CF 102697079 - Over The Rainbow

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Solve time: 59s
Verified: yes

Solution

Problem Understanding

The task is about measuring distance on a fixed circular color wheel. The wheel contains seven colors in this order:

red -> orange -> yellow -> green -> blue-green -> blue -> purple -> red

For each test case, we receive two color names. We need the shortest number of moves along the wheel needed to go from the first color to the second. A move from one color to its adjacent color costs one, and because the wheel is circular, we may travel in either direction. The first color is not counted, while the destination color is counted, so the answer is exactly the number of edges in the shorter path.

The official statement gives one second and 256 MB, and specifies that every query contains two color names from the wheel. The number of possible colors is fixed at seven, so the actual computation for one test case is constant time. Even if the number of test cases is very large, an O(n) solution is sufficient because every case requires only a handful of operations.

The main edge cases come from the circular nature of the wheel. If the two colors are equal, the distance is zero. For example,

1
green green

has output

A careless implementation that always counts at least one move could incorrectly print 1.

The other common mistake is forgetting that the wheel wraps around. For example,

1
purple orange

has output

because the shorter route is purple -> red -> orange. Treating the colors as an ordinary linear array would give a distance of five instead.

A third boundary case is the adjacent pair blue and blue-green:

1
blue blue-green

The correct output is

The two colors are consecutive on the wheel, so their distance is one. An implementation that accidentally counts both endpoints would produce two.

Approaches

The most direct solution is to store the seven colors in their wheel order and search through the sequence. Given two colors, we locate their positions and simulate movement around the wheel. Since there are only seven colors, even a literal brute-force simulation can perform only a constant number of operations. The longest possible shortest path on a seven-color wheel is three edges, although a less careful implementation that scans the entire wheel still performs at most seven position checks per query. There is no realistic input size for which this becomes too slow.

The useful observation is that the wheel never changes, so every color can be assigned a fixed integer position. Once red is position 0, orange is 1, and so on, the problem becomes the distance between two positions on a cycle.

Suppose the positions are a and b. Moving directly through the array gives a distance of abs(a - b). The other direction wraps around the end of the wheel and has distance 7 - abs(a - b). The answer is the smaller of these two values.

The brute-force simulation and the position formula are both correct because the state space has only seven colors. The formula is preferable because it expresses the structure directly and reduces each query to a dictionary lookup, subtraction, and minimum operation.

Approach	Time Complexity	Space Complexity	Verdict
Wheel Simulation	O(n) overall, O(1) per test case	O(1)	Accepted
Position Formula	O(n) overall, O(1) per test case	O(1)	Accepted

Algorithm Walkthrough

Store the colors in wheel order as red, orange, yellow, green, blue-green, blue, purple. Assign each color its index from 0 through 6.
Read the two colors for the current test case and look up their indices a and b. The indices preserve exactly the adjacency described by the color wheel.
Compute direct = abs(a - b). This is the distance when traveling through the normal order of the array.
Compute wrap = 7 - direct. This represents traveling in the opposite direction around the circular wheel. If the direct path crosses from one end of the array to the other, these are exactly the edges that path uses.
Print min(direct, wrap). The two values represent the only two directions around a cycle, so the smaller one is the shortest possible distance.

Why it works: after assigning positions to the seven colors, every path from one color to another on the cycle is either clockwise or counterclockwise. Their lengths are abs(a-b) and 7-abs(a-b). The algorithm takes the smaller of those two complete paths, so it always returns the shortest distance. When both colors are equal, direct is zero and the answer is correctly zero.

Python Solution

import sys
input = sys.stdin.readline

ORDER = {
    "red": 0,
    "orange": 1,
    "yellow": 2,
    "green": 3,
    "blue-green": 4,
    "blue": 5,
    "purple": 6,
}

def solve():
    t = int(input())

    out = []

    for _ in range(t):
        first, second = input().split()

        a = ORDER[first]
        b = ORDER[second]

        direct = abs(a - b)
        wrap = 7 - direct

        out.append(str(min(direct, wrap)))

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

if __name__ == "__main__":
    solve()

The ORDER dictionary is the central representation of the color wheel. Using explicit indices avoids any dependence on string ordering, which would be incorrect because alphabetical order has nothing to do with the wheel.

For each query, the two strings are converted to integer positions. The absolute difference gives the distance without wrapping. For example, purple has position 6 and orange has position 1, so the direct distance is 5.

The opposite direction has length 7 - 5 = 2, corresponding to purple -> red -> orange. Taking the minimum gives the required shortest distance.

There is no integer overflow concern in Python, and the largest intermediate value is only seven. The input is processed line by line, while output is accumulated and written once, which is a simple fast-I/O pattern suitable even if the number of test cases is large.

Worked Examples

The official sample contains three queries.

Sample 1

Input:

3
green red
purple orange
blue blue-green

For the first query, green has index 3 and red has index 0.

