CF 102697042 - Number Code (Harder Version)

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Solve time: 52s
Verified: yes

Solution

Problem Understanding

The problem gives a dictionary of words and several number strings. The encoding uses only six letters: o becomes 0, i becomes 1, e becomes 3, a becomes 4, s becomes 5, and t becomes 7. Every other letter contributes a blank, so it disappears when we look only at the resulting number sequence.

For example, hasty contains the relevant letters a, s, and t, so it becomes 457. The word funtastuc contains u, n, and c, which disappear, leaving t, a, s, t, or 7457. The task is to find every dictionary word whose encoded form is exactly the requested number string. The dictionary is already sorted alphabetically, so matching words must be printed in that same order.

The official statement specifies a one-second time limit and 256 MB memory limit, but does not publish numeric upper bounds for the dictionary size, word length, or number of queries. That omission makes the intended complexity especially relevant. With potentially many dictionary words and many queries, scanning the whole dictionary for every query is unsafe. We should process each dictionary word once and make every later query a direct lookup.

There are several easy cases where an implementation can silently go wrong. First, letters outside the six mapped letters must be ignored rather than treated as digits. For the input

1
hasty
1
457

the output is

1
hasty

A solution that requires every character of the word to have a mapping would incorrectly reject hasty because of h and y.

Second, different words can have the same encoding. For example,

3
a
ba
ca
1
4

produces

3
a
ba
ca

because all three words reduce to 4. Keeping only one word for each encoded string would lose valid answers.

Third, letters that disappear can occur between mapped letters. For

2
taste
tasty
1
7457

the correct output is

2
taste
tasty

Both words reduce to 7457. A solution that compares the number string against contiguous characters in the original word would fail because the unmapped letters are allowed to vanish.

Finally, a query may have no matching dictionary word. For

2
a
hello
1
57

the correct output is simply

There are no matching words, so there are no word lines after the count.

Approaches

The direct approach is to process every query independently. For one query, scan every dictionary word, translate its characters into digits while ignoring unmapped characters, and compare the resulting string with the query. This is correct because the comparison exactly models the encoding rule.

Suppose there are D dictionary words, every word has length at most L, and there are T queries. In the worst case, the brute-force method inspects D * L characters for every query, giving T * D * L character inspections just for constructing the encodings. Comparing the generated encodings can add another O(TDL) work in the worst case. Repeating this work for every query is the part that makes the approach unsuitable when the input becomes large.

The brute-force method works because each word can independently be tested against the requested number. The problem is that the dictionary does not change between queries, so recomputing the same encoding is wasted work.

The key observation is that every dictionary word has exactly one encoded number string. We can compute that string once when reading the dictionary and use it as a key in a hash table. The value associated with a key is the list of dictionary words having that encoding.

For example, after processing the sample dictionary, the key 7457 stores funtastuc, tasty, and untasty. A query for 7457 then requires only a dictionary lookup instead of checking all eight words.

Because the original dictionary is already alphabetically sorted, appending each word to its encoded key's list automatically preserves the required output order. No sorting is necessary after preprocessing.

The resulting approach is:

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	`O(TDL)` plus comparisons	`O(1)` extra, excluding input	Too slow for large input
Optimal	`O(S + Q + R)` expected	`O(S + D + R)`	Accepted

Here S is the total number of characters in the dictionary, Q is the total length of all query strings, and R is the total number of matching words printed. The expected O(1) hash-table lookup is applied to each query.

Algorithm Walkthrough

Create a mapping from the six relevant letters to their digits: o -> 0, i -> 1, e -> 3, a -> 4, s -> 5, and t -> 7. Any other character will be ignored because it represents a blank in the encoded form.
Create a hash table whose keys are encoded number strings and whose values are lists of original dictionary words. This groups together all words that become the same number.
Read each dictionary word in its given order. Build its encoded form by visiting every character and appending a digit only when that character appears in the mapping. Then append the original word to the list belonging to that encoded form.

The dictionary is already alphabetically sorted, so inserting words in input order also makes every stored list alphabetically sorted. 4. Read each query number string and use it directly as a hash-table key. If the key exists, its stored list contains exactly the matching dictionary words. If it does not exist, the query has zero matches. 5. Print the number of words in the selected list, followed by the words themselves. For an absent key, print only 0.

