1930. Unique Length-3 Palindromic Subsequences
Description
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.
Solutions
Solution: Hash Table
- Time complexity: O(n*26)
- Space complexity: O(26)
JavaScript
js
/**
* @param {string} s
* @return {number}
*/
const countPalindromicSubsequence = function (s) {
const BASE_CHAR_CODE = 'a'.charCodeAt(0);
let result = 0;
for (let code = 0; code < 26; code++) {
const char = String.fromCharCode(BASE_CHAR_CODE + code);
const start = s.indexOf(char);
if (start === -1) continue;
const end = s.lastIndexOf(char);
if (start >= end) continue;
const isVisited = Array.from({length: 26}).fill(false);
let count = 0;
for (let index = start + 1; index < end; index++) {
const charCode = s.charCodeAt(index) - BASE_CHAR_CODE;
if (isVisited[charCode]) continue;
count += 1;
isVisited[charCode] = true;
}
result += count;
}
return result;
};