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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.

Clarifying Questions

  1. What constitutes a valid palindromic subsequence?
    • A palindromic subsequence is a sequence that reads the same forwards and backwards. Since the string consists only of ‘a’ and ‘b’, subsequences like ‘a’, ‘b’, ‘aba’, and ‘bbb’ are palindromic.
  2. Do the characters in the string need to be adjacent for it to be considered a subsequence?
    • No, characters in a subsequence do not need to be adjacent.
  3. Examples for clarification:
    • Example 1:
      • Input: s = "ababa"
      • Output: 1
      • Explanation: The entire string is a palindrome, so it can be removed in one step.
    • Example 2:
      • Input: s = "abb"
      • Output: 2
      • Explanation: We can remove “bb” in one step (leaving ‘a’), and then remove ‘a’.

Strategy

  1. Check if the entire string is a palindrome:
    • If the string is a palindrome, we can remove it in one step.
  2. If not a palindrome:
    • Since the string consists of only ‘a’ and ‘b’, we can always remove all ‘a’s in one step and all ‘b’s in another step. Therefore, in this case, it will take 2 steps.

Time Complexity

Code

def removePalindromeSub(s: str) -> int:
    # Helper function to check if a string is a palindrome
    def is_palindrome(x):
        return x == x[::-1]

    # If the string is empty, no steps are needed
    if not s:
        return 0
    
    # If the string is a palindrome, we can remove it in one step
    if is_palindrome(s):
        return 1
    
    # Otherwise, we need two steps
    return 2

Explanation

  1. Helper Function is_palindrome: This function checks if the given string x is a palindrome.
  2. Empty String Check: If the input string s is empty, we return 0 because no steps are needed.
  3. Palindrome Check: If the string s is a palindrome, it can be removed in one step, so return 1.
  4. Otherwise: We need to remove the palindromic subsequences in two steps (all ‘a’s in one step and all ‘b’s in another), hence return 2.

This solution efficiently handles the problem within optimal time complexity.

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