Overview

This problem asks you to determine whether each search query string is contained as a “subsequence” of a book’s title, and count the number of bookshelves that have at least one matching book.

Analysis

What is a Subsequence?

A subsequence is a string obtained by removing zero or more characters from the original string while preserving the order of the remaining characters. Unlike a “substring,” the characters do not need to be contiguous.

For example, for the string "programming": - "pgm" → p…r…o…g…r…a…m can be found in order, so it is a subsequence ✓ - "prg" → p, r are found, and g appears after r, so it is a subsequence ✓ - "gp" → p does not appear after g, so it is not a subsequence ✗

Is a Naive Approach Sufficient?

Let’s check the constraints: - Number of bookshelves \(N \leq 100\) - Total number of books \(\sum K_i \leq 5000\) - Total characters in titles \(\leq 10^5\) - Number of queries \(Q \leq 200\) - Total characters in queries \(\leq 10^5\)

Even with a naive approach that checks the subsequence relationship between each query and every book title, a single check takes \(O(|S| + |T|)\). In the worst case we perform \(Q \times \sum K_i\) checks, but the cost of each check is the sum of the title length and query length, and the constraints are small enough that this is not a problem.

Algorithm

Subsequence Check (Greedy Method)

Whether string \(T\) is a subsequence of string \(S\) can be determined in \(O(|S|)\) using the following greedy method:

1. Scan \(S\) from the beginning, one character at a time
2. If the current character matches the “next unmatched character” of \(T\), advance to the next character of \(T\)
3. If all characters of \(T\) have been matched, then \(T\) is a subsequence

In Python, you can create an iterator with it = iter(s) and write all(c in it for c in t) for a concise implementation. Since c in it advances the iterator from its current position to search for c, the ordering is automatically preserved.

Overall Flow

1. Read the input and group book titles by bookshelf
2. For each query \(T_j\):
   • Scan all bookshelves, and if at least one book on a bookshelf contains \(T_j\) as a subsequence, count that bookshelf
   • Once a match is found, move on to the next bookshelf (early termination)
3. Output the count results

Complexity

• Time complexity: \(O\Bigl(Q \times \sum_{i,j} |S_{i,j}| + Q \times \sum_j |T_j|\Bigr)\)
  • Considering the worst case where all titles are scanned for each query, the cost per query is \(O(\text{total title characters} + |T|)\)
  • Under the given constraints, this is approximately \(200 \times 10^5 = 2 \times 10^7\), which is fast enough
• Space complexity: \(O\Bigl(\sum_{i,j} |S_{i,j}| + \sum_j |T_j|\Bigr)\) (for storing the input)

Implementation Notes

• Subsequence check using Python iterators: all(c in it for c in t) is extremely concise and correct. When the in operator is used on an iterator, it searches sequentially from the current position, so the ordering condition for subsequences is naturally satisfied.

• Early termination: By using break to move to the next bookshelf as soon as one book on a bookshelf matches, unnecessary checks are avoided.

• Bulk input reading: Reading all input at once with sys.stdin.buffer.read() and splitting into tokens is faster than processing input line by line.

  Source Code

import sys
from functools import lru_cache

def is_subsequence(t, s):
    """Check if t is a subsequence of s"""
    it = iter(s)
    return all(c in it for c in t)

def main():
    input_data = sys.stdin.buffer.read().decode()
    tokens = input_data.split()
    idx = 0
    
    N = int(tokens[idx]); idx += 1
    
    shelves = []
    for i in range(N):
        K = int(tokens[idx]); idx += 1
        books = []
        for j in range(K):
            books.append(tokens[idx]); idx += 1
        shelves.append(books)
    
    Q = int(tokens[idx]); idx += 1
    
    results = []
    for _ in range(Q):
        query = tokens[idx]; idx += 1
        count = 0
        for books in shelves:
            found = False
            for title in books:
                if is_subsequence(query, title):
                    found = True
                    break
            if found:
                count += 1
        results.append(count)
    
    print('\n'.join(map(str, results)))

if __name__ == '__main__':
    main()

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