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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

Solution 1

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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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