343. Integer Break

Problem

Divides a positive integer into the sum of at least two positive integers, which maximizes the product of these integers and gives the maximum product.

Input:10
Output:36
Explanation:10 = 3 + 3 + 4, 3x3x4 = 36

Ideas

When PQ > P + q, as long as p and Q are greater than 1 satisfied. In a positive integer, two is 2 o'clock pq==p+q, so as long as a minimum of 2, the other is at least 3, that is, p+q at least 5, then as long as a number greater than or equal to 5, you can divide this number into p+q, so that PQ > P + q, if p or Q there are more than or equal to 5 of the number, The P or Q continues to decompose, and continues to decompose until the factor is less than 5.

Because to decompose to less than 5, the final factor must be 2 or 3 (or 4), because dividing 4 and dividing 2 is the same (two 2 and one 4 are the same product), so we consider how to allocate 3 and 2.

Consider a number with M 3 and N 2 added, i.e. \ (x = 3m + 2n\), then the product equals M 3 and N 2 multiplied, define a function,\ (f = 3^m * 2^n = 3^m * 2^{\frac{x-3m}{2}}\), you can Getting $ln F $ is a function that increments m, as shown in the following formula. So the larger the greater the better, M represents 3 of the number, that is, when the decomposition of the time as much as possible to divide a few 3, divided into 2 can not be divided.
\[\begin{split}ln\ f &= m * ln3 + \frac{x-3m}{2} * LN2 \ &= \frac{x}{2} ln \ 2 + (ln\ 3-\frac{3}{2}ln\ 2) *m \e Nd{split}\]

Time complexity O (n), Spatial complexity O (1)

Code

class Solution(object):    def integerBreak(self, n):        """        :type n: int        :rtype: int        """        if(n == 2 or n == 3):            return n-1        res = 1        while(n>=5):            n -= 3            res *= 3        res *= n        return res     

For the following reasons, you can also write this.
If n%3 = = 0, can be decomposed into N/3 3, their product is \ (3^{n/3}\).
if n%3 = = 1, the description and form of N/3 * 3 + 1, one of 3 and a separate 1 is rewritten to 2 2, their product is \ (3^{n/3-1} * 2 * 2\).
if n%3 = = 2, the description and form of N/3 * 3 + 2, their product is \ (3^{N/3} * 2\).

class Solution(object):    def integerBreak(self, n):        """        :type n: int        :rtype: int        """        if n>=4:            if n%3 == 0:                return 3**(n/3)            if n%3 == 1:                return 3**(n/3-1)*4            else:                return 3**(n/3)*2        return n-1

343. Integer Break