Given a string s
, partition s
such that every substring of the partition is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s
.
Example 1:
Input: s = "aab" Output: 1 Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.
Example 2:
Input: s = "a" Output: 0
Example 3:
Input: s = "ab" Output: 1
Constraints:
1 <= s.length <= 2000
s
consists of lower-case English letters only.struct Solution;
use std::collections::HashMap;
impl Solution {
fn min_cut(s: String) -> i32 {
let n = s.len();
let s: Vec<char> = s.chars().collect();
let mut memo: HashMap<(usize, usize), i32> = HashMap::new();
Self::dp(0, n, &mut memo, &s)
}
fn dp(start: usize, end: usize, memo: &mut HashMap<(usize, usize), i32>, s: &[char]) -> i32 {
if let Some(&res) = memo.get(&(start, end)) {
return res;
}
let res = if Self::is_palindrome(start, end, s) {
0
} else {
let mut res = std::i32::MAX;
for i in start + 1..end {
if Self::is_palindrome(start, i, s) {
res = res.min(1 + Self::dp(i, end, memo, s));
}
}
res
};
memo.insert((start, end), res);
res
}
fn is_palindrome(start: usize, end: usize, s: &[char]) -> bool {
!s[start..end]
.iter()
.zip(s[start..end].iter().rev())
.any(|(a, b)| a != b)
}
}
#[test]
fn test() {
let s = "aab".to_string();
let res = 1;
assert_eq!(Solution::min_cut(s), res);
let s = "coder".to_string();
let res = 4;
assert_eq!(Solution::min_cut(s), res);
}