CF 102697044 - Yoda Translator

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Solve time: 58s
Verified: yes

Solution

Problem Understanding

The task is deliberately simple. We receive one sentence as a single line, and the sentence contains an even number of words. We have to split the sequence of words into two equal halves, then place the second half before the first half.

For example, the words

reverse this sentence you have to

are divided as

reverse this sentence and you have to

so the translated sentence becomes

you have to reverse this sentence.

The official problem specifies a 1 second time limit and 256 MB of memory, and guarantees that the number of words is even. The statement does not provide a numerical upper bound on the sentence length, so the safest design is linear in the number of words. Reading, splitting, and rebuilding the sentence all naturally take O(n) time, where n is the number of words. Any approach that repeatedly scans or shifts a large portion of the sentence can unnecessarily become quadratic.

There are a few small cases that can expose careless implementations. With only two words, such as hello world, each half contains one word, so the correct output is world hello. An implementation that assumes both halves contain several words could mishandle the midpoint.

Another boundary case is a sentence with repeated words. For input go go stay stay, the correct output is stay stay go go. The two halves must be treated according to their positions, not according to distinct word values, so using a set or any operation that removes duplicates would be incorrect.

A final common mistake is forgetting that the input is a complete line rather than a sequence of integers. For input reverse this sentence you have to, calling input().split() is appropriate because it separates the line into words, while processing only one token would lose the rest of the sentence.

Approaches

A brute-force way to think about the transformation is to construct every possible ordering of the words and search for the ordering in which the second half appears first. This is correct because the required answer is one of those permutations, but it is completely impractical. With n words, there are n! possible permutations, so even n = 10 already gives 3,628,800 candidates, and the number grows far too quickly to be useful.

The brute-force approach works because it eventually considers the desired arrangement, but fails because the problem does not ask us to discover an unknown permutation. The positions that have to move are already completely determined by the midpoint.

The key observation is that the required output is simply a rotation of the word sequence by half its length. If the words are w[0], w[1], ..., w[n-1], the answer is

w[n/2], ..., w[n-1], w[0], ..., w[n/2-1].

Once the words are stored in a list, Python can obtain the two halves with slicing and concatenate them. Every word is copied a constant number of times, giving linear time.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n · n!)	O(n) per candidate	Too slow
Optimal	O(n)	O(n)	Accepted

Algorithm Walkthrough

Read the complete input line and split it into a list of words. We need the words themselves because the transformation operates on word positions rather than individual characters.
Compute the midpoint as n // 2, where n is the number of words. The statement guarantees an even number of words, so this midpoint divides the sentence into two equally sized parts.
Take the words from the midpoint through the end. These form the second half and must appear first in the translated sentence.
Take the words from the beginning through the position immediately before the midpoint. These form the first half and must appear after the second half.
Concatenate the two parts with a space between them and print the resulting sentence. Joining the list rather than manually adding spaces avoids both missing spaces and an unwanted trailing space.

Why it works: the invariant is that the two slices contain exactly the original first half and original second half, in their original internal order. The algorithm outputs the second slice followed by the first slice, which is precisely the required transformation. Since every original word belongs to exactly one half, appears exactly once, and keeps its position relative to the other words in the same half, no word can be lost, duplicated, or reordered incorrectly.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    words = input().split()
    mid = len(words) // 2

    answer = words[mid:] + words[:mid]
    print(" ".join(answer))

if __name__ == "__main__":
    solve()

The first line reads the entire sentence and split() converts it into the sequence of words used by the algorithm. Multiple spaces, if present, are handled naturally by split(), while the required output is reconstructed with exactly one space between consecutive words.

mid = len(words) // 2 is safe because the problem guarantees an even number of words. The expression words[mid:] starts exactly at the first word of the second half, while words[:mid] contains exactly the first half.

The two lists are concatenated in the required order. Finally, " ".join(answer) creates the output sentence without a trailing space. There are no integer-overflow concerns because the only arithmetic operation is the division of the word count by two.

Worked Examples

The official sample contains six words. The midpoint is therefore three.

Step	Words	Midpoint	Second half	First half	Output
Read input	reverse this sentence you have to	3
Split	reverse, this, sentence, you, have, to	3	you, have, to	reverse, this, sentence
Rearrange	reverse, this, sentence, you, have, to	3	you, have, to	reverse, this, sentence	you have to reverse this sentence

This demonstrates the main transformation directly. The internal order of reverse this sentence is unchanged, and the internal order of you have to is also unchanged. Only the two halves exchange positions.

For a second example, consider the two-word sentence hello world.

Step	Words	Midpoint	Second half	First half	Output
Read input	hello world	1
Split	hello, world	1	world	hello
Rearrange	hello, world	1	world	hello	world hello

This example confirms the smallest meaningful case. Each half contains exactly one word, so the operation is simply a swap.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Splitting, slicing, concatenating, and joining process O(n) total words.
Space	O(n)	The word list and resulting list contain O(n) words.

The time limit is 1 second and the memory limit is 256 MB. Since the input itself contains n words, O(n) processing is asymptotically optimal: an algorithm must at least read those words. The implementation therefore scales directly with the size of the input rather than introducing unnecessary repeated scans or permutation generation.

Test Cases

import sys
import io

def solve():
    words = input().split()
    mid = len(words) // 2
    answer = words[mid:] + words[:mid]
    print(" ".join(answer))

def run(inp: str) -> str:
    global input
    old_stdin = sys.stdin
    old_input = input

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

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

# Provided sample
assert run("reverse this sentence you have to\n") == \
       "you have to reverse this sentence\n", "sample 1"

# Minimum-size input
assert run("hello world\n") == \
       "world hello\n", "minimum size"

# All words equal
assert run("go go go go\n") == \
       "go go go go\n", "all equal values"

# Boundary case with exactly two words in each half
assert run("one two three four\n") == \
       "three four one two\n", "half boundary"

# Larger input
large_words = [f"w{i}" for i in range(10000)]
large_input = " ".join(large_words) + "\n"
large_expected = " ".join(large_words[5000:] + large_words[:5000]) + "\n"
assert run(large_input) == large_expected, "large input"

Test input	Expected output	What it validates
`hello world`	`world hello`	Smallest even number of words and midpoint handling
`go go go go`	`go go go go`	Repeated values are preserved rather than deduplicated
`one two three four`	`three four one two`	Correct boundary between the two halves
10,000 distinct words	Second half followed by first half	Linear processing on a large input

Edge Cases

The smallest case is hello world. The list has length two, so mid = 2 // 2 = 1. The slice words[1:] is ["world"], and words[:1] is ["hello"]. Their concatenation is ["world", "hello"], producing world hello. This catches an off-by-one error where an implementation might accidentally use n // 2 - 1 or start the second half one position too early.

Repeated words require no special handling. For go go go go, the midpoint is two, giving ["go", "go"] for both halves. Concatenating them produces the same four-word sequence. The algorithm works because it operates on positions, not on word identities.

For the boundary case one two three four, the midpoint is two. The second half is three four, and the first half is one two, so the answer is three four one two. This confirms that the word at index mid belongs to the second half, while the word at index mid - 1 belongs to the first half.

For a large sentence containing 10,000 words, the midpoint is 5,000. The algorithm copies the two halves and joins them once, so its work grows proportionally with the input size. A permutation-based brute-force strategy would require considering up to 10,000! arrangements and is not remotely feasible, while the direct rotation remains linear.