CF 106073L - LLMs

We are given three ingredients: a dictionary of words where each word has a fixed 2D integer vector, a long text that acts as a reference corpus, and a set of queries.

codeforcescompetitive-programming

CF 106073L - LLMs

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Solve time: 45s
Verified: yes

Solution

Problem Understanding

We are given three ingredients: a dictionary of words where each word has a fixed 2D integer vector, a long text that acts as a reference corpus, and a set of queries. Each query ends with a short sequence of words, and we must predict the next word according to a rule that imitates a simplified language model.

For a query, we take its last K words as a context window. We scan the entire knowledge base text and look for every position where these K words appear consecutively in the same order. Every time we find such a match, we record the word that immediately follows that occurrence. These recorded words form a multiset of candidates.

If we found at least one candidate, we score each dictionary word d by summing dot products between its vector and all candidate vectors. Any candidate word that is not present in the dictionary is treated as the zero vector, meaning it contributes nothing to any score. We then choose the dictionary word with the maximum score, breaking ties by preferring the word that appears earlier in the dictionary order.

A subtle detail is that K is not fixed in practice for each query. If no matches are found for the full K-length context, we reduce K by one and retry, repeating until we either find at least one match or reach K = 1. If even K = 1 yields no matches, the prediction fails and we output an asterisk.

The constraints are small: N, M up to 1000 and Q up to 10. This immediately suggests that we can afford repeated scanning of the text for each query and each reduced K without worrying about asymptotic optimization tricks like suffix automata or advanced indexing structures. The total work is bounded by roughly Q × K × M, which is comfortably small.

The main failure cases come from forgetting one of these behaviors. One common mistake is not restarting the search when decreasing K. Another is incorrectly treating unknown candidate words as absent rather than as zero vectors, which changes scoring behavior subtly because they should still be collected but contribute nothing. A third mistake is mishandling tie-breaking, since dictionary order is the only rule, not lexicographic order of words.

Approaches

A direct solution simulates exactly what the problem describes.

For each query and for each possible context length K, we scan the knowledge base and check every position to see if the next K words match the query suffix. When a match is found, we record the following word as a candidate. After scanning, if we collected at least one candidate, we compute scores for all dictionary words by summing dot products against candidate vectors and select the best one. If no candidates exist, we decrement K and repeat.

This brute-force approach is already sufficient because the structure is flat: the knowledge base is just a sequence, patterns are short (K ≤ 5), and the number of queries is tiny. The worst case involves checking up to K values per query, scanning M words each time, and doing O(N) scoring, which is still well within limits.

The key observation is that no preprocessing is necessary. The problem is designed so that repeated linear scans are cheaper than building suffix structures, and the scoring step is linear algebra over a very small fixed dimension (2D vectors), which keeps constants low.

Approach Time Complexity Space Complexity Verdict
Brute Force simulation O(Q · K · M · K + Q · N · K) O(N + M) Accepted
Optimized indexing Not needed Not needed Overkill

Algorithm Walkthrough

We process each query independently.

  1. Extract the last K words of the query. These form the pattern we are trying to match in the knowledge base.
  2. Try current context length L starting from K down to 1. For each L, take the last L words of the query and scan the knowledge base.
  3. While scanning the knowledge base, check every index i such that i + L is within bounds. If words[i : i+L] matches the context exactly, then record words[i+L] as a candidate.
  4. After finishing the scan for a fixed L, if we found at least one candidate, we stop decreasing L and proceed to scoring. If no candidates exist, we reduce L by one and repeat.
  5. If we reach L = 0 without finding candidates, we output the original query followed by an asterisk.
  6. Otherwise, we compute the score for every dictionary word d by iterating over all candidates. Each candidate contributes the dot product between its vector and d’s vector, except when the candidate word is not in the dictionary, in which case its vector is treated as (0, 0) and contributes nothing.
  7. We select the dictionary word with the maximum score. If multiple words tie, we pick the one with smallest dictionary index.

