Solution 1

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]
}

TypeScript

function integerBreak(n: number): number {
    let dp = new Array(n + 1).fill(1);
    for (let i = 3; i <= n; i++) {
        for (let j = 1; j < i; j++) {
            dp[i] = Math.max(dp[i], j * (i - j), j * dp[i - j]);
        }
    }
    return dp.pop();
}

Rust

impl Solution {
    pub fn integer_break(n: i32) -> i32 {
        if n < 4 {
            return n - 1;
        }
        let count = (n - 2) / 3;
        (3i32).pow(count as u32) * (n - count * 3)
    }
}

C

int integerBreak(int n) {
    if (n < 4) {
        return n - 1;
    }
    int count = (n - 2) / 3;
    return pow(3, count) * (n - count * 3);
}

Problem