2059. Unique Length 3 Palindromic Subsequences¶

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2059. Unique Length-3 Palindromic Subsequences

Medium

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.

Solution¶

class Solution {
    public int countPalindromicSubsequence(String s) {
        int n = s.length(), res = 0;
        for (char c = 'a'; c <= 'z'; c++) {
            int first = s.indexOf(c);
            int last = s.lastIndexOf(c);
            if (first != -1 && last != -1 && first < last) {
                Set<Character> temp = new HashSet<>();
                for (int i = first + 1; i < last; i++) temp.add(s.charAt(i));
                res += temp.size();
            }
        }
        return res;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here