CF 102697097 - Neon Lights
We have several candidate strings that could be written on a neon sign. Each string contains lowercase English letters and spaces. For every candidate, we care only about which different letters appear in it.
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Solve time: 49s
Verified: yes
Solution
Problem Understanding
We have several candidate strings that could be written on a neon sign. Each string contains lowercase English letters and spaces. For every candidate, we care only about which different letters appear in it. Spaces do not count as letters, and repeated occurrences of the same letter count only once.
The required output is the candidate containing the largest number of distinct lowercase letters. If several candidates contain the same number of distinct letters, the earliest one in the input must be printed. The original statement gives a one second time limit and 256 MB memory limit.
The key constraint is actually the alphabet size rather than a large numerical bound. There are only 26 possible lowercase letters. We can inspect every character once and record whether its letter has already appeared. Since the statement does not expose a useful upper bound on the string lengths, the safest complexity target is linear in the total number of input characters. A quadratic scan of each string can become too slow when a string is long, while an approach proportional to the input size is appropriate for a one second limit.
There are several small cases where careless code can choose the wrong string. A string containing only spaces has zero distinct letters, for example 3 followed by abc, a spaces-only line, and ab; the answer is abc, because it has three distinct letters. A solution that counts every character instead of every different letter would incorrectly favor a string such as aaaaaaaa over abcdef.
Spaces also need to be excluded. For example, with input 2, followed by abc and a b, both strings contain exactly three distinct letters, so the first string must remain the answer because ties go to the earliest input. A careless implementation that treats the space as another character would incorrectly give a b a count of four.
Finally, ties must not replace the current answer. For input 3 with abcde, edcba, and abc, both of the first two strings contain five distinct letters, so the correct output is abcde. Replacing the answer whenever the count is greater than or equal to the best count would incorrectly output edcba.
Approaches
The most direct brute-force method is to determine the distinct letters of each string by comparing every character with all earlier characters in that same string. For a string of length L, this can perform roughly L(L−1)/2 character comparisons in the worst case, giving O(L 2 ) time. Across all candidates, the worst-case work is ∑ i O(L i 2 ). The method is correct because a character is counted precisely when no equal character appeared before it, but repeated comparisons are unnecessary.
The brute-force approach works because it explicitly answers the question "has this letter appeared before?" The observation that there are only 26 possible letters lets us store that answer directly. We keep a Boolean array of size 26, indexed by ord(c) - ord('a'). When a letter is encountered for the first time, we mark it and increase the distinct-letter count. Every later occurrence costs constant time and does not change the count.
This changes the work for a string of length L from quadratic to linear. We then maintain the best string seen so far and replace it only when the new distinct-letter count is strictly larger. Using a strict comparison automatically preserves the earliest string on ties.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(∑L i 2 ) | O(1) | Too slow for long strings |
| Optimal | O(∑L i ) | O(1) | Accepted |
The official problem page gives the three sample cases and confirms that spaces are part of the input strings while the objective counts different letters.
Algorithm Walkthrough
- