# Leetcode 343. Integer Break

## Description

Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.

Example 1:

Input: 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.

Example 2:

Input: 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

Note: You may assume that n is not less than 2 and not larger than 58.

## Solution

Breaking the array into several numbers, multiply them to maximize the product of those integers.

In a straightforward way, we search with memorization array.

### Solution1. DFS, Time: O(n2), Space: O(n2), 1ms

java

class Solution {
public int integerBreak(int n) {
return dfs(n, 0, new int[n + 1][n + 1]);
}

private int dfs(int cur, int count, int[][] memo) {
if (cur == 0 && count >= 2) return 1;
if (cur <= 0 && count <= 1) return 0;
if (memo[cur][count] > 0) return memo[cur][count];
int res = 0;
int product = 1;
for (int i = 1; i <= cur; i++) {
res = Math.max(res, product* i *dfs(cur - i, count + 1, memo));
}
memo[cur][count] = res;
return res;
}
}

Why is O(n2?)

Because we will do add for n + n - 1 + n - 2 + ... + 1 times.

Python3

import functools

class Solution:
def integerBreak(self, n: int) -> int:

@functools.lru_cache
def dfs(cur, count):
if cur == 0 and count >= 2:
return 1
if cur == 0 and count <= 1:
return 0
res, product = 0, 1
for i in range(1, cur + 1):
res = max(res, product * i * dfs(cur - i, count + 1))
return res
return dfs(n, 0)

Obviously, the straightforward way is too slow. We can speed up in a mathmetic solution.

Considering two tips:

• when n > 4, (n – 3) * 3 > n.
• 2 * 2 * 2 < 3 * 3

We need multiply as many as 3s.

So we will get an accepted, O(n) solution:

### Solution2. Time: O(n), Space: O(1), 0ms

Java

class Solution {
public int integerBreak(int n) {
if (n == 2) return 1;
if (n == 3) return 2;
int product = 1;
while (n > 4) {
product *= 3;
n -= 3;
}
return product * n;
}
}

Python3

class Solution:
def integerBreak(self, n: int) -> int:
if n <= 3:
return n - 1
product = 1
while n > 4:
product *= 3
n -= 3
return product * n