343. Integer Break
Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.
Example 1:
Input: 2 Output: 1 Explanation: 2 = 1 + 1, 1 × 1 = 1.
Example 2:
Input: 10 Output: 36 Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.
Note: You may assume that n is not less than 2 and not larger than 58.
Rust Solution
struct Solution;
use std::collections::HashMap;
impl Solution {
fn integer_break(n: i32) -> i32 {
let mut memo: HashMap<(i32, bool), i32> = HashMap::new();
Self::dp(n, true, &mut memo)
}
fn dp(n: i32, split: bool, memo: &mut HashMap<(i32, bool), i32>) -> i32 {
if let Some(&res) = memo.get(&(n, split)) {
return res;
}
let mut res = if split { 0 } else { n };
for i in 1..n {
res = res.max(i * Self::dp(n - i, false, memo));
}
memo.insert((n, split), res);
res
}
}
#[test]
fn test() {
let n = 2;
let res = 1;
assert_eq!(Solution::integer_break(n), res);
let n = 10;
let res = 36;
assert_eq!(Solution::integer_break(n), res);
}
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