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: Dynamic Programming

We define $f[i]$ as the maximum product that can be obtained by splitting the positive integer $i$, with an initial condition of $f[1] = 1$. The answer is $f[n]$.

Consider the last number $j$ split from $i$, where $j \in [1, i)$. For the number $j$ split from $i$, there are two cases:

1. Split $i$ into the sum of $i - j$ and $j$, without further splitting, where the product is $(i - j) \times j$;
2. Split $i$ into the sum of $i - j$ and $j$, and continue splitting, where the product is $f[i - j] \times j$.

Therefore, we can derive the state transition equation:

$$ f[i] = \max(f[i], f[i - j] \times j, (i - j) \times j) \quad (j \in [0, i)) $$

Finally, returning $f[n]$ will suffice.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the given positive integer.

Python3

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

Java

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

C++

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

Go

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

TypeScript

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

Rust

impl Solution {
    pub fn integer_break(n: i32) -> i32 {
        let n = n as usize;
        let mut f = vec![0; n + 1];
        f[1] = 1;
        for i in 2..=n {
            for j in 1..i {
                f[i] = f[i].max(f[i - j] * j).max((i - j) * j);
            }
        }
        f[n] as i32
    }
}

JavaScript

/**
 * @param {number} n
 * @return {number}
 */
var integerBreak = function (n) {
    const f = Array(n + 1).fill(1);
    for (let i = 2; i <= n; ++i) {
        for (let j = 1; j < i; ++j) {
            f[i] = Math.max(f[i], f[i - j] * j, (i - j) * j);
        }
    }
    return f[n];
};

C#

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

C

#define max(a, b) (((a) > (b)) ? (a) : (b))

int integerBreak(int n) {
    int* f = (int*) malloc((n + 1) * sizeof(int));
    f[1] = 1;
    for (int i = 2; i <= n; ++i) {
        f[i] = 0;
        for (int j = 1; j < i; ++j) {
            f[i] = max(f[i], max(f[i - j] * j, (i - j) * j));
        }
    }
    return f[n];
}

Solution 1: Mathematics

When $n < 4$, since the problem requires splitting into at least two integers, $n - 1$ yields the maximum product. When $n \geq 4$, we split into as many $3$s as possible. If the last segment remaining is $4$, we split it into $2 + 2$ for the maximum product.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

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;
}

Rust

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

JavaScript

/**
 * @param {number} n
 * @return {number}
 */
var integerBreak = function (n) {
    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;
};

C#

public 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

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