Solution Explanation

This problem asks for the minimum number of steps to make a string empty by removing palindromic subsequences. The key observation is that the string consists only of 'a' and 'b' characters.

Understanding the Constraints:

Because the string only contains 'a' and 'b', there are only a few possibilities:

1. The string is already a palindrome: In this case, you can remove the entire string in one step.

2. The string is not a palindrome: You can always remove all the 'a's in one step (forming a palindromic subsequence of 'a's), and then remove all the 'b's in another step (forming a palindromic subsequence of 'b's). Therefore, at most two steps are required.

Algorithm:

The solution efficiently checks if the input string s is a palindrome. If it is, it returns 1 (one step needed). If it's not a palindrome, it returns 2 (two steps needed).

Time and Space Complexity:

• Time Complexity: O(n), where n is the length of the string. This is because the palindrome check takes linear time (we iterate through half the string).

• Space Complexity: O(1). The algorithm uses a constant amount of extra space, regardless of the input string size.

Code Explanation (Python):

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

This concise Python code directly implements the logic. s[::-1] reverses the string. If the reversed string is equal to the original string, it's a palindrome (returns 1); otherwise, it's not (returns 2).

Code Explanation (Other Languages):

The code in other languages (Java, C++, Go, TypeScript, Rust) all follow the same basic principle: they efficiently check if the input string is a palindrome and return 1 or 2 accordingly. They use a two-pointer approach (or equivalent) to iterate the string only once, further optimizing the time complexity. For instance, the Java code:

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;
    }
}

iterates from both ends of the string inwards. If it finds any characters that don't match, it immediately knows the string is not a palindrome and returns 2. If the loop completes without finding any mismatches, the string is a palindrome and 1 is returned. The other languages' implementations have analogous logic, optimized for that specific language's syntax.