README_EN.md

343. Integer Break

中文文档

Description

Given an integer n, break it into the sum of k positive integers, where k >= 2, and maximize the product of those integers.

Return the maximum product you can get.

 

Example 1:

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

Example 2:

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

 

Constraints:

• 2 <= n <= 58

Solutions

Dynamic programming.

Python3

class Solution:
    def integerBreak(self, n: int) -> int:
        dp = [1] * (n + 1)
        for i in range(2, n + 1):
            for j in range(1, i):
                dp[i] = max(dp[i], dp[i - j] * j, (i - j) * j)
        return dp[n]

Java

class Solution {
    public int integerBreak(int n) {
        int[] dp = new int[n + 1];
        dp[1] = 1;
        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j < i; ++j) {
                dp[i] = Math.max(Math.max(dp[i], dp[i - j] * j), (i - j) * j);
            }
        }
        return dp[n];
    }
}

C++

class Solution {
public:
    int integerBreak(int n) {
        vector<int> dp(n + 1);
        dp[1] = 1;
        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j < i; ++j) {
                dp[i] = max(max(dp[i], dp[i - j] * j), (i - j) * j);
            }
        }
        return dp[n];
    }
};

Go

func integerBreak(n int) int {
	dp := make([]int, n+1)
	dp[1] = 1
	for i := 2; i <= n; i++ {
		for j := 1; j < i; j++ {
			dp[i] = max(max(dp[i], dp[i-j]*j), (i-j)*j)
		}
	}
	return dp[n]
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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