CF 102697020 - Codebusters

CF 102697020 - Codebusters

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Solve time: 1m 10s
Verified: yes

Solution

Problem Understanding

The problem asks us to build a very small encryption transformation. Each input word represents a sequence of characters, and the encryption rule treats characters in pairs. Inside every adjacent pair, the first character moves after the second character, which is the same as swapping every odd-positioned character with the following even-positioned character. If the word has an unpaired final character, it stays unchanged. The first input value tells us how many words must be transformed, and the output is each transformed word on its own line.

The main constraint is the length of the words. Since the operation only needs to inspect each character once, a linear solution is the natural target. Any approach that repeatedly searches, rebuilds, or compares large parts of a word would add unnecessary work. If a word contained hundreds of thousands of characters, an algorithm slower than O(length) would risk becoming too expensive because it would perform repeated operations on data that only needs a single pass.

The tricky cases come from positions near the end of the word. A word with an odd number of characters has a final character without a partner, so it must not be moved. For example, with input:

1
ABCDE

the correct output is:

BADCE

A careless implementation that swaps every index with the next index without checking the boundary may try to access a character after E and fail or accidentally modify memory in other languages.

Another edge case is a one-character word. For example:

1
Z

the correct output is:

Z

There is no pair to swap, so the algorithm must leave the character untouched. An implementation that assumes every character has a partner would produce an incorrect result.

Repeated characters also need no special handling. For example:

1
AAAA

the correct output is:

AAAA

The swaps still happen, but equal characters hide the movement. The algorithm should rely on positions rather than character values.

Approaches

The straightforward approach is to simulate the encryption exactly as described. Starting from the first character, swap it with the next character, then move forward by two positions and repeat. This is correct because every affected pair is independent from the others. For a word of length n, the algorithm performs about n / 2 swaps, so it runs in O(n) time.

There is no meaningful brute force search for this problem because the transformation is already deterministic. A slower implementation might repeatedly create new strings after every swap. While still logically correct, doing this in languages where strings are immutable can turn every swap into an O(n) copy operation. For a word of length n, that could become O(n²), performing roughly n² / 2 character movements in the worst case.

The key observation is that swaps do not affect future decisions. After positions 0 and 1 are exchanged, positions 2 and 3 can be handled independently. This lets us process the string from left to right while changing each pair exactly once. Building a mutable list of characters and joining it at the end gives a direct O(n) implementation.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(n)	Too slow for large inputs
Optimal	O(n)	O(n)	Accepted

Algorithm Walkthrough

1. Read each word and convert it into a mutable sequence of characters. Python strings cannot be changed in place, so using a list allows individual characters to be swapped.
2. Start from index 0 and move through the word two positions at a time. The current index represents the first character of a pair, and the next index represents the second character.
3. If the next index exists, exchange the two characters. This applies the encryption rule to the current pair without affecting any other pair.
4. Continue until all pairs have been processed. If the length is odd, the final character is never selected as the first element of a complete pair, so it remains unchanged.
5. Join the character sequence back into a string and print the result.

Why it works:

The algorithm maintains the invariant that every pair before the current position has already been encrypted exactly once, and every pair after the current position is still in its original state. Since the encryption rule only exchanges characters inside each adjacent pair, processing one pair cannot change the correct result of another pair. The only possible leftover character is the last character of an odd-length word, and leaving it unchanged matches the required transformation.

Python Solution

import sys
input = sys.stdin.readline

def transform(word):
    chars = list(word)
    i = 0
    n = len(chars)

    while i + 1 < n:
        chars[i], chars[i + 1] = chars[i + 1], chars[i]
        i += 2

    return ''.join(chars)

def main():
    t = int(input())
    ans = []

    for _ in range(t):
        word = input().strip()
        ans.append(transform(word))

    sys.stdout.write("\n".join(ans))

if __name__ == "__main__":
    main()

The transform function contains the whole algorithm. Converting the string into a list is necessary because Python strings are immutable, while list elements can be swapped directly.

The loop condition i + 1 < n is the boundary check that protects odd-length words. When the final character has no partner, the loop stops before accessing outside the string. Increasing i by two skips directly to the next independent pair, avoiding unnecessary checks.

The solution does not need special treatment for character values, duplicate letters, or alphabet ordering. The encryption depends only on positions, so swapping indexes is enough.

Worked Examples

For the input word ABCDEFG, the processing is:

Current index	Pair processed	State after swap
0	AB → BA	BACDEFG
2	CD → DC	BADC EFG
4	EF → FE	BADCFEG

The final output is:

BADCFEG

This example demonstrates the normal case where every character except the final one belongs to a pair.

For the input word HELLO, the processing is:

Current index	Pair processed	State after swap
0	HE → EH	EHLLO
2	LL → LL	EHLLO

The final character O has no partner, so the output remains:

EHLLO

This example shows the odd-length boundary case and confirms that the last character is preserved.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each character is visited at most once during the pair swaps.
Space	O(n)	A list containing the characters is created so swaps can be performed.

The algorithm scales linearly with the total number of characters across all test cases. It avoids repeated string rebuilding, which keeps the implementation efficient even when the input contains many long words.

Test Cases

import sys
import io

def solve(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    def transform(word):
        chars = list(word)
        i = 0
        while i + 1 < len(chars):
            chars[i], chars[i + 1] = chars[i + 1], chars[i]
            i += 2
        return ''.join(chars)

    t = int(input())
    out = []
    for _ in range(t):
        out.append(transform(input().strip()))
    return "\n".join(out)

assert solve("""4
ABCDEFG
GAME
RAMBOOK
EHLLO
""") == """BADCFEG
AGEM
ARBMOOK
HELLO""", "samples"

assert solve("""1
A
""") == "A", "single character"

assert solve("""1
AAAA
""") == "AAAA", "all equal values"

assert solve("""3
AB
ABC
ABCDE
""") == """BA
BAC
BADCE""", "odd length boundaries"

assert solve("""1
ZYXWVUT
""") == "YZXWVU T".replace(" ", ""), "long alternating pattern"

Test input	Expected output	What it validates
A	A	Minimum size and no available pair
AAAA	AAAA	Equal characters and position-based swapping
AB, ABC, ABCDE	BA, BAC, BADCE	Odd-length handling and pair boundaries
ZYXWVUT	YZXWVUT	Multiple consecutive swaps

Edge Cases

For a one-character input, the algorithm starts at index 0, but the condition i + 1 < n is false immediately. No swap occurs, so:

Input:
1
Z

Output:
Z

The loop boundary prevents an invalid access and matches the rule that there is no pair to exchange.

For an odd-length word, the final character is intentionally ignored after all complete pairs are processed. With:

Input:
1
ABCDE

Output:
BADCE

the algorithm swaps A with B, then C with D. The character E remains at the end because the loop stops when only one unprocessed character remains.

For repeated characters:

Input:
1
AAAA

Output:
AAAA

the algorithm still performs two swaps, but each swap exchanges identical values. This confirms that the implementation does not rely on finding different characters, only on applying the positional transformation.