README_EN.md

1332. Remove Palindromic Subsequences

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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

Python3

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

Java

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;
    }
}

TypeScript

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

C++

class Solution {
public:
    int removePalindromeSub(string s) {
        if (s.empty())
            return 0;
        string t = s;
        reverse(s.begin(), s.end());
        if (s == t)
            return 1;
        return 2;
    }
};

Go

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)
}

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