# 1332. Remove Palindromic Subsequences

## Description

You are given a string s consisting only of letters 'a' and 'b'. In a single step you can remove one palindromic subsequence from s.

Return the minimum number of steps to make the given string empty.

A string is a subsequence of a given string if it is generated by deleting some characters of a given string without changing its order. Note that a subsequence does not necessarily need to be contiguous.

A string is called palindrome if is one that reads the same backward as well as forward.

Example 1:

Input: s = "ababa"
Output: 1
Explanation: s is already a palindrome, so its entirety can be removed in a single step.


Example 2:

Input: s = "abb"
Output: 2
Explanation: "abb" -> "bb" -> "".
Remove palindromic subsequence "a" then "bb".


Example 3:

Input: s = "baabb"
Output: 2
Explanation: "baabb" -> "b" -> "".
Remove palindromic subsequence "baab" then "b".


Constraints:

• 1 <= s.length <= 1000
• s[i] is either 'a' or 'b'.

## Solutions

• class Solution {
public int removePalindromeSub(String s) {
for (int i = 0, j = s.length() - 1; i < j; ++i, --j) {
if (s.charAt(i) != s.charAt(j)) {
return 2;
}
}
return 1;
}
}

• class Solution {
public:
int removePalindromeSub(string s) {
for (int i = 0, j = s.size() - 1; i < j; ++i, --j) {
if (s[i] != s[j]) {
return 2;
}
}
return 1;
}
};

• class Solution:
def removePalindromeSub(self, s: str) -> int:
return 1 if s[::-1] == s else 2


• func removePalindromeSub(s string) int {
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
if s[i] != s[j] {
return 2
}
}
return 1
}

• function removePalindromeSub(s: string): number {
for (let i = 0, j = s.length - 1; i < j; ++i, --j) {
if (s[i] !== s[j]) {
return 2;
}
}
return 1;
}


• impl Solution {
pub fn remove_palindrome_sub(s: String) -> i32 {
let mut l = 0;
let mut r = s.len() - 1;
let s: Vec<char> = s.chars().collect();
while l < r {
if s[l] != s[r] {
return 2;
}
l += 1;
r -= 1;
}
1
}
}