CF 102697039 - Number Code (Easier Version)

CF 102697039 - Number Code (Easier Version)

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Solve time: 53s
Verified: yes

Solution

Problem Understanding

The problem defines a fixed substitution code between six digits and six letters:

0 represents o, 1 represents i, 3 represents e, 4 represents a, 5 represents s, and 7 represents t.

Each test case is a sequence of space-separated number strings. Every number string represents one word, and every digit in that string can be translated independently using the code. Since the easier version guarantees that the original words contain only letters represented by the code, there is no ambiguity. We simply decode every number string and preserve the spaces between them.

For example, 74573 becomes taste: 7 -> t, 4 -> a, 5 -> s, 7 -> t, and 3 -> e.

The official limits are a 1 second time limit and 256 MB of memory, but the statement does not give numerical bounds for the number of test cases or the length of each sequence. The safe interpretation is that the algorithm should be linear in the amount of input. Any approach that explores possible words instead of directly translating the supplied digits can become exponentially more expensive as a number string grows. A direct scan of every input character is optimal because no character needs to be examined more than once. The problem page confirms the six mappings, the input format, and the 1 second, 256 MB limits.

There are a few small cases that can expose careless implementations. A single digit is already a complete word. For input

1
0

the correct output is

o

An implementation that assumes every word contains at least two digits would fail here.

A word can contain repeated digits. For input

1
55 555

the correct output is

ss sss

A careless implementation that converts only distinct digits, or accidentally removes repeated characters, would produce the wrong words.

The digit 0 is also significant because it decodes to the letter o. For input

1
101

the correct output is

ioi

Treating the number as an integer instead of as a string can silently remove leading zeroes. For example, 010 must decode to oio, not io.

Finally, spaces separate words and must remain spaces in the output. For input

1
50 50

the correct output is

so so

Decoding the entire line character by character without treating spaces separately can accidentally concatenate the two words.

Approaches

The most literal brute-force approach would be to treat each encoded number as an unknown word and try all possible words over the six allowed letters. A number with length L would have 6^L possible words. We could encode each candidate word and check whether its encoding matches the given number. This is correct because the original word must consist only of the six letters represented by the code, so enumerating all such words eventually includes the answer.

The problem is that this search grows exponentially. A single 20-digit number would already have 6^20 = 3,656,158,440,062,976 possible candidate words. Even tiny input lines become impossible under a 1 second limit. The brute force is solving a much harder problem than the input actually asks us to solve.

The key observation is that the mapping is one-to-one. Every possible input digit has exactly one corresponding letter, and the statement guarantees that no other letters occur in the original words. There is consequently no need to guess anything. When we read a 5, the corresponding letter has to be s; when we read a 7, it has to be t. Every character has exactly one answer.

So the optimal algorithm scans each input line from left to right. A digit is replaced using a six-entry mapping, while a space is copied unchanged. Equivalently, we can split the line into number strings, decode each string independently, and join the decoded words with spaces. Either implementation takes constant work per input character.

The brute-force approach works because it considers every possible word, but fails because the number of possible words is exponential. The observation that every encoded character has exactly one possible letter removes the search entirely and reduces the problem to a direct substitution.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(6^L) for a word of length L	O(L) per candidate	Too slow
Optimal	O(N)	O(N) for the output	Accepted

Here N is the total number of characters in the input sequences. The exponential bound for brute force applies to the longest individual encoded word, while the optimal method processes the complete input linearly.

Algorithm Walkthrough

1. Read the number of test cases. Each following line represents one encoded sentence, so the complete line must be preserved rather than reading only one number.
2. Store the six digit-to-letter mappings in a dictionary, with 0 -> o, 1 -> i, 3 -> e, 4 -> a, 5 -> s, and 7 -> t. A dictionary makes each translation a constant-time lookup.
3. For each test case, read the entire line and remove only its trailing newline. Keeping the internal spaces is necessary because they separate words.
4. Scan every character in the line. If it is a space, append the space directly. Otherwise, look up the corresponding letter in the mapping and append that letter.
5. Print the resulting decoded line. Since every input character has exactly one translation, the output is uniquely determined.

Why it works

The invariant is that after processing any prefix of an input line, the constructed output is exactly the decoded version of that prefix. A space is copied unchanged, so word boundaries are preserved. For every allowed digit, the mapping gives exactly the letter specified by the encoding rule, so every decoded character is correct. Extending this invariant one character at a time means that after the final character, the complete output line is exactly the required sequence of decoded words.

Python Solution

import sys
input = sys.stdin.readline

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

    t = int(input())

    for _ in range(t):
        line = input().rstrip('\n')
        decoded = []

        for ch in line:
            if ch == ' ':
                decoded.append(' ')
            else:
                decoded.append(mapping[ch])

        print(''.join(decoded))

if __name__ == "__main__":
    solve()

The mapping dictionary represents the code directly, so there is no arithmetic conversion or special case for any of the six digits.

The program reads an entire test case with input() rather than repeatedly reading individual tokens. This preserves the spaces between words. rstrip('\n') removes the line ending while leaving every meaningful space untouched.

The loop then applies exactly the transformation described in the algorithm. Spaces are copied, while digits are translated through the dictionary. The input guarantees that every non-space character belongs to the six allowed digits, so no fallback case is necessary.

