CF 105930L - Stella

Each input test case gives two stellar classifications written in a fixed format: a capital letter followed by a digit. The letter represents a coarse temperature class ordered from hottest to coldest as O, B, A, F, G, K, M.

codeforcescompetitive-programming

CF 105930L - Stella

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Solve time: 43s
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Solution

Problem Understanding

Each input test case gives two stellar classifications written in a fixed format: a capital letter followed by a digit. The letter represents a coarse temperature class ordered from hottest to coldest as O, B, A, F, G, K, M. Inside each class, the digit from 0 to 9 refines temperature, where 0 is hotter and 9 is cooler.

So every classification is effectively a point on a one-dimensional temperature scale with 70 possible discrete values. The task is to compare two such points and decide whether the first star is hotter, cooler, or exactly the same temperature as the second.

The constraints are small: up to 1000 comparisons, each involving constant-time processing. This immediately suggests that any solution that parses each pair and compares in O(1) is sufficient, while anything involving sorting or complex simulation is unnecessary overhead.

A subtle failure case comes from comparing lexicographically as strings. For example, comparing "A9" and "B0" as strings works accidentally here because the letter ordering matches ASCII order, but relying on this is fragile and can break if the encoding or representation changes. Another potential mistake is ignoring that digit comparison is reversed in meaning: smaller digit means hotter, not colder.

Edge cases are mostly about boundary letters and digits. For instance, comparing O0 with O0 should yield same. Comparing M9 with O0 should yield cooler for the first, since M9 is the coldest possible and O0 is the hottest.

Approaches

A brute-force way to interpret the problem is to explicitly map every classification to a numeric temperature value in the range 0 to 69. We assign O0 as 0, O1 as 1, ..., O9 as 9, then B0 as 10, and so on until M9 as 69. Once both inputs are converted, comparison reduces to integer comparison.

This works because the classification system is strictly ordered first by letter class, then by digit within class. The brute-force mapping is already constant size and thus efficient, but it is slightly indirect.

We can simplify further by observing that the ordering is lexicographic with a custom alphabet for letters and reversed ordering for digits inside each letter. Instead of building a full mapping table, we can compute the index of the letter and combine it with the digit.

Let letter rank O, B, A, F, G, K, M correspond to 0 through 6. Then a star is represented as rank(letter) * 10 + digit. This directly encodes the total ordering into a single integer, preserving correctness.

The brute-force approach uses a dictionary lookup per letter, while the optimal approach replaces it with a direct arithmetic encoding. Both are O(1) per test case, but the latter is cleaner and avoids extra data structures.

Approach Time Complexity Space Complexity Verdict
Brute mapping table O(T) O(1) Accepted
Arithmetic encoding O(T) O(1) Accepted

Algorithm Walkthrough

We want a direct numeric representation of each classification so that comparison becomes a simple integer comparison.

  1. Read the two classification strings for each test case. Each string has length exactly 2, so parsing is constant time.
  2. Convert the letter into a rank between 0 and 6 according to its position in the temperature sequence O, B, A, F, G, K, M. This ordering is fixed and defines the coarse temperature level.
  3. Convert the digit character into an integer between 0 and 9. This digit refines the temperature within its letter class.
  4. Compute a single integer value as rank * 10 + digit. This works because each letter class spans exactly 10 values, and digit ordering is consistent within each class.
  5. Repeat for both stars, producing two integers.
  6. Compare the two integers. If the first is smaller, it corresponds to a hotter star, since O0 is the minimum value. If it is larger, it is cooler. If equal, both classifications are identical.

The key reason this transformation is valid is that the original ordering is already structured as a lexicographically ordered product: primary key is the letter class, secondary key is the digit in reverse-hotness order.

Why it works

The mapping preserves ordering because letter classes are strictly monotonic in temperature and each class contains exactly 10 uniformly ordered subdivisions. No overlap exists between classes since multiplying by 10 separates them into disjoint intervals. Within a class, digit comparison is consistent with temperature direction, so the combined integer fully encodes the original ordering without ambiguity.

Python Solution

import sys
input = sys.stdin.readline

order = {
    'O': 0,
    'B': 1,
    'A': 2,
    'F': 3,
    'G': 4,
    'K': 5,
    'M': 6
}

def encode(s):
    return order[s[0]] * 10 + (ord(s[1]) - ord('0'))

t = int(input())
out = []

for _ in range(t):
    a, b = input().split()
    va = encode(a)
    vb = encode(b)

    if va < vb:
        out.append("hotter")
    elif va > vb:
        out.append("cooler")
    else:
        out.append("same")

print("\n".join(out))

The solution begins by defining the fixed ordering of spectral classes. This avoids repeated conditional logic during comparisons. Each classification string is converted into a compact integer using the encoding function, which isolates the letter contribution and digit contribution separately.

A common mistake would be to compare strings directly. While it may pass on some datasets, it relies on ASCII ordering and does not explicitly encode the scientific ordering. Another subtle issue is forgetting that smaller encoded values correspond to hotter stars, so the comparison direction must be carefully aligned with the meaning of the mapping.

Worked Examples

Consider the input:

O2 O7
Step Left Right Encoded Left Encoded Right Result
Parse O2 O7 - - -
Encode letter O→0 O→0 0*10+2=2 0*10+7=7 -
Compare 2 7 2 < 7 hotter hotter

This shows that within the same class, lower digit means hotter.

Now consider:

A0 B9
Step Left Right Encoded Left Encoded Right Result
Parse A0 B9 - - -
Encode letter A→2 B→1 2*10+0=20 1*10+9=19 -
Compare 20 19 20 > 19 cooler cooler

This demonstrates cross-class ordering, where any B-class star is hotter than any A-class star, regardless of digit.

Complexity Analysis

Measure Complexity Explanation
Time O(T) Each test case performs constant-time parsing and comparison
Space O(1) Only fixed mappings and output storage are used

The solution comfortably fits the constraints since even 1000 operations is negligible under strict time limits. Each operation is simple arithmetic and string indexing.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    order = {'O':0,'B':1,'A':2,'F':3,'G':4,'K':5,'M':6}

    def encode(s):
        return order[s[0]] * 10 + (ord(s[1]) - ord('0'))

    t = int(input())
    out = []
    for _ in range(t):
        a, b = input().split()
        va = encode(a)
        vb = encode(b)
        if va < vb:
            out.append("hotter")
        elif va > vb:
            out.append("cooler")
        else:
            out.append("same")

    return "\n".join(out)

# provided samples
assert run("5\nO2 O7\nM9 M2\nG2 G2\nA0 B9\nF8 K1") == "hotter\ncooler\nsame\ncooler\nhotter"

# custom cases
assert run("1\nO0 O0") == "same"
assert run("1\nO0 M9") == "hotter"
assert run("1\nM9 O0") == "cooler"
assert run("1\nG9 G0") == "cooler"
Test input Expected output What it validates
O0 O0 same identical extreme boundary
O0 M9 hotter full-range comparison
M9 O0 cooler reversed full-range
G9 G0 cooler within-class digit ordering reversal

Edge Cases

One edge case is comparing identical stars such as M9 M9. Encoding gives both as 6*10+9=69, and the algorithm correctly outputs same since equality is preserved exactly.

Another edge case is the boundary between classes, such as B0 A9. Encoding yields B0 = 1_10+0=10 and A9 = 2_10+9=29, so B0 is correctly identified as hotter despite having a smaller digit range advantage in the other class.

A final case is intra-class reversal like F8 F1. Encoding gives 38 and 31 respectively, so 38 > 31 and the output is cooler for the first star, matching the fact that higher digit means colder within a class.