CF 102697080 - 1337 5p34k

CF 102697080 - 1337 5p34k

Rating: -
Tags: -
Solve time: 41s
Verified: yes

Solution

Problem Understanding

The task is a character-by-character translation into a basic form of leet speak. The input is one string containing arbitrary text. For each lowercase letter that has a specified leet replacement, we substitute the corresponding digit. Every other character, including spaces, punctuation, digits, and letters not present in the conversion table, remains unchanged.

The required conversions are a -> 4, b -> 8, e -> 3, i -> 1, l -> 1, o -> 0, s -> 5, t -> 7, and z -> 2. For example, the word test becomes 7357, because t, e, s, and t are independently translated.

The input contains a single string, so there is no test-case count and no need for any parsing beyond reading one complete line. The official problem gives a 1 second time limit and 256 MB of memory. Since the input can be treated as a string of length n, an O(n) solution is the natural target. Even without a useful explicit maximum length in the statement, reading and processing each character once is optimal because every character potentially affects the output.

There are several small cases where an implementation can silently go wrong. Consider the input a. The correct output is 4. An implementation that only handles characters appearing in the sample sentence may accidentally leave it unchanged.

For the input all, the correct output is 411. Both occurrences of l must become 1, and the two replacements happen independently.

For the input hello!, the correct output is h3110!. The exclamation mark must survive unchanged. A careless implementation that reconstructs words rather than processing the original characters can accidentally lose spaces or punctuation.

For the input 123, the correct output is 123. Existing digits are not translated again. In particular, the 1 in the input is not related to the rule i -> 1, because the rule applies to the letter i, not to the digit 1.

Approaches

A direct brute-force implementation could examine every input character and, for each character, compare it against all nine source letters until a match is found. This is correct because every character is tested against the complete conversion table, and characters without a match are copied unchanged. In the worst case, if the input has length n, this performs up to 9n character comparisons. For n = 100000, that is at most 900000 comparisons, which would still pass comfortably here, but the repeated searches are unnecessary.

The key observation is that the conversion rule is a fixed mapping from one character to one character. There is no dependency between adjacent characters, no need to remember previous input, and no transformation that changes the length of the string. We can store the nine replacements in a dictionary and perform an average O(1) lookup for each input character.

The brute-force works because the conversion table is tiny, but it repeatedly searches information that never changes. The observation that every character has an independent fixed replacement lets us reduce each character's work to one dictionary lookup. The complete translation is consequently linear in the input length.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(9n) = O(n)	O(1)	Accepted, but unnecessary work
Optimal	O(n)	O(n) for the output	Accepted

Algorithm Walkthrough

1. Read the complete input line without removing internal spaces. The newline added by standard input should be removed, but spaces inside the text are part of the string and must remain in the output.
2. Create a mapping containing the nine specified letter-to-digit conversions. A dictionary is a natural representation because the current input character directly identifies the lookup key.
3. Visit every character in the input string exactly once. If the character exists in the mapping, append its replacement to the output. Otherwise append the original character.
4. Print the constructed output string. Since every input character contributes exactly one output character, the output has the same length as the input.

Why it works

The invariant is that after processing the first k input characters, the output contains exactly the correct leet translation of those same k characters. Initially, no characters have been processed, so the invariant holds trivially. When processing the next character, the algorithm uses the prescribed replacement if one exists and otherwise copies the character unchanged, which is exactly the required translation rule. Thus the invariant remains true after every character, and after the final character the complete output is correct.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    s = input().rstrip('\n')

    mapping = {
        'a': '4',
        'b': '8',
        'e': '3',
        'i': '1',
        'l': '1',
        'o': '0',
        's': '5',
        't': '7',
        'z': '2',
    }

    result = ''.join(mapping.get(ch, ch) for ch in s)
    print(result)

if __name__ == "__main__":
    solve()

The first line of solve reads the entire sentence. Using rstrip('\n') removes only the line terminator, rather than calling strip(), which could remove meaningful whitespace from the beginning or end of the input.

The dictionary contains exactly the nine rules from the problem. The get call is useful because it provides the original character as the default when the character is not one of the nine keys.

