CF 102697111 - Word Subsitution

CF 102697111 - Word Subsitution

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Solve time: 56s
Verified: yes

Solution

Problem Understanding

The problem asks us to perform one of two inverse text substitutions. We are given a message, a single-character key, and a word key. The single character is guaranteed not to appear anywhere inside the key word.

The message determines which direction of the substitution we should perform. If the message contains the key word, we treat the message as encrypted text and replace every occurrence of the key word with the key character. Otherwise, we treat the message as plain text and replace every occurrence of the key character with the key word. The required output is the transformed message. The official statement confirms this rule and gives the two examples hello world -> hebruhbruho worbruhd and back again.

There are no explicit size bounds in the published statement, so the safest interpretation is to make the solution linear in the length of the message, apart from the unavoidable cost of constructing the resulting string. With a one-second limit and 256 MB of memory, repeatedly searching and rebuilding the message would be unnecessarily expensive if the input were large. Python's built-in string replacement operations are implemented efficiently enough for this simple substitution, and the key word cannot contain the key character, which removes an important source of ambiguity.

The first edge case is a message containing the key word but no key character. For example:

bruh
l
bruh

The message contains bruh, so this is a decryption operation. The correct output is:

l

A careless solution that checks only whether the key character occurs might incorrectly attempt encryption.

The second edge case is a message containing neither the key word nor the key character. For example:

hello
l
xyz

Since xyz does not occur, the operation is encryption. The only l is replaced, giving:

hexyz

The fact that the key word is absent is what determines the direction, not whether a replacement can actually be made.

The third edge case is multiple adjacent occurrences. For example:

ll
l
ab

The message does not contain ab, so both occurrences of l must be replaced. The result is:

abab

Replacing only the first occurrence would silently give the wrong answer.

Approaches

The direct brute-force approach is to scan the message character by character and manually look for the key word whenever encryption or decryption is needed. This is correct because every replacement is determined entirely by the chosen direction and the current substring. However, if the key word has length k and the message has length n, checking every possible starting position character by character can require O(nk) comparisons. In the worst case, with a long message consisting mostly of repeated characters and a similarly long key word, this can approach O(n^2) operations.

The structure of the problem lets us avoid implementing any pattern matching ourselves. We only need to replace all occurrences of one fixed substring with another fixed string. Python already provides exactly this operation with str.replace.

The brute-force solution works because each occurrence can be treated independently, but it fails to exploit the fact that the replacement target is fixed throughout the entire message. Once we determine the direction by checking whether the key word occurs, the whole transformation reduces to one call to replace.

The guarantee that the key character is not contained in the key word is also useful. During encryption, replacing a character with the key word cannot create another occurrence of the key character inside the inserted word. During decryption, replacing the key word with the key character cannot create another copy of the key word. The transformation can consequently be performed in one pass without worrying about replacements recursively triggering more replacements.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(nk)	O(n)	Too slow for large strings
Optimal	O(n + output)	O(n + output)	Accepted

Algorithm Walkthrough

1. Read the entire message without stripping internal spaces. The message is a line of text, so using strip() is unnecessary and can accidentally alter meaningful leading or trailing spaces.
2. Read the key character and key word, removing only the newline from those two lines. They are separate input values and do not contain spaces.
3. Check whether the key word occurs anywhere in the message. If it does, the message is considered encrypted and we must decrypt it.
4. In the decryption case, replace every occurrence of the key word with the key character. Calling replace without a count replaces all occurrences, which is exactly what the problem requires.
5. If the key word does not occur, the message is considered plaintext and we must encrypt it. Replace every occurrence of the key character with the key word.
6. Print the resulting message exactly as produced. Spaces and all other characters remain unchanged.

The correctness follows from the decision rule in the statement. The algorithm first chooses decryption exactly when the key word occurs, and encryption otherwise. In the decryption branch, every occurrence of the key word is replaced by the key character, which is precisely the specified inverse operation. In the encryption branch, every occurrence of the key character is replaced by the key word. Since the key character does not occur inside the key word, an inserted key word cannot accidentally introduce another key character that would need further replacement. Thus the resulting message is exactly the required transformation.

Python Solution

import sys
input = sys.stdin.readline

message = input().rstrip('\n')
key_letter = input().rstrip('\n')
key_word = input().rstrip('\n')

if key_word in message:
    answer = message.replace(key_word, key_letter)
else:
    answer = message.replace(key_letter, key_word)

print(answer)

The first input reads the complete message, including spaces. Using rstrip('\n') removes only the line terminator, so the actual message is preserved.

