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Formatted question description: https://leetcode.ca/all/1332.html

# 1332. Remove Palindromic Subsequences (Easy)

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.

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

Example 1:

Input: s = "ababa"
Output: 1


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".


Example 4:

Input: s = ""
Output: 0


Constraints:

• 0 <= s.length <= 1000
• s only consists of letters 'a' and 'b'

Related Topics:
String

## Solution 1.

• class Solution {
public int removePalindromeSub(String s) {
if (s == null || s.length() == 0)
return 0;
int low = 0, high = s.length() - 1;
while (low < high) {
if (s.charAt(low) == s.charAt(high)) {
low++;
high--;
} else
return 2;
}
return 1;
}
}

############

class Solution {
public int removePalindromeSub(String s) {
if (s.length() == 0) {
return 0;
}
if (new StringBuilder(s).reverse().toString().equals(s)) {
return 1;
}
return 2;
}
}

• // OJ: https://leetcode.com/problems/remove-palindromic-subsequences/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int removePalindromeSub(string s) {
if (s.empty()) return 0;
bool isPalindrome = true;
for (int i = 0, j = s.size() - 1; i < j && isPalindrome; ++i, --j) {
isPalindrome = s[i] == s[j];
}
return isPalindrome ? 1 : 2;
}
};

• class Solution:
def removePalindromeSub(self, s: str) -> int:
if not s:
return 0
if s[::-1] == s:
return 1
return 2


• func removePalindromeSub(s string) int {
if len(s) == 0 {
return 0
}
if s == reverse(s) {
return 1
}
return 2
}

func reverse(s string) string {
r := []byte(s)
for i, j := 0, len(r)-1; i < j; i, j = i+1, j-1 {
r[i], r[j] = r[j], r[i]
}
return string(r)
}

• function removePalindromeSub(s: string): number {
if (s.length == 0) return 0;
if (s == s.split('').reverse().join('')) return 1;
return 2;
}


• 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
}
}