CF 102697084 - Fine Tuned

Rating: -
Tags: -
Solve time: 1m 6s
Verified: yes

Solution

Problem Understanding

The problem models transposing a piece of music represented by note names. Each note belongs to the twelve-note chromatic scale, written using the names A, A#, B, C, C#, D, D#, E, F, F#, G, G#. The first input value tells us how many semitones every note should move. A positive value moves upward, while a negative value moves downward. The second input line contains the sequence of notes in the piece.

The required output is the same sequence after applying the same semitone shift independently to every note. The relative distances between notes never change. For example, moving E upward by three semitones gives G, because the sequence goes E -> F -> F# -> G.

There is no need to simulate actual guitar tuning. The guitar description only motivates the operation. Computationally, the problem is a transformation of symbols from a fixed alphabet of size twelve.

The constraints shown by the problem do not give a restrictive bound on the number of notes in the sequence, so the safe target is linear time in the number of input notes. A solution that processes every note once is effectively optimal, since every output note has to be produced. The semitone value itself can also be handled without repeatedly applying the shift, even if its absolute value is large.

Several edge cases are easy to mishandle. A negative shift must wrap around the beginning of the chromatic scale. For example,

-1
A

produces

G#

because moving one semitone below A reaches G#. An implementation that only subtracts one from an array index and forgets cyclic wrapping would access an invalid position or produce the wrong note.

A positive shift can cross the end of the scale. For example,

1
G#

produces

because the scale is cyclic. Using (index + shift) % 12 handles this naturally.

A shift of twelve or any multiple of twelve must leave every note unchanged. For example,

12
C# E G

produces

C# E G

because twelve semitones form one complete octave. A solution that only handles shifts smaller than twelve would fail here.

The notes themselves are symbolic strings rather than individual characters. In particular, # belongs to the note name, so G# must be treated as one token. Splitting the line by whitespace handles this correctly.

Approaches

The most direct brute-force solution is to repeatedly move every note by one semitone. For a shift of k semitones and m notes, this performs |k| * m note transitions in the worst case. The approach is correct because every transition corresponds exactly to one semitone movement, and performing the required number of transitions produces the requested pitch. However, if |k| is large, this repeats the same twelve-state cycle many times and does unnecessary work.

The key observation is that the chromatic scale contains exactly twelve distinct positions before the pattern repeats at the next octave. Shifting a note by twelve semitones returns to the same note, so only the remainder of the shift modulo twelve matters. Instead of applying a shift of, say, 1,000,000 semitones one step at a time, we can replace it with 1,000,000 mod 12, which is only four semitones.

We can assign every note an integer position from zero through eleven. Then transposing a note becomes an ordinary modular-index operation. If a note has position p and the required shift is k, its new position is (p + k) % 12. Python's modulo operation already gives a nonnegative result for a negative dividend, so negative transpositions require no special branch.

The brute-force method works because it explicitly simulates every semitone movement, but fails when the shift is large. The observation that the note system repeats every twelve semitones lets us collapse all complete octaves and process each note in constant time.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(	k	m)
Optimal	O(m)	O(m)	Accepted

Here, m is the number of notes and k is the requested semitone shift.

Algorithm Walkthrough

Store the twelve chromatic notes in their natural cyclic order, starting with A. This gives every possible note a unique index from 0 to 11.
Build a mapping from each note name to its index. This lets us convert an input note such as F# into an integer position in constant time.
Read the semitone shift and reduce it modulo twelve. A complete group of twelve semitones changes nothing, so only the remainder can affect the final note.
Read all note names from the second input line. Each whitespace-separated token represents one note, including tokens such as C# and G#.
For every input note, find its current index and add the reduced shift. Apply modulo twelve to wrap around the chromatic scale. The resulting index identifies the transposed note.
Print the transformed notes separated by spaces, preserving their original order.

Why it works

For every note, its position in the twelve-note cycle completely determines its pitch modulo one octave. Twelve semitones return to the same position, so replacing the original shift k by k mod 12 does not change the requested result. For an input note at position p, moving it upward or downward by k semitones places it at position (p + k) mod 12. The algorithm computes exactly that position and converts it back to the corresponding note name. Since every note is transformed independently using the same shift, the entire output sequence is correct.

Python Solution

import sys
input = sys.stdin.readline

NOTES = [
    "A", "A#", "B", "C", "C#", "D",
    "D#", "E", "F", "F#", "G", "G#"
]

INDEX = {note: i for i, note in enumerate(NOTES)}

def solve():
    shift = int(input())
    notes = input().split()

    shift %= 12

    answer = []
    for note in notes:
        pos = INDEX[note]
        answer.append(NOTES[(pos + shift) % 12])

    print(*answer)

if __name__ == "__main__":
    solve()

The NOTES array defines the cyclic order used by the entire solution. Starting at A is arbitrary as long as the order correctly represents adjacent semitones.

