README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0343.Integer%20Break/README_EN.md	Math Dynamic Programming

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

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);
}

Solution 2

Python3

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

Java

class Solution {
    public int integerBreak(int n) {
        if (n < 4) {
            return n - 1;
        }
        if (n % 3 == 0) {
            return (int) Math.pow(3, n / 3);
        }
        if (n % 3 == 1) {
            return (int) Math.pow(3, n / 3 - 1) * 4;
        }
        return (int) Math.pow(3, n / 3) * 2;
    }
}

C++

class Solution {
public:
    int integerBreak(int n) {
        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;
    }
};

Go

func integerBreak(n int) int {
	if n < 4 {
		return n - 1
	}
	if n%3 == 0 {
		return int(math.Pow(3, float64(n/3)))
	}
	if n%3 == 1 {
		return int(math.Pow(3, float64(n/3-1))) * 4
	}
	return int(math.Pow(3, float64(n/3))) * 2
}

TypeScript

function integerBreak(n: number): number {
    if (n < 4) {
        return n - 1;
    }
    const m = Math.floor(n / 3);
    if (n % 3 == 0) {
        return 3 ** m;
    }
    if (n % 3 == 1) {
        return 3 ** (m - 1) * 4;
    }
    return 3 ** m * 2;
}