343. Integer Break

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:

Split $i$ into the sum of $i - j$ and $j$, without further splitting, where the product is $(i - j) \times j$;
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.

Python3JavaC++GoTypeScriptRustJavaScriptC#C

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]

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];
    }
}

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];
    }
};

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]
}

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];
}

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
    }
}

/**
 * @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];
};

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];
    }
}

#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 $O(1)$.