CF 102697069 - Word Rotation

CF 102697069 - Word Rotation

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Solve time: 43s
Verified: yes

Solution

Problem Understanding

The task is to rotate every character of a lowercase word independently. For the character at position i, the input gives an integer shift k[i]. A positive shift moves the character forward through the alphabet, while a negative shift moves it backward. The alphabet is cyclic, so moving past z continues at a, and moving before a continues at z.

For example, rotating a by 5 gives f, while rotating a by -5 gives v. A shift does not have to be between -25 and 25, so a shifted by 30 is the same as a shifted by 4, giving e.

The first input line is the word itself. The second line contains one integer for every character, with the integer at position i describing how far that character should move. The output is the resulting word after all rotations have been applied.

The statement available for this problem does not specify a numerical upper bound for the word length or for the rotation values. The time limit is 1 second and the memory limit is 256 MB. Regardless of the missing explicit bounds, the intended solution should process each character a constant number of times. An algorithm proportional to the magnitude of a shift would be unsafe because a single shift can be much larger than 26. The alphabet has only 26 positions, so every shift can first be reduced modulo 26.

The first edge case is a negative shift. For example,

a
-5

produces

v

A careless implementation using only ord(c) + k and converting directly back to a character can produce a character outside the lowercase alphabet. The modulo operation must also work correctly for negative values.

The second edge case is a shift larger than one complete alphabet. For example,

a
30

produces

e

because 30 mod 26 = 4. Repeatedly moving the character 30 positions is unnecessary and becomes especially inefficient for very large shifts.

The third edge case occurs when the shift lands exactly on an alphabet boundary. For example,

z
1

produces

a

The character position must be computed modulo 26, rather than clamped to z.

The fourth edge case is a negative shift that crosses the beginning of the alphabet. For example,

b
-3

produces

y

This catches implementations that attempt to handle negative positions with a separate subtraction branch but forget the wraparound.

Approaches

A direct brute-force approach would literally perform one character movement at a time. For a character c and shift k, we could repeatedly increment or decrement its alphabet position until |k| movements had been performed. This is correct because every individual movement represents exactly one step around the cyclic alphabet.

The problem is that this work depends on the magnitude of the shift rather than on the size of the input. If the word has length N and the largest absolute shift is K, the worst case takes O(NK) individual movements. For example, if there are 100,000 characters and every shift has magnitude 10^9, this would require up to 10^14 character movements. That is far beyond what a 1-second solution can perform.

The key observation is that moving through an alphabet of 26 characters is periodic. After 26 forward movements, a character is exactly where it started. The same is true for 26 backward movements. Consequently, a shift k has exactly the same effect as k mod 26.

Once the shift has been reduced modulo 26, we can convert the current character to a number from 0 through 25, add the reduced shift, and take the result modulo 26. This gives the final alphabet position immediately, regardless of how large or negative the original shift was.

The brute-force works because every single movement is simulated, but fails when the shifts are large. The observation that the alphabet repeats every 26 positions lets us replace an arbitrary number of movements with one modular arithmetic operation.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(NK) where `K = max(	shift[i]	)`
Optimal	O(N)	O(N)	Accepted

Algorithm Walkthrough

1. Read the word and the sequence of shifts. There is one shift for every character, so the character at index i must be paired with shift[i].
2. For each character, convert it into a zero-based alphabet position using ord(c) - ord('a'). Thus a becomes 0, b becomes 1, and z becomes 25.
3. Reduce the corresponding shift modulo 26. Only the remainder matters because 26 positions form one complete cycle of the alphabet.
4. Add the reduced shift to the original alphabet position and take the result modulo 26. This handles both positive and negative rotations while keeping the result in the range 0 through 25.
5. Convert the resulting alphabet position back into a lowercase character using chr(position + ord('a')).
6. Append the resulting character to an output list and join the list after processing all characters. Building a list and joining it avoids repeatedly constructing increasingly large strings.

Why it works

For every character, its position is represented modulo 26. A shift of k changes that position to (position + k) mod 26. Since two shifts that differ by a multiple of 26 produce the same position, replacing k by k mod 26 does not change the result. The algorithm applies exactly this transformation independently to every character, so every output character is the character required by its corresponding rotation.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    s = input().strip()
    shifts = list(map(int, input().split()))

    base = ord('a')
    result = []

    for c, shift in zip(s, shifts):
        position = ord(c) - base
        position = (position + shift) % 26
        result.append(chr(position + base))

    print(''.join(result))

if __name__ == "__main__":
    solve()

The first line reads the word without its trailing newline. The second line is split into integers so that each value can be paired with the character at the same index.

