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:
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 ∈ [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) × j;
- Split i into the sum of i - j and j, and continue splitting, where the product is f[i - j] × 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.
PythonJavaC++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](code-box)
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];
}
}(code-box)
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];
}
};(code-box)
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]
}(code-box)
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];
}(code-box)
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
}
}(code-box)
/**
* @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];
};(code-box)
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];
}
}(code-box)
#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];
}(code-box)
Solution 1: Mathematics
When n < 4, since the problem requires splitting into at least two integers, n - 1 yields the maximum product. When n ≥ 4, we split into as many 3s 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).
PythonJavaC++GoTypeScriptRustJavaScriptC#C
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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,
}
}
}(code-box)
/**
* @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;
};(code-box)
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;
}
}(code-box)
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;
}(code-box)
