# Solving Leetcode Interviews in Seconds with AI: Unique Length-3 Palindromic Subsequences


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1930" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given a string s, return the number of unique palindromes of length three that are a subsequence of s. Note that even if there are multiple ways to obtain the same subsequence, it is still only counted once. A palindrome is a string that reads the same forwards and backwards. A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.  For example, "ace" is a subsequence of "abcde".    Example 1:  Input: s = "aabca" Output: 3 Explanation: The 3 palindromic subsequences of length 3 are: - "aba" (subsequence of "aabca") - "aaa" (subsequence of "aabca") - "aca" (subsequence of "aabca")  Example 2:  Input: s = "adc" Output: 0 Explanation: There are no palindromic subsequences of length 3 in "adc".  Example 3:  Input: s = "bbcbaba" Output: 4 Explanation: The 4 palindromic subsequences of length 3 are: - "bbb" (subsequence of "bbcbaba") - "bcb" (subsequence of "bbcbaba") - "bab" (subsequence of "bbcbaba") - "aba" (subsequence of "bbcbaba")    Constraints:  3 <= s.length <= 105 s consists of only lowercase English letters.  

	# Explanation
	Here's a breakdown of the solution and the code:

*   **High-Level Approach:**
    *   Iterate through all possible middle characters (a to z).
    *   For each middle character, find the first and last occurrence in the string.
    *   Count the unique characters between the first and last occurrences of the middle character. These represent the outer characters of the palindrome.

*   **Complexity:**
    *   Runtime: O(26 \* n) which simplifies to O(n), where n is the length of the string.
    *   Space: O(1) - Constant space used to store the unique characters.

	
	# Code
	```python
	def count_palindromic_subsequences(s: str) -> int:
    """
    Given a string s, return the number of unique palindromes of length three that are a subsequence of s.
    """
    count = 0
    for char_code in range(ord('a'), ord('z') + 1):
        char = chr(char_code)
        first_occurrence = -1
        last_occurrence = -1

        for i in range(len(s)):
            if s[i] == char:
                if first_occurrence == -1:
                    first_occurrence = i
                last_occurrence = i

        if first_occurrence != -1 and last_occurrence != first_occurrence:
            unique_chars = set()
            for i in range(first_occurrence + 1, last_occurrence):
                unique_chars.add(s[i])
            count += len(unique_chars)

    return count
	```