How do you solve Remove Palindromic Subsequences on LeetCode?

Remove Palindromic Subsequences (LeetCode #1332) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: A string with only two characters can be efficiently processed by removing all instances of one character type in one step.

What is the brute force solution for Remove Palindromic Subsequences?

The brute force approach for Remove Palindromic Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking if the entire string is a palindrome. If it is, we can remove it in one step. If not, we can remove individual characters or groups of characters until the string is empty. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove Palindromic Subsequences?

Remove Palindromic Subsequences uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Remove Palindromic Subsequences?

Always clarify the constraints and properties of the input string before diving into the solution. Consider edge cases, such as strings that are already palindromes or consist of a single character. Practice recognizing patterns in problems, as this can often lead to simpler and more efficient solutions.

What are common mistakes when solving Remove Palindromic Subsequences?

Students often forget to check if the string is a palindrome before deciding the number of steps. Some may try to overcomplicate the solution by attempting to remove characters one by one instead of recognizing the pattern.

#1332

Remove Palindromic Subsequences

Easy

Two PointersStringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Since the string consists only of 'a' and 'b', we can always remove all characters in at most 2 steps: one for all 'a's and one for all 'b's. This is efficient because we don't need to check for palindromes.

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Algorithm

2 steps

1Step 1: Return 1 if the string is a palindrome.
2Step 2: If not, return 2 because we can remove all 'a's in one step and all 'b's in another.

solution.py2 lines

1def removePalindromeSub(s):
2    return 1 if s == s[::-1] else 2

ℹ

Complexity note: The time complexity is O(n) because we only need to check if the string is a palindrome once. The space complexity is O(1) since we are using a constant amount of space.

1A string with only two characters can be efficiently processed by removing all instances of one character type in one step.
2Checking for palindromes can be done in linear time, making it feasible to determine the number of steps needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.