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.

// 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;
    }
};

Java

  • 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;
        }
    }
    
  • // 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;
        }
    };
    
  • print("Todo!")
    

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