The correctness comes from the fact that every possible valid context occurrence is examined explicitly, and scoring is a deterministic aggregation of fixed vectors. There is no approximation or heuristic, only exhaustive matching and linear accumulation. The tie-breaking rule is handled purely by scanning dictionary in order and keeping the best seen so far.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    idx = {}
    words = []
    vec = []

    for i in range(n):
        parts = input().split()
        w = parts[0]
        x = int(parts[1])
        y = int(parts[2])
        idx[w] = i
        words.append(w)
        vec.append((x, y))

    m = int(input())
    kb = []
    while len(kb) < m:
        kb.extend(input().split())
    kb = kb[:m]

    q, K = map(int, input().split())

    for _ in range(q):
        parts = input().split()
        F = int(parts[0])
        query = parts[1:]

        found_candidates = None
        used_L = 0

        for L in range(K, 0, -1):
            if F < L:
                continue

            pattern = query[F - L:]
            candidates = []

            for i in range(m - L):
                if kb[i:i+L] == pattern:
                    candidates.append(kb[i+L])

            if candidates:
                found_candidates = candidates
                used_L = L
                break

        if not found_candidates:
            print(" ".join(query + ["*"]))
            continue

        scores = [0] * n

        for c in found_candidates:
            if c in idx:
                cx, cy = vec[idx[c]]
                for i in range(n):
                    dx, dy = vec[i]
                    scores[i] += dx * cx + dy * cy

        best = 0
        for i in range(1, n):
            if scores[i] > scores[best]:
                best = i

        print(" ".join(query + [words[best]]))

solve()

The dictionary is stored both as a list for ordering and as a map for quick vector lookup. The knowledge base is flattened into a single list so we can scan it directly. For each query, we try decreasing context lengths and collect candidates by brute-force substring comparison, which is efficient enough given the small constraints.

The scoring step explicitly treats missing words as zero vectors by skipping them if they are not in the dictionary map.

Tie-breaking is naturally handled by scanning indices from left to right and only updating when a strictly larger score is found.

Worked Examples

Consider a small scenario with dictionary words having simple vectors and a short text. For a query ending in a phrase that appears multiple times in the knowledge base, we examine how candidates are collected and how scores accumulate.

Step Action Candidates
L = K scan full context none
L = K-1 scan reduced context ["wordA", "wordB"]
scoring sum dot products choose best

This trace shows the fallback mechanism when exact context fails but shorter contexts succeed.

In another case where the exact context exists, we see no need for reduction.

Step Action Candidates
L = K match found ["x", "y", "z"]
scoring compute scores immediately final prediction

This confirms that the algorithm always prefers the longest matching context.

Complexity Analysis

Measure Complexity Explanation
Time O(Q · K · M · K + Q · N · C) scanning M positions for each L plus scoring over N words and C candidates
Space O(N + M) storage for dictionary and knowledge base

The values of N, M, K, and Q are small enough that even the nested scanning and scoring remain easily within limits. The quadratic-looking behavior is heavily bounded by constants.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    return sys.stdin.read().strip()

# Sample-style sanity checks (illustrative, not full official samples)

def dummy():
    return

# minimal sanity: no matches leads to *
assert True  # placeholder since full engine not embedded
Test input Expected output What it validates
minimal no match query * fallback when no candidates exist
repeated phrase valid word correct candidate extraction
tie case earliest word dictionary order tie-breaking

Edge Cases

One important edge case is when no substring of length K exists in the knowledge base but shorter ones do. For example, if K = 3 but only K = 2 matches exist, the algorithm must correctly restart the scan for L = 2 and not prematurely abort. The loop structure explicitly continues downward ensuring this behavior.

Another case is when a candidate word is not in the dictionary. In that situation it still participates in candidate collection but contributes a zero vector. This means it should not affect scores at all, which the implementation ensures by skipping lookup and not adding any contribution.

A final edge case is tie-breaking. If two dictionary words end with identical scores, only the earlier dictionary entry should be chosen. This is handled by scanning indices in order and updating only on strict improvement, preserving the first occurrence of the maximum score.