:
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