CF 105562A - Alphabetical Aristocrats

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Solve time: 52s
Verified: yes

Solution

Problem Understanding

We are given a collection of surnames written as free-form strings. Each surname may contain uppercase letters, lowercase letters, spaces, and apostrophes. The task is to sort these surnames using a custom comparison rule rather than standard lexicographic order over the full string.

The sorting key is defined by ignoring everything in the string before the first uppercase letter. Starting from that first capital letter, we take the remainder of the string exactly as it appears and use that substring as the comparison key. These keys are compared lexicographically using ASCII ordering, which means uppercase letters come before lowercase letters, spaces are significant characters, and apostrophes also participate in ordering.

The output is simply the original surnames reordered according to this rule, with all original formatting preserved.

The constraint n ≤ 1000 and string length ≤ 50 implies that any O(n² log n) or O(n log n) solution is easily sufficient. Even if we build keys for each string and sort them directly, we are working with at most 1000 short strings, so both preprocessing and sorting remain trivial in cost.

A subtle aspect is correctly identifying the first uppercase letter. It is guaranteed that such a letter exists in each string, and the substring starting there is unique across all inputs. This uniqueness ensures the sort order is well-defined and stable ties are unnecessary.

Edge cases that matter are mostly about parsing the key correctly. For example, a string like "van 't Hek" should use "Hek" onward starting from the capital H, not from the beginning. Another case is "DeN bRAnD hEeK", where only the first capital D is relevant, so the key becomes the full string from that D onward, preserving mixed casing.

Approaches

A brute-force approach would explicitly simulate comparisons between every pair of strings during sorting, and for each comparison, scan from the first capital letter onward to compare character by character. Each comparison costs O(L) where L is at most 50, and sorting n items costs O(n log n) comparisons, so overall this becomes O(n log n · L). With n = 1000 and L = 50, this is around 1000 × 10 × 50 = 500,000 character checks, which is already small, but the implementation would repeatedly recompute the “start of key” and rescan strings, causing unnecessary overhead.

The key observation is that the comparison key for each surname depends only on a fixed substring. We can preprocess each string once, extract the suffix starting from its first uppercase character, and then sort using this precomputed key. This turns repeated recomputation inside comparisons into a single linear preprocessing pass.

Once each string has its key, sorting becomes a standard lexicographic sort on these keys while still outputting the original strings.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n log n · L)	O(1) extra	Accepted
Optimal (precompute keys)	O(n L + n log n)	O(n L)	Accepted

Algorithm Walkthrough

We transform each surname into a pair consisting of its sorting key and the original string, then sort by the key.

For each input string, scan from left to right until we find the first uppercase letter. This position defines where the meaningful comparison begins. This is necessary because everything before it is irrelevant under the problem rule.
Extract the substring starting at that uppercase character through the end of the string. This substring is the sorting key. We do not modify casing or remove spaces, since ASCII lexicographic order depends on the exact characters.
Store a pair of (key, original string) for every input line. This preserves the original output requirement while giving us a clean comparison basis.
Sort all pairs by the key using standard lexicographic ordering.
Output the original strings in sorted order, ignoring the keys.

Why it works is based on the fact that every comparison between two surnames depends only on their suffixes starting at the first uppercase letter. By precomputing this suffix once, we ensure that every comparison during sorting is consistent with the problem’s rule. Since sorting only depends on these fixed keys, and the keys are identical to what a manual comparator would generate on demand, the resulting order is identical to the intended order.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input().strip())
    items = []

    for _ in range(n):
        s = input().rstrip("\n")
        start = 0
        for i, ch in enumerate(s):
            if 'A' <= ch <= 'Z':
                start = i
                break
        key = s[start:]
        items.append((key, s))

    items.sort(key=lambda x: x[0])

    for _, s in items:
        print(s)

if __name__ == "__main__":
    solve()

The core of the implementation is the preprocessing loop that finds the first uppercase letter. This is a simple linear scan per string, which is safe because strings are short.

We then slice the string once to form the key. Python’s slicing creates a new string, but given the constraints, this overhead is negligible.

Sorting is done using Python’s built-in timsort with a key function, which compares only the precomputed keys.

Worked Examples

Consider the first sample ordering scenario with a small subset:

Input:

Bakker
van der Steen
Groot Koerkamp

We compute keys:

String	First uppercase position	Key
Bakker	0	Bakker
van der Steen	4	Steen
Groot Koerkamp	0	Groot Koerkamp

After sorting by key:

Key	Original
Bakker	Bakker
Groot Koerkamp	Groot Koerkamp
Steen	van der Steen

Output becomes:

Bakker
Groot Koerkamp
van der Steen

This shows how leading lowercase prefixes are ignored entirely and only the capitalized suffix governs ordering.

Now consider a case with mixed casing:

Input:

DeN bRAnD hEeK
Brand 'Heek
van Brand heek

Keys:

String	Key
DeN bRAnD hEeK	DeN bRAnD hEeK
Brand 'Heek	Brand 'Heek
van Brand heek	Brand heek

Sorting lexicographically compares character by character starting from ASCII values, where space, uppercase, and lowercase differences matter. This produces a deterministic order consistent with raw ASCII comparison.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n + nL)	scanning each string once plus sorting n items by key
Space	O(nL)	storing extracted keys alongside originals

With n ≤ 1000 and L ≤ 50, both bounds are extremely small, and the solution runs comfortably within limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from sys import stdout
    out = []
    input = sys.stdin.readline

    n = int(input().strip())
    items = []
    for _ in range(n):
        s = input().rstrip("\n")
        start = 0
        for i, ch in enumerate(s):
            if 'A' <= ch <= 'Z':
                start = i
                break
        items.append((s[start:], s))

    items.sort(key=lambda x: x[0])
    return "\n".join(s for _, s in items)

# provided-style samples
assert run("3\nBakker\nvan der Steen\nGroot Koerkamp\n") == "Bakker\nGroot Koerkamp\nvan der Steen"

# minimum size
assert run("1\nAaa\n") == "Aaa"

# all same key prefix but different tails
assert run("3\naA\nbA\ncA\n") == "aA\nbA\ncA"

# punctuation and spaces
assert run("2\nvan 't Hek\nvan 't Aa\n") == "van 't Aa\nvan 't Hek"

# mixed casing stability check
assert run("2\nDeN bRAnD hEeK\nden brandHeek\n") == "DeN bRAnD hEeK\nden brandHeek"

Test input	Expected output	What it validates
single element	itself	base case correctness
same suffix structure	sorted order	lexicographic key handling
punctuation cases	deterministic order	ASCII-sensitive comparison
mixed casing	consistent ordering	correct key extraction

Edge Cases

A common pitfall is starting comparison from the beginning of the string instead of the first uppercase letter. For example, in "van der Steen", the correct key is "Steen", not "van der Steen". The algorithm explicitly scans until 'S' and slices from there, ensuring correctness.

Another edge case is strings where the first uppercase letter is not the first character, such as "fakederSteenOfficial". Here the first uppercase is 'S', so the key becomes "SteenOfficial". The preprocessing loop guarantees this position is found before slicing, so no incorrect prefix leaks into the comparison.

Finally, mixed-case sequences like "DeN bRAnD hEeK" rely on raw ASCII ordering after extraction. Since we do not normalize case, uppercase and lowercase differences are preserved exactly as required, and sorting remains faithful to the specification.