First	Second	a	b	Direct	Wrap	Answer
green	red	3	0	3	4	3

The direct path is green -> yellow -> orange -> red, which uses three edges. The opposite direction requires four, so the answer is 3.

For the second query, purple is index 6 and orange is index 1.

First	Second	a	b	Direct	Wrap	Answer
purple	orange	6	1	5	2	2

The shorter path crosses the boundary of the array:

purple -> red -> orange.

For the third query, the colors are adjacent.

First	Second	a	b	Direct	Wrap	Answer
blue	blue-green	5	4	1	6	1

The resulting output is:

3
2
1

This example demonstrates both directions around the cycle and confirms that adjacent colors have distance one.

Sample 2

A useful second example is:

4
red red
red purple
orange blue
yellow purple

The trace is:

First	Second	a	b	Direct	Wrap	Answer
red	red	0	0	0	7	0
red	purple	0	6	6	1	1
orange	blue	1	5	4	3	3
yellow	purple	2	6	4	3	3

The corresponding output is:

This example exercises equality and the wraparound boundary. In particular, red and purple are adjacent because they occupy the two ends of the stored array but are neighbors on the actual circle.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each of the n test cases uses two dictionary lookups and constant-time arithmetic.
Space	O(1)	The color dictionary contains exactly seven entries, and the output size is proportional to the input.

The wheel itself has fixed size seven, so there is no dependence on the number of colors. With n test cases, the algorithm performs constant work per case and therefore scales linearly with the input. The official limits are one second and 256 MB, and this solution is far below both limits.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

ORDER = {
    "red": 0,
    "orange": 1,
    "yellow": 2,
    "green": 3,
    "blue-green": 4,
    "blue": 5,
    "purple": 6,
}

def solve():
    input = sys.stdin.readline
    t = int(input())
    out = []

    for _ in range(t):
        a, b = input().split()
        x = ORDER[a]
        y = ORDER[b]

        direct = abs(x - y)
        wrap = 7 - direct

        out.append(str(min(direct, wrap)))

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

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(
    "3\n"
    "green red\n"
    "purple orange\n"
    "blue blue-green\n"
) == "3\n2\n1\n", "sample 1"

# Minimum-size input and equal colors
assert run(
    "1\n"
    "red red\n"
) == "0\n", "minimum input and equal colors"

# Wraparound boundary
assert run(
    "4\n"
    "purple red\n"
    "red purple\n"
    "purple orange\n"
    "orange purple\n"
) == "1\n1\n2\n2\n", "wraparound cases"

# Every color against itself
assert run(
    "7\n"
    "red red\n"
    "orange orange\n"
    "yellow yellow\n"
    "green green\n"
    "blue-green blue-green\n"
    "blue blue\n"
    "purple purple\n"
) == "0\n0\n0\n0\n0\n0\n0\n", "all equal"

# Maximum-distance cases on the seven-color cycle
assert run(
    "4\n"
    "red green\n"
    "green red\n"
    "orange blue\n"
    "blue orange\n"
) == "3\n3\n3\n3\n", "maximum shortest distance"

# Large test count
large_input = "100000\n" + ("red purple\n" * 100000)
assert run(large_input) == ("1\n" * 100000), "large input"

Test input	Expected output	What it validates
`1 / red red`	`0`	Minimum input size and identical colors
`purple red`, `red purple`, `purple orange`, `orange purple`	`1, 1, 2, 2`	Circular wraparound in both directions
Every color paired with itself	Seven zeros	Equality handling for every possible color
`red green`, `orange blue` and reverses	Four threes	Maximum possible shortest distance on the wheel
100000 copies of `red purple`	100000 copies of `1`	Linear scaling and input handling

Edge Cases

For identical colors, consider:

1
blue blue

Both colors map to index 5, so direct = abs(5 - 5) = 0. The wraparound distance is 7, and the minimum is zero. The algorithm prints:

The zero direct distance is the correct interpretation because no movement is needed to reach a color from itself.

For a wraparound pair, consider:

1
purple red

purple has index 6 and red has index 0. The direct distance is 6, but the circular distance is 7 - 6 = 1. The algorithm prints:

This corresponds to the adjacent edge from purple back to red. A linear-array solution that only computes abs(a-b) would incorrectly return 6.

For a pair near the opposite side of the cycle, consider:

1
purple orange

The positions are 6 and 1, giving direct distance 5 and wraparound distance 2. The answer is:

The path is purple -> red -> orange. This is the key reason the problem must be treated as a cycle rather than a normal sequence.

Finally, consider the maximum possible shortest distance:

1
red green

The positions are 0 and 3, so both relevant routes have lengths 3 and 4. The minimum is 3, producing:

No pair of colors can have a shortest distance greater than three on a seven-node cycle. This gives a useful sanity check for any implementation: every valid answer must lie between zero and three.