The dictionary INDEX reverses that array, allowing an input string to be converted to its numeric position in constant time. Without this mapping, searching through the twelve notes for every input token would still be fast in practice, but the dictionary makes the intended constant-time transformation explicit.

The line shift %= 12 removes complete octaves before any notes are processed. This is especially useful for large positive or negative shifts. Python's modulo behavior also makes expressions such as (-1) % 12 equal to 11, which is exactly the index needed to move one semitone below A to G#.

The expression (pos + shift) % 12 performs the cyclic wraparound. There is no separate case for crossing from G# to A or from A downward to G#, so there are fewer boundary conditions where an off-by-one error can occur.

The input uses split() rather than reading individual characters because notes such as C# consist of two characters but represent one musical note. The output is generated with print(*answer), which inserts exactly one space between consecutive notes.

Worked Examples

Consider the first sample:

-3
E A D G B E

The shift is reduced to 9 modulo twelve. Moving downward by three semitones is equivalent to moving upward by nine semitones.

Note	Original index	Shift	New index	Result
E	7	9	4	C#
A	0	9	9	F#
D	5	9	2	B
G	10	9	7	E
B	2	9	11	G#
E	7	9	4	C#

The resulting sequence is C# F# B E G# C#. The trace shows how a negative shift can be handled entirely through modulo arithmetic. No special downward-wrapping case is necessary.

For the second sample:

3
E A D G B E

Here the shift is already smaller than twelve, so it remains 3.

Note	Original index	Shift	New index	Result
E	7	3	10	G
A	0	3	3	C
D	5	3	8	F
G	10	3	1	A#
B	2	3	5	D
E	7	3	10	G

The output is G C F A# D G. The trace demonstrates that the transformation is applied independently to every note while preserving their order.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(m)	Every input note is converted and shifted once.
Space	O(m)	The output sequence contains one transformed token for every input note.

The algorithm performs a constant amount of work per note and does not depend on the magnitude of the semitone shift after reducing it modulo twelve. Since the output itself contains m notes, linear time is optimal up to constant factors. The fixed twelve-note lookup table and dictionary use only constant additional space, while the answer requires O(m) space.

Test Cases

The statement provides two samples, so both are included below. The custom cases cover a single note, octave-sized shifts, negative wraparound, positive wraparound, and repeated identical notes.

import sys
import io

NOTES = [
    "A", "A#", "B", "C", "C#", "D",
    "D#", "E", "F", "F#", "G", "G#"
]

INDEX = {note: i for i, note in enumerate(NOTES)}

def solve():
    shift = int(input())
    notes = input().split()
    shift %= 12

    answer = [
        NOTES[(INDEX[note] + shift) % 12]
        for note in notes
    ]

    print(*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:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        input = old_input

# Provided samples
sys.stdout = io.StringIO()
assert run("-3\nE A D G B E\n") == "C# F# B E G# C#\n", "sample 1"
assert run("3\nE A D G B E\n") == "G C F A# D G\n", "sample 2"

# Minimum-size input
assert run("0\nA\n") == "A\n", "zero shift on one note"

# Complete octave
assert run("12\nC# E G\n") == "C# E G\n", "one octave"

# Negative wraparound
assert run("-1\nA B C\n") == "G# A# B\n", "negative boundary"

# Positive wraparound and repeated values
assert run("13\nG# G# G#\n") == "A A A\n", "positive boundary and repeated notes"

# Restore normal stdout for interactive use of this test file.
sys.stdout = sys.__stdout__

Test input	Expected output	What it validates
`0` with `A`	`A`	Minimum-size input and zero shift
`12` with `C# E G`	`C# E G`	Complete octave leaves notes unchanged
`-1` with `A B C`	`G# A# B`	Negative wraparound at the beginning of the cycle
`13` with `G# G# G#`	`A A A`	Positive wraparound, modulo reduction, and repeated values

Edge Cases

A negative shift crossing the beginning of the note cycle is handled by modular arithmetic. For

-1
A

the input note A has index 0. After adding the shift, its raw index is -1, and Python evaluates -1 % 12 as 11. Index 11 is G#, so the output is correctly G#.

A positive shift crossing the end of the cycle behaves symmetrically. For

1
G#

G# has index 11. Adding one gives 12, and 12 % 12 is 0, which maps back to A. The output is therefore A.

A shift containing one or more complete octaves is reduced before processing. For

24
F# A C

the shift becomes 0 because 24 % 12 = 0. Every note keeps its original position, so the output is F# A C. This avoids performing twenty-four unnecessary transitions per note.

A note containing # must remain a single token. For

2
A# C# G#

the three tokens are looked up as A#, C#, and G#, rather than being split into individual characters. Their indices are 1, 4, and 11, and adding two gives indices 3, 6, and 1, producing C D# A#.

The input may contain any number of notes on the second line, so the implementation processes every token returned by split() rather than assuming the sequence contains exactly the six strings of standard guitar tuning. The guitar tuning in the examples is a particular input sequence, not a restriction on the transformation itself.