# Python solution (40ms) with explanation

• ``````class Solution(object):
def integerBreak(self, n):
"""
:type n: int
:rtype: int
"""
if n == 2:
return 1
if n == 3:
return 2
list_3 = [3] * (n/3) # generate a list of 3
mod_3 = n%3
if mod_3 == 1: # if a 1 is left, then add it to the first element to get a 4
list_3[0] += 1
if mod_3 == 2: # if a 2 is left, then put it into the list
list_3.append(2)
return reduce(lambda a, b: a*b, list_3)
``````

The key for this problem is that we need to break the number to `2`s, `3`s and `4`s.
First we need to know a fact that,`if a,b > 3, |a-b| <= 1, then a*b>=a+b`.

So, `if n = a + b, a = a1+a2, b=b1+b2`, we should break n to `a1+a2+b1+b2, |a1-a2|<1 and |b1-b2|<1` instead of `a + b`, because `a1*a2>a, b1*b2>b`. However, we shall stop when we get a `3` or `2`, so what we shall do is to find the list of `3` and `2`.

You may have noticed why the `4` appeared. 'Cause if we break `4`, we get `2+2`, and `2+2 = 2*2`, so it's the same with the condition that we get two `2`s.

• I like your solution. I would write like this instead of those one liner :)

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