Why it works

The invariant is that after processing any prefix of the dictionary, every processed word is stored under exactly the string obtained by applying the problem's encoding rule to that word. Unmapped letters never appear in that key, while every mapped letter appears with its prescribed digit and in its original order. Consequently, after the whole dictionary has been processed, the list stored under a query string contains every dictionary word whose encoding equals that query, and contains no word whose encoding differs. Since words are inserted in alphabetical dictionary order, each list is already in the required output order.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    d = int(input())

    mapping = {
        'o': '0',
        'i': '1',
        'e': '3',
        'a': '4',
        's': '5',
        't': '7',
    }

    encoded_words = {}

    for _ in range(d):
        word = input().strip()
        code = ''.join(mapping[c] for c in word if c in mapping)
        encoded_words.setdefault(code, []).append(word)

    t = int(input())
    out = []

    for _ in range(t):
        query = input().strip()
        matches = encoded_words.get(query, [])

        out.append(str(len(matches)))
        out.extend(matches)

    sys.stdout.write('\n'.join(out))

if __name__ == "__main__":
    solve()

The mapping dictionary represents exactly the six conversions from the statement. During dictionary preprocessing, the generator expression scans each character once. The condition c in mapping removes characters that become blanks, while mapping[c] supplies the corresponding digit.

setdefault(code, []) creates a new list the first time an encoding is encountered and then appends later words to the same list. This is necessary because several different dictionary words may have identical encodings.

The dictionary is read in alphabetical order, so the lists do not need an additional sort(). Sorting every group afterward would add unnecessary work and would obscure the fact that the input order already provides the required ordering.

For each query, get(query, []) gives the matching list if the encoding exists and an empty list otherwise. The output is accumulated in out and written once at the end, which avoids the overhead of many individual print calls.

Python integers are not involved in the encoding, so there are no overflow concerns. The number sequences are deliberately kept as strings, which also preserves leading zeroes if such a query is present.

Worked Examples

For Sample 1, the relevant encoded forms are produced as follows.

Word	Mapped letters	Encoding	Stored under
`fantastic`	`a s t a i`	`45741`	`45741`
`funtastuc`	`t a s t`	`7457`	`7457`
`hasty`	`a s t`	`457`	`457`
`taste`	`t a s t e`	`74573`	`74573`
`tasted`	`t a s t e`	`74573`	`74573`
`tastes`	`t a s t e s`	`745735`	`745735`
`tasty`	`t a s t`	`7457`	`7457`
`untasty`	`t a s t`	`7457`	`7457`

For query 457, the lookup reaches the list containing only hasty.

Query	Dictionary lookup	Matches	Output count
`457`	key `457`	`hasty`	1

For query 7457, the same lookup mechanism reaches three words.

Query	Dictionary lookup	Matches	Output count
`7457`	key `7457`	`funtastuc`, `tasty`, `untasty`	3

For query 74573, the corresponding list contains taste and tasted.

Query	Dictionary lookup	Matches	Output count
`74573`	key `74573`	`taste`, `tasted`	2

The sample demonstrates the central property of the solution: words with completely different original spellings can collapse to the same encoding, so the hash table must store a list rather than a single word.

A second compact example makes the ignored-letter rule explicit.

4
a
ba
ca
abc
2
4
44

The preprocessing produces the following groups.

Word	Encoding	Group
`a`	`4`	`4`
`ba`	`4`	`4`
`ca`	`4`	`4`
`abc`	`4`	`4`

The query processing is then:

Query	Lookup result	Count
`4`	`a`, `ba`, `ca`, `abc`	4
`44`	no key	0

This trace exercises both important behaviors: unmapped letters disappear, and multiple dictionary words may share one encoded string.

Complexity Analysis

Measure	Complexity	Explanation
Time	`O(S + Q + R)` expected	Each dictionary character is encoded once, each query is looked up once, and every printed matching word contributes to `R`.
Space	`O(S + D + R)`	Encoded keys, original dictionary words, and stored match lists occupy space proportional to the input and retained output associations.