The loop uses zip(s, shifts) to make the correspondence between characters and rotations explicit. For each pair, ord(c) - ord('a') changes the character into a number from 0 to 25.

Python's % operator handles negative operands in a way that is useful here. For example, (-3) % 26 is 23, so a character at position 1, representing b, becomes (1 - 3) % 26 = 24, which represents y. No special negative-shift branch is needed.

The modulo operation is applied after adding the shift, so the resulting position is always between 0 and 25. There is no integer-overflow concern in Python, and the arithmetic remains constant time even when the shifts are very large.

Worked Examples

Example 1

The sample input is:

agjnp
7 -2 54 -80 25

The character and shift state evolves as follows.

Index	Character	Original position	Shift	New position	Result
0	a	0	7	7	h
1	g	6	-2	4	e
2	j	9	54	11	l
3	n	13	-80	11	l
4	p	15	25	14	o

The final output is:

hello

The third character demonstrates why large positive shifts can be reduced modulo 26, since 54 % 26 = 2. The fourth character demonstrates the same idea for a large negative shift, since -80 % 26 = 24.

Example 2

Consider a word containing wraparound in both directions:

azb
1 -1 -3

The state is:

Index	Character	Original position	Shift	New position	Result
0	a	0	1	1	b
1	z	25	-1	24	y
2	b	1	-3	24	y

The final output is:

byy

The first character checks forward wraparound in the general formula, while the second and third characters check backward movement across a.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N)	Each character and its corresponding shift are processed once.
Space	O(N)	The output characters are stored before being joined.

The algorithm does not depend on the magnitude of the shifts, because every shift is handled with one modulo operation. This makes it suitable even when the input contains very large positive or negative rotation values. The 256 MB memory limit is also easily sufficient for storing a word and its output.

Test Cases

The original statement provides one sample, so the first assertion below uses that sample. Since the published statement does not expose a numerical maximum for N, the stress case uses 100000 characters as a representative large input rather than claiming it is the official maximum.

import sys
import io

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

    s = input().strip()
    shifts = list(map(int, input().split()))

    base = ord('a')
    result = []

    for c, shift in zip(s, shifts):
        position = (ord(c) - base + shift) % 26
        result.append(chr(position + base))

    print(''.join(result))

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided sample
assert run(
    "agjnp\n"
    "7 -2 54 -80 25\n"
) == "hello\n", "sample 1"

# Minimum-size input
assert run(
    "a\n"
    "0\n"
) == "a\n", "minimum size"

# All-equal values
assert run(
    "aaaaa\n"
    "1 1 1 1 1\n"
) == "bbbbb\n", "all equal characters"

# Boundary wrapping in both directions
assert run(
    "azb\n"
    "1 -1 -3\n"
) == "byy\n", "alphabet boundaries"

# Very large positive and negative shifts
assert run(
    "abcde\n"
    "1000000000000 -1000000000000 26 52 -78\n"
) == "abcde\n", "large shifts"

# Large input stress case
n = 100000
large_input = "a" * n + "\n" + " ".join(["1"] * n) + "\n"
assert run(large_input) == "b" * n + "\n", "large input"

Test input	Expected output	What it validates
a with shift 0	a	Minimum-size input and zero rotation
aaaaa with five shifts of 1	bbbbb	Independent processing of repeated characters
azb with 1 -1 -3	byy	Forward and backward alphabet boundaries
abcde with shifts around 10^12 and multiples of 26	abcde	Reduction of very large shifts
100000 a characters with shift 1	100000 b characters	Linear-time behavior on a large input

Edge Cases

For the negative-shift case,

a
-5

the character position starts at 0. The algorithm computes (0 - 5) % 26 = 21, and position 21 is v. The output is therefore v. No special handling for negative values is necessary because modular arithmetic represents the wraparound directly.

For a shift larger than one alphabet,

a
30

the computed position is (0 + 30) % 26 = 4, which corresponds to e. The algorithm performs the same constant amount of work for this shift as it does for a shift of 4.

For forward wraparound,

z
1

the position is (25 + 1) % 26 = 0, giving a. The modulo operation is exactly what moves the result from the end of the alphabet back to its beginning.

For backward wraparound,

b
-3

the position is (1 - 3) % 26 = 24, giving y. A manually written implementation that assumes the intermediate position must stay nonnegative could easily get this boundary case wrong, while the modular formula handles it directly.

The sample itself combines several of these cases in one input. The shift 54 tests a value larger than 26, -80 tests a large negative value, and the resulting letters confirm that each rotation is applied to the character at the matching index.