CF 102697061 - Half Winds

A compass can be divided into sixteen principal and intermediate directions. The four cardinal directions are North, East, South, and West, while the four diagonal directions are Northeast, Southeast, Southwest, and Northwest.

codeforcescompetitive-programming

CF 102697061 - Half Winds

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Solve time: 51s
Verified: yes

Solution

Problem Understanding

A compass can be divided into sixteen principal and intermediate directions. The four cardinal directions are North, East, South, and West, while the four diagonal directions are Northeast, Southeast, Southwest, and Northwest. The remaining eight directions lie halfway between adjacent directions. These are the half winds, such as NNE for north-northeast and ESE for east-southeast.

The input is exactly one three-character acronym naming one of these eight half winds. The task is simply to expand that acronym into its lowercase English name, with a hyphen between the two words. For example, ESE becomes east-southeast.

There is no large numeric input here. The input size is fixed at three characters, so even an approach that checks every possible direction is effectively constant time. The one-second time limit and 256 MB memory limit are far more than enough for any direct implementation. The main concern is correctness rather than asymptotic performance.

The eight valid acronyms are NNE, ENE, ESE, SSE, SSW, WSW, WNW, and NNW. Since the input is guaranteed to be one of these valid half winds, there is no need to handle malformed strings.

A common boundary case is NNW. It sits between NW and N, so its name is north-northwest. An implementation that tries to generate names by treating the compass directions as a simple linear sequence can accidentally mishandle the transition from NW back to N.

Another easy mistake is reversing the two parts of a name. For example, ENE is east-northeast, not northeast-east. The acronym's first letter identifies the cardinal direction that comes first in the English name.

Repeated letters are also normal. For example, the input NNE contains two N characters and must produce north-northeast. An implementation that assumes the three characters must all be distinct would incorrectly reject this valid direction.

Approaches

The most direct brute-force solution stores the eight possible acronyms and their corresponding names, then checks them one by one until the input matches. Since there are only eight possibilities, the worst case performs eight comparisons before finding the answer. This is already constant time, and there is no realistic performance problem even with the simplest implementation.

The cleaner approach is to use a dictionary mapping each acronym directly to its full name. The input itself is the key, so the answer can be retrieved immediately without explicitly traversing the eight possibilities. The useful observation is that the problem has a closed, fixed set of valid translations. There is no calculation to perform and no pattern that needs to be inferred at runtime, so a direct lookup expresses the problem exactly.

The brute-force method works because the input domain contains only eight values, but a lookup table is easier to verify and maintain. It also avoids special handling for the cyclic boundary between northwest and north.

Approach Time Complexity Space Complexity Verdict
Brute Force O(1) O(1) Accepted
Optimal O(1) average O(1) Accepted

Algorithm Walkthrough

  1. Read the three-character acronym from standard input. Removing surrounding whitespace is sufficient because the input contains only the acronym.
  2. Store the eight valid acronym-to-name translations in a dictionary. Each key is one possible half wind, and its value is exactly the required lowercase output.
  3. Use the input acronym as the dictionary key and print the associated full name. Since the input is guaranteed to be valid, a missing-key case does not need to be handled.

Why it works

The dictionary contains exactly one entry for every valid half wind, and each entry maps the acronym to its required English name. The input is guaranteed to be one of those eight acronyms, so the lookup always selects the unique correct translation. No geometric calculation or interpretation is necessary because the compass naming is already completely represented by the fixed mapping.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    direction = input().strip()

    names = {
        "NNE": "north-northeast",
        "ENE": "east-northeast",
        "ESE": "east-southeast",
        "SSE": "south-southeast",
        "SSW": "south-southwest",
        "WSW": "west-southwest",
        "WNW": "west-northwest",
        "NNW": "north-northwest",
    }

    print(names[direction])

if __name__ == "__main__":
    solve()

The first line of solve reads the acronym and uses strip() so the trailing newline from standard input does not become part of the dictionary key.

The names dictionary is the complete translation table. The spelling and hyphenation are stored exactly as required by the output format, which avoids constructing strings dynamically and eliminates several possible naming mistakes.

