1312. Minimum Insertion Steps to Make a String Palindrome

Given a string `s`. In one step you can insert any character at any index of the string.

Return the minimum number of steps to make `s` palindrome.

Palindrome String is one that reads the same backward as well as forward.

Example 1:

```Input: s = "zzazz"
Output: 0
Explanation: The string "zzazz" is already palindrome we don't need any insertions.
```

Example 2:

```Input: s = "mbadm"
Output: 2
Explanation: String can be "mbdadbm" or "mdbabdm".
```

Example 3:

```Input: s = "leetcode"
Output: 5
Explanation: Inserting 5 characters the string becomes "leetcodocteel".
```

Example 4:

```Input: s = "g"
Output: 0
```

Example 5:

```Input: s = "no"
Output: 1
```

Constraints:

• `1 <= s.length <= 500`
• All characters of `s` are lower case English letters.

1312. Minimum Insertion Steps to Make a String Palindrome
``````struct Solution;

use std::collections::HashMap;

impl Solution {
fn min_insertions(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 {
let n = end - start;
if n < 2 {
return 0;
}
if let Some(&res) = memo.get(&(start, end)) {
return res;
}
let res = if s[start] == s[end - 1] {
Self::dp(start + 1, end - 1, memo, s)
} else {
1 + Self::dp(start, end - 1, memo, s).min(Self::dp(start + 1, end, memo, s))
};
memo.insert((start, end), res);
res
}
}

#[test]
fn test() {
let s = "zzazz".to_string();
let res = 0;
assert_eq!(Solution::min_insertions(s), res);
let s = "mbadm".to_string();
let res = 2;
assert_eq!(Solution::min_insertions(s), res);
let s = "leetcode".to_string();
let res = 5;
assert_eq!(Solution::min_insertions(s), res);
let s = "g".to_string();
let res = 0;
assert_eq!(Solution::min_insertions(s), res);
let s = "no".to_string();
let res = 1;
assert_eq!(Solution::min_insertions(s), res);
}
``````