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Leetcode 1332. Remove Palindromic Subsequences

Problem Statement

We are given a string s consisting only of characters ‘a’ and ‘b’. A subsequence is a sequence that can be derived from s by deleting some or no characters without changing the order of the remaining characters.

The task is to remove all characters from the string s using the minimum number of operations, where in each operation you can choose a palindromic subsequence and remove it from s.

Clarifying Questions

  1. What is a palindromic subsequence?
    • A palindromic subsequence is a subsequence that reads the same backward as forward. For example, “aba” is a palindromic subsequence.
  2. Can the string s be empty?
    • Yes, the string s can be empty. If s is empty, it doesn’t need any operations since it’s already empty.
  3. What is the length limit of s?
    • The typical constraint for problems like this on LeetCode is 1≤ s ≤1000, where s is the length of the string.

With these understandings, we can move towards solving the problem.

Strategy

  1. Check if the string is empty: If s is empty, return 0 since no operations are needed.

  2. Check if the string is a palindrome: If the string s is a palindrome, you can remove the entire string in one operation because the whole string itself is a palindromic subsequence.

  3. General Case: If s is not a palindrome, we can always remove all ‘a’s in one operation and all ‘b’s in another operation. Thus, the answer will be 2.

Code

#include <string>

class Solution {
public:
    int removePalindromeSub(std::string s) {
        if (s.empty()) {
            return 0;
        }
        if (isPalindrome(s)) {
            return 1;
        }
        return 2;
    }

private:
    bool isPalindrome(const std::string& s) {
        int left = 0, right = s.size() - 1;
        while (left < right) {
            if (s[left] != s[right]) {
                return false;
            }
            ++left;
            --right;
        }
        return true;
    }
};

Time Complexity

  1. Checking if a string is a palindrome: This operation takes O(n) time where n is the length of the string s since we need to compare each character with its corresponding character from the end.

  2. Overall Complexity: The overall time complexity of the removePalindromeSub function is O(n) due to the palindrome check.

This solution ensures we efficiently tackle the problem with optimal operations based on whether the string is a palindrome or not.

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