The next two reads obtain the key letter and key word. Since these values occupy their own lines, removing the newline is sufficient.

The membership test key_word in message implements the exact rule that determines whether the operation is encryption or decryption. There is no need to count occurrences because one occurrence is enough to select the decryption branch.

The calls to replace deliberately omit the optional count argument. A limited replacement such as replace(old, new, 1) would modify only the first occurrence and would fail when the same key appears several times.

Python integers are irrelevant here, so there are no overflow concerns. There are also no indexing boundaries to manage because the built-in substring replacement handles them internally.

Worked Examples

Sample 1

For the first sample, the message is hello world, the key letter is l, and the key word is bruh.

Message	Key Letter	Key Word	Key Word Present?	Operation	Result
hello world	l	bruh	No	l -> bruh	hebruhbruho worbruhd

The word bruh is absent, so the algorithm chooses encryption. Both occurrences of l are replaced independently. The first l in hello becomes bruh, and the second l becomes another bruh.

Sample 2

For the second sample, the message is hebruhbruho worbruhd, with the same keys.

Message	Key Letter	Key Word	Key Word Present?	Operation	Result
hebruhbruho worbruhd	l	bruh	Yes	bruh -> l	hello world

This time the key word occurs several times, so the algorithm chooses decryption. Every bruh is converted back into l, reconstructing the original message.

The two examples together demonstrate the central invariant: the presence of the key word selects the reverse operation, and every occurrence is transformed in that direction.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n + output)	The message is scanned to find the key word and processed to construct the replaced string.
Space	O(n + output)	Python stores the input message and the resulting string.

Here n is the length of the message. The published problem has a one-second time limit and 256 MB memory limit. The solution performs only a constant number of string operations and avoids any explicit quadratic substring search, so it is easily appropriate for the intended constraints.

Test Cases

import sys
import io

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

    message = input().rstrip('\n')
    key_letter = input().rstrip('\n')
    key_word = input().rstrip('\n')

    if key_word in message:
        return message.replace(key_word, key_letter)
    return message.replace(key_letter, key_word)

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

# Provided sample 1
assert run("hello world\nl\nbruh\n") == "hebruhbruho worbruhd\n"

# Provided sample 2
assert run("hebruhbruho worbruhd\nl\nbruh\n") == "hello world\n"

# Minimum-size message
assert run("a\na\nb\n") == "b\n"

# Key word occurs repeatedly and adjacently
assert run("abab\nx\nab\n") == "xx\n"

# Message contains neither key word nor key letter
assert run("hello\nx\nxyz\n") == "hello\n"

# Multiple key-letter occurrences
assert run("aaaa\na\nlong\n") == "longlonglonglong\n"

# Large message
large = "a" * 10000
assert run(large + "\na\nxyz\n") == ("xyz" * 10000) + "\n"

Test input	Expected output	What it validates
a / a / b	b	Minimum-size input and a single replacement
abab / x / ab	xx	Multiple adjacent key-word occurrences
hello / x / xyz	hello	Neither key appears, so encryption changes nothing
aaaa / a / long	longlonglonglong	Every occurrence must be replaced
a...a / a / xyz	Repeated xyz	Large input and linear processing

Edge Cases

The first edge case is when the message itself is exactly the key word:

bruh
l
bruh

The membership test succeeds immediately. The algorithm takes the decryption branch and evaluates bruh.replace(bruh, l), producing:

l

There is no special case needed for an exact match.

The second edge case is when the key word is absent and the key letter is also absent:

hello
x
xyz

The test xyz in hello is false, so the algorithm selects encryption. Since x does not occur, hello.replace(x, xyz) leaves the message unchanged:

hello

This is correct because the problem chooses encryption solely from the absence of the key word.

The third edge case contains adjacent occurrences:

abab
x
ab

The key word ab occurs twice, starting at positions zero and two. The decryption operation replaces both occurrences simultaneously, giving:

xx

A character-by-character implementation that advances too far after finding a match could accidentally skip the second occurrence. str.replace avoids that indexing error.

The fourth edge case has repeated key letters:

aaaa
a
long

The key word long is absent, so encryption is selected. Each of the four a characters is replaced, resulting in:

longlonglonglong

The key word does not contain a, as guaranteed by the statement, so no newly inserted a characters appear and the transformation does not need to repeat itself.