# Question

Formatted question description: https://leetcode.ca/all/343.html

 343	Integer Break

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.

@tag-dp
@tag-math


# Algorithm

### Iteration

Positive integers start from 1, but 1 cannot be split into the sum of two positive integers, so it cannot be used as an input.

Then 2 can only be split into 1+1, so the product is also 1.

The number 3 can be split into 2+1 or 1+1+1. Obviously, the product of the first split method is two.

The number 4 is divided into 2+2, and the product is the largest, which is 4.

The number 5 is divided into 3+2, and the product is the largest, which is 6.

The number 6 is divided into 3+3, and the product is the largest, which is 9.

The number 7 is divided into 3+4, and the product is the largest, which is 12.

The number 8 is divided into 3+3+2, and the product is the largest, which is 18.

The number 9 is divided into 3+3+3, and the product is the largest, 27.

The number 10 is divided into 3+3+4, the product is the largest, 36.

Starting from 5, all numbers need to be divided into 3 first, until the remaining number is 2 or 4, because there is no need to divide the remaining 4, it does not make sense to divide into two 2 and not divide, and 4 cannot be divided into 3+1, which will be smaller than the product of 2+2.

So, first preprocess the case where n is 2 and 3, and then initialize the result res to 1, and then start the loop when n is greater than 4, multiply res by 3, and subtract 3 from n. According to the previous analysis, when out of the loop, n only Can be 2 or 4, then multiply by res to return.

### DP

dp[i] represents the maximum product of the sum of at least two positive integers divided from i int[] dp = new int[n + 1];

And formula:

dp[i] = Math.max(
dp[i],
Math.max(
j * (i - j),
j * dp[i - j]
)


# Code

Java

• 
public class Integer_Break {

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

}
}

class Solution_dp {
public int integerBreak(int n) {

// Among them, dp[i] represents the maximum product of the sum of at least two positive integers divided from i
int[] dp = new int[n + 1];
dp = 1;
dp = 1;
dp = 1;

for (int i = 3; i <= n; ++i) {
for (int j = 1; j < i; ++j) {
dp[i] = Math.max(
dp[i],
Math.max(
j * (i - j),
j * dp[i - j]
)
);
}
}
return dp[n];
}
}
}

• // OJ: https://leetcode.com/problems/integer-break/
// Time: O(N * sqrt(N))
// Space: O(N)
class Solution {
public:
int integerBreak(int n) {
vector<int> memo(59, 0);
memo = 1;
for (int i = 2; i <= n; ++i)
for (int j = 1, b = ceil(sqrt(i)); j <= b; ++j)
memo[i] = max(memo[i], max(j, memo[j]) * max(i - j, memo[i - j]));
return memo[n];
}
};

• class Solution:
def integerBreak(self, n: int) -> int:
if n < 4:
return n - 1
if n % 3 == 0:
return pow(3, n // 3)
if n % 3 == 1:
return pow(3, n // 3 - 1) * 4
return pow(3, n // 3) * 2

############

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