# 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

Dynamic programming.

• 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;
}
}

• 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;
}
};

• 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

• 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
}

• 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;
}

• 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)
}
}

• /**
* @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];
}
}