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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.
@tagdp
@tagmath
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[0] = 1; dp[1] = 1; dp[2] = 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/integerbreak/ // Time: O(N * sqrt(N)) // Space: O(N) class Solution { public: int integerBreak(int n) { vector<int> memo(59, 0); memo[1] = 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