# Java 0ms DP Solution

• ``````public int integerBreak(int n) {
int[] res = new int[n+1];
res[1] =1;
// For numbers from 2 to 6
for (int i = 2; i <= 6 && i <= n; i++){
res[i] = (int)i/2 * ( i - (int)i/2);
}
// For numbers larger than 6, apply DP.
for (int i = 7; i <= n; i++) {
res[i] = Math.max(res[i-2] * 2, res[i-3] * 3);
}
return res[n];
}``````

• Nice solution. The first for loop actually works till 7

• @jguo11 What is the logic ? .. Why does the second loop work?

• @kireeti2 In order to get the maximum, we need to break the integer into as many 2 or 3 as we can. For example, for 15, the max comes from 3+3+3+3+3; for 16, it will be 3+3+3+3+2+2, etc.

• @jguo11 said in Java 0ms DP Solution:

yes i got that ..but why would that work. any specific reason ..or just an observation?

• @kireeti2 For now, those are just my observation and there must be some math deductions can be done to prove it.

• @kireeti2 The key for this problem is that we need to break the number to `2`s(or `4`s), and `3`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.

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.