The output is assembled in a list and joined once. This avoids repeatedly constructing larger strings with +=, which can cause unnecessary copying in some string-building patterns.

Python integers are not involved at all. Treating the encoded values as strings is also necessary because a leading 0 is part of the encoded word and must not disappear.

Worked Examples

Sample 1

The first sample contains four encoded sentences.

Input line	Character or word	Decoded state
50 50	50	so
50 50	second 50	so so
45 15	45	as
45 15	second 15	as is
1 473 4 74573	1	i
1 473 4 74573	473	i ate
1 473 4 74573	4	i ate a
1 473 4 74573	74573	i ate a taste
17 15 70 517	17	it
17 15 70 517	15	it is
17 15 70 517	70	it is to
17 15 70 517	517	it is to sit

The resulting output is

so so
as is
i ate a taste
it is to sit

The trace shows that spaces never participate in the digit mapping. They simply remain as word separators, while every digit is translated independently.

Sample 2

Consider a sentence containing a leading zero and repeated digits.

2
010 555
473 74573

The first line is processed as follows.

Character	Mapping	Output so far
0	o	o
1	i	oi
0	o	oio
space	unchanged	oio
5	s	oio s
5	s	oio ss
5	s	oio sss

The second line follows the same rule.

Word	Decoded word
473	ate
74573	taste

The output is

oio sss
ate taste

This example demonstrates why the input must be treated as text. The first word begins with 0, and converting it to an integer before decoding would lose that character.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N)	Every input character is examined exactly once.
Space	O(N)	The decoded output for a test case is stored before printing.

Here N is the total number of characters in the encoded test cases. Since the algorithm performs only a constant-time dictionary lookup or a space copy for each character, it scales linearly with the actual input size. The 1 second time limit and 256 MB memory limit are comfortably compatible with this approach because there is no search, recursion, or nested loop over the encoded words.

Test Cases

The official statement provides one sample containing four test cases. The problem does not specify a numerical maximum for the number of characters, so the stress test below uses a large input rather than claiming a nonexistent formal maximum.

import sys
import io

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

    t = int(input())

    for _ in range(t):
        line = input().rstrip('\n')
        print(''.join(
            mapping[ch] if ch != ' ' else ' '
            for ch in line
        ))

def run(inp: str) -> str:
    global input

    old_stdin = sys.stdin
    old_input = input

    try:
        sys.stdin = io.StringIO(inp)
        input = sys.stdin.readline

        old_stdout = sys.stdout
        sys.stdout = io.StringIO()
        try:
            solve()
            return sys.stdout.getvalue()
        finally:
            sys.stdout = old_stdout
    finally:
        sys.stdin = old_stdin
        input = old_input

# Provided sample
sample = """4
50 50
45 15
1 473 4 74573
17 15 70 517
"""

assert run(sample) == """so so
as is
i ate a taste
it is to sit
""", "provided sample"

# Minimum-size input
assert run("1\n0\n") == "o\n", "single digit"

# All digits and repeated values
assert run("1\n013457\n") == "oieast\n", "all six mappings"

# Leading zero and word boundaries
assert run("1\n010 55 707\n") == "oio ss tot\n", "leading zero and spaces"

# All equal values
assert run("1\n777 0000 11111\n") == "ttt oooo iiiii\n", "repeated digits"

# Large stress-sized input
large_word = "013457" * 10000
expected_large = "oieast" * 10000 + "\n"
assert run("1\n" + large_word + "\n") == expected_large, "large input"

Test input	Expected output	What it validates
1 / 0	o	Minimum-size input and single-character decoding
1 / 013457	oieast	Every mapping is used exactly once and leading 0 is preserved
1 / 010 55 707	oio ss tot	Leading zeroes and word boundaries
1 / 777 0000 11111	ttt oooo iiiii	Repeated characters and all-equal values
One 60,000-character encoded word	Corresponding 60,000-character decoded word	Linear-time behavior on a large input

The helper replaces standard input with an in-memory stream and captures standard output, allowing the same solve() function used by the submission to be tested directly. The large test is deliberately constructed from repeated valid digits, so it checks both performance and the absence of accidental special handling for particular values.

Edge Cases

A one-digit word such as

1
0

must produce

o

The algorithm processes the single 0, performs one dictionary lookup, and immediately prints o. There is no assumption about word length, so the smallest possible input is handled naturally.

A leading zero must remain part of the encoded word. For

1
010

the algorithm visits 0, 1, and 0, producing o, i, and o, so the output is

oio

Because the program never converts the encoded sequence to an integer, the leading zero cannot be lost.

Repeated digits require repeated output characters. For

1
555

each of the three 5 characters independently maps to s, giving

sss

The algorithm has no set-based or deduplication operation, so repeated input characters remain repeated in the output.

Spaces represent boundaries between words. For

1
50 50

the first 50 becomes so, the space is copied, and the second 50 becomes so. The final output is

so so

This is why the solution reads each complete line rather than treating the entire test case as one undifferentiated number.

Finally, consider a long sequence such as

1
013457013457013457

Every character is handled once, producing

oieastoieastoieast

There is no dependence on the number of possible words, so the running time grows only with the length of the actual input. That linear behavior is the central reason the solution remains fast even when the encoded sequences are large.