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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
Explanation: String is already palindrome

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
    }
}

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