1930. Unique Length-3 Palindromic Subsequences

Problem

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

/**
 * @param {string} s
 * @return {number}
 */
var countPalindromicSubsequence = function(s) {
    var a = 'a'.charCodeAt(0);
    var res = 0;
    for (var i = 0; i < 26; i++) {
        var char = String.fromCharCode(i + a);
        var first = s.indexOf(char);
        var last = s.lastIndexOf(char);
        if (last - first < 2) continue;
        var map = {};
        for (var j = first + 1; j < last; j++) {
            if (!map[s[j]]) {
                res += 1;
                map[s[j]] = true;
            }
        }
    }
    return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).