Leetcode/1930. Unique Length-3 Palindromic Subsequences.java at main · chakrabortyrajarshi2005/Leetcode · GitHub

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1930. Unique Length-3 Palindromic Subsequences
Solved
Medium
Topics
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Hint
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.class Solution {
    public int countPalindromicSubsequence(String s) {
        Set<Character> letters = new HashSet();
        for (Character c: s.toCharArray()) {
            letters.add(c);
        }

        int ans = 0;
        for (Character letter : letters) {
            int i = -1;
            int j = 0;

            for (int k = 0; k < s.length(); k++) {
                if (s.charAt(k) == letter) {
                    if (i == -1) {
                        i = k;
                    }

                    j = k;
                }
            }

            Set<Character> between = new HashSet();
            for (int k = i + 1; k < j; k++) {
                between.add(s.charAt(k));
            }

            ans += between.size();
        }

        return ans;
    }
}