0962MediumHash TableStringBit ManipulationPrefix Sum

Three-Letter Mirrored Picks

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Trains the technique from

LeetCode 1930Unique Length-3 Palindromic Subsequences

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A tape reads s, a string of lowercase letters. A pick takes three letters from it, keeping their order, and the pick is mirrored when its first and last letters are the same.

Return how many different mirrored picks the tape holds, counting two picks as the same when they read the same.

Examples

Example 1

Input: s = "bcabcaba"
Output: 8

Every one of a, b and c appears at least twice, and the distinct letters lying between the first and last place of each supply the middles, which comes to this many different picks.

Example 2

Input: s = "abc"
Output: 0

No letter appears twice, so no pick can have matching ends.

Example 3

Input: s = "aaa"
Output: 1

The only pick is "aaa".

Constraints

3 <= s.length <= 10^5
s consists of lowercase English letters only

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_palindromic_subsequence(s: str) -> int:

Java

public int countPalindromicSubsequence(String s)

The intended approach

Time O(A * n) for a tape of length n over an alphabet of size A.Space O(A) for the first and last places.

A mirrored pick is settled by two letters: the one at both ends and the one in the middle. So for each of the twenty-six letters that could sit at the ends, the middles available are exactly the distinct letters appearing strictly between that letter's first and last place on the tape, because any pair of that letter can be widened out to the first and last without losing any middle. So find each letter's first and last place in one pass, then for each letter that appears at least twice, count the distinct letters in the stretch between them. Adding those counts over the letters gives the answer, and the whole thing is a bounded number of passes over the tape.

Pythonentry point: count_palindromic_subsequence

def count_palindromic_subsequence(s):
    n = len(s)
    first = {}
    last = {}
    for i in range(n):
        ch = s[i]
        if ch not in first:
            first[ch] = i
        last[ch] = i
    total = 0
    for ch in first:
        lo = first[ch]
        hi = last[ch]
        if hi - lo < 2:
            continue
        middles = set()
        for i in range(lo + 1, hi):
            middles.add(s[i])
        total += len(middles)
    return total

The obvious version

Take every three positions in order, keep the picks whose outer letters match, and collect the resulting three-letter strings in a set. Cubic in the tape length and it reasons about nothing.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_palindromic_subsequence(s):
    n = len(s)
    seen = set()
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if s[i] == s[k]:
                    seen.add(s[i] + s[j] + s[k])
    return len(seen)

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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