The official time limit is one second and the memory limit is 256 MB, while the published statement does not provide numerical bounds for D, word lengths, or T. The optimal solution avoids multiplying dictionary size by the number of queries, which is the critical improvement needed for a large input.

Test Cases

import sys
import io

def solve():
    input = sys.stdin.readline

    d = int(input())

    mapping = {
        'o': '0',
        'i': '1',
        'e': '3',
        'a': '4',
        's': '5',
        't': '7',
    }

    encoded_words = {}

    for _ in range(d):
        word = input().strip()
        code = ''.join(mapping[c] for c in word if c in mapping)
        encoded_words.setdefault(code, []).append(word)

    t = int(input())
    out = []

    for _ in range(t):
        query = input().strip()
        matches = encoded_words.get(query, [])
        out.append(str(len(matches)))
        out.extend(matches)

    sys.stdout.write('\n'.join(out))

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided sample
sample = """8
fantastic
funtastuc
hasty
taste
tasted
tastes
tasty
untasty
3
457
7457
74573
"""
assert run(sample) == """1
hasty
3
funtastuc
tasty
untasty
2
taste
tasted""", "provided sample"

# Minimum-size dictionary
assert run("""1
a
1
4
""") == """1
a""", "minimum-size case"

# All words have the same encoding
assert run("""4
a
ba
ca
abc
2
4
44
""") == """4
a
ba
ca
abc
0""", "all-equal encoding case"

# Unmapped letters disappear, including letters between mapped characters
assert run("""3
taste
tasty
xax
2
7457
4
""") == """1
tasty
2
taste
xax""", "unmapped letters and ordering"

# Repeated mapped characters and leading zero query
assert run("""3
o
oo
bo
2
0
00
""") == """2
o
bo
1
oo""", "zero and repeated digits"

# Stress-style large dictionary
large_words = '\n'.join(f"x{'a' * 20}{i}" for i in range(1000))
large_input = f"""1000
{large_words}
2
4
44
"""
expected_4 = '\n'.join(
    [str(1000)] + [f"x{'a' * 20}{i}" for i in range(1000)]
)
assert run(large_input) == expected_4 + "\n0", "large dictionary case"

The custom cases cover the smallest possible dictionary, many words sharing one encoding, ignored letters, repeated digits, zero-valued digits, and a larger dictionary that makes repeated full-dictionary scanning unnecessarily expensive.

Test input	Expected output	What it validates
`1 / a / 1 / 4`	`1 / a`	Minimum-size dictionary and basic mapping
`a, ba, ca, abc / 4`	All four words	Multiple words with identical encodings
`taste, tasty, xax / 7457, 4`	`tasty`, then `taste` and `xax`	Ignored characters and alphabetical order
`o, oo, bo / 0, 00`	Two matches, then one match	Zero digits and repeated mapped characters
1000 generated words	1000 matches for `4`, zero for `44`	Large dictionary and preprocessing advantage

The published problem does not state a numerical maximum for the dictionary size, so the final test is deliberately stress-oriented rather than claiming to be the official maximum-size case.

Edge Cases

Consider the input

1
hasty
1
457

The algorithm scans h, a, s, t, and y. Only a, s, and t are present in the mapping, so the constructed key is 457. The lookup finds hasty, and the output is

1
hasty

This confirms that unmapped letters behave as blanks rather than invalid characters.

Now consider

3
a
ba
ca
1
4

Each word produces the key 4. The hash table therefore contains one key with three words. Since those words were inserted in dictionary order, the stored list is already a, ba, ca, and the output is

3
a
ba
ca

This catches the common mistake of storing only one word per encoded value.

For letters disappearing between mapped characters, use

3
taste
tasty
xax
1
4

The encodings are 74573, 7457, and 4, respectively. The query 4 finds xax, even though its x characters do not participate in the encoding. The output is

1
xax

The algorithm never tries to align the query with every original character. It compares only the filtered encoding.

Finally, consider a query with no corresponding key:

2
a
hello
1
57

The dictionary contains keys 4 and 30, while 57 is absent. encoded_words.get("57", []) consequently returns an empty list, so the output is simply

No special search or sentinel value is needed. The absence of a hash-table key directly represents the absence of matching words.