The generator processes characters from left to right and produces exactly one output character for each input character. ''.join(...) combines them into the final string efficiently instead of repeatedly concatenating strings inside the loop.

There are no index calculations or explicit boundaries, so there is no off-by-one issue. Python integers are irrelevant to the computation because the replacements are handled as characters rather than numeric arithmetic.

Worked Examples

The official sample contains one sentence. We can trace the character-level transformation of its meaningful portions as follows.

Input character	Mapping result	Output so far
t	7	7
h	h	7h
i	1	7h1
s	5	7h15
space	space	7h15
i	1	7h15 1
s	5	7h15 15
space	space	7h15 15
a	4	7h15 15 4
space	space	7h15 15 4
t	7	7h15 15 4 7
e	3	7h15 15 4 73
s	5	7h15 15 4 735
t	7	7h15 15 4 7357

Continuing through the remaining characters gives 7h15 15 4 7357 1npu7, which matches the official output.

A second example is hello!.

Input character	Mapping result	Output so far
h	h	h
e	3	h3
l	1	h31
l	1	h311
o	0	h3110
!	!	h3110!

The example demonstrates two properties of the algorithm. Each occurrence of a replaceable letter is handled independently, and characters outside the mapping are preserved exactly.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Every input character is processed once and dictionary lookup is constant time for this fixed mapping.
Space	O(n)	The translated string contains one output character for every input character.

The algorithm performs a single pass over the input and does not allocate any structure proportional to the number of possible character mappings. With the stated 1 second time limit and 256 MB memory limit, this linear approach is comfortably within the intended resource bounds.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    s = input().rstrip('\n')

    mapping = {
        'a': '4',
        'b': '8',
        'e': '3',
        'i': '1',
        'l': '1',
        'o': '0',
        's': '5',
        't': '7',
        'z': '2',
    }

    return ''.join(mapping.get(ch, ch) for ch in s)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    try:
        sys.stdin = io.StringIO(inp)
        return solve() + '\n'
    finally:
        sys.stdin = old_stdin

# Provided sample
assert run("this is a test input\n") == "7h15 15 4 7357 1npu7\n", "sample 1"

# Minimum-size input
assert run("a\n") == "4\n", "single replaceable character"

# All characters that have replacements
assert run("abeilos tz\n") == "4831105 72\n", "all mappings"

# Characters that must remain unchanged
assert run("123! XYZ\n") == "123! XYZ\n", "non-mapped characters"

# Repeated mappings and punctuation
assert run("all!!!\n") == "411!!!\n", "repeated l and punctuation"

# Large input
large_input = "a" * 100000 + "\n"
assert run(large_input) == "4" * 100000 + "\n", "large input"

Test input	Expected output	What it validates
a	4	Minimum-size input and a direct replacement
abeilos tz	4831105 72	All replacement rules and preservation of the space
123! XYZ	123! XYZ	Digits, punctuation, and unmapped letters remain unchanged
all!!!	411!!!	Repeated replacements and trailing punctuation
100000 copies of a	100000 copies of 4	Linear processing on a large input

Edge Cases

For a single-character input such as a, the algorithm performs one dictionary lookup, finds 4, and produces 4. There is no special handling for length one, which is useful because it means the general loop already covers the minimum case correctly.

For repeated replacements, consider all. The first a becomes 4, and both l characters independently become 1, giving 411. The algorithm never treats consecutive equal characters as one unit, so no occurrence is lost.

For characters outside the mapping, consider 123!. None of these characters is a dictionary key, so every lookup falls back to the original character. The result remains 123!. This also prevents an existing digit such as 1 from being accidentally transformed because another rule happens to produce the digit 1.

For whitespace and punctuation, consider hello! world. The letters are translated while the space and exclamation mark are copied. The use of rstrip('\n') rather than strip() preserves the actual text content, and the character-by-character transformation never modifies whitespace in the middle of the input.

For a large input, the algorithm still performs exactly one translation attempt per character. A string of 100000 a characters consequently requires 100000 dictionary lookups and produces 100000 4 characters, with no nested search or repeated string reconstruction.