The final lookup uses names[direction]. Because every legal input appears as a key, this operation cannot fail for a valid test case. Python dictionaries provide average constant-time lookup.

There are no integer calculations, so integer overflow is irrelevant. There are also no array indices or ranges, so there are no numerical off-by-one boundaries to manage.

Worked Examples

For Sample 1, the input is ESE.

Step Direction Lookup result Output
1 ESE east-southeast east-southeast

The dictionary directly identifies ESE as east-southeast. This demonstrates the ordinary lookup case without any compass-cycle boundary.

For Sample 2, the input is NNE.

Step Direction Lookup result Output
1 NNE north-northeast north-northeast

The repeated N characters cause no special case because the entire three-character acronym is treated as a single dictionary key. The result also demonstrates that the cardinal component, north, comes first in the English name.

Complexity Analysis

Measure Complexity Explanation
Time O(1) The input contains exactly three characters and the dictionary contains only eight entries.
Space O(1) The translation table has a fixed number of entries.

The input size is fixed, so the solution uses only a tiny constant amount of work and memory. It is comfortably within the one-second time limit and 256 MB memory limit.

Test Cases

The official problem has two samples, ESE and NNE. Since the input domain is fixed to valid three-character half-wind acronyms, there is no meaningful valid test containing an all-equal three-character acronym such as NNN, because such a string is not a half wind. Likewise, the maximum input size is always three characters.

The following tests cover the official samples, the smallest valid input shape, repeated characters, the cyclic NNW boundary, and every possible half wind.

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

def solve():
    direction = input().strip()

    names = {
        "NNE": "north-northeast",
        "ENE": "east-northeast",
        "ESE": "east-southeast",
        "SSE": "south-southeast",
        "SSW": "south-southwest",
        "WSW": "west-southwest",
        "WNW": "west-northwest",
        "NNW": "north-northwest",
    }

    print(names[direction])

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 samples
assert run("ESE\n") == "east-southeast\n", "sample 1"
assert run("NNE\n") == "north-northeast\n", "sample 2"

# Custom cases
assert run("NNW\n") == "north-northwest\n", "cyclic compass boundary"
assert run("SSW\n") == "south-southwest\n", "repeated S character"
assert run("WNW\n") == "west-northwest\n", "repeated W character"
assert run("ENE\n") == "east-northeast\n", "repeated E character"

# All eight valid values together
all_input = "NNE\nENE\nESE\nSSE\nSSW\nWSW\nWNW\nNNW\n"
# The original problem accepts one acronym per test invocation, so these
# are checked individually rather than as one multi-case input.
expected = {
    "NNE": "north-northeast",
    "ENE": "east-northeast",
    "ESE": "east-southeast",
    "SSE": "south-southeast",
    "SSW": "south-southwest",
    "WSW": "west-southwest",
    "WNW": "west-northwest",
    "NNW": "north-northwest",
}
for acronym, answer in expected.items():
    assert run(acronym + "\n") == answer + "\n", acronym
Test input Expected output What it validates
ESE east-southeast Official sample and ordinary lookup
NNE north-northeast Official sample and repeated character handling
NNW north-northwest Boundary between northwest and north
SSW south-southwest Repeated S character
WNW west-northwest Repeated W character
ENE east-northeast Repeated E character

Edge Cases

The NNW boundary is handled by an explicit dictionary entry. For the input

NNW

the lookup finds north-northwest, which is the direction between northwest and north. A generated solution that simply walks forward through a linear list of directions can accidentally miss this wraparound case, while the lookup table has no such dependency.

The repeated-character case is handled without any character-by-character assumptions. For

NNE

the three-character string is used as one key, producing

north-northeast

The fact that N appears twice is irrelevant to the lookup. The same reasoning applies to ENE, SSE, and WNW.

The minimum and maximum input sizes coincide because every legal input contains exactly three characters. An input such as

ESE

already has both the minimum and maximum valid length. The algorithm performs one dictionary lookup regardless of which of the eight legal acronyms is supplied.

An all-equal acronym such as

NNN

is not a valid test case under the problem's input guarantees. It does not represent one of the eight half winds, so the solution is deliberately not required to assign it a result. This is preferable to inventing an output for an input the problem never permits.