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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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  }
  

• 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;
      }
  };
  

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

• 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
  }
  

• 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;
  }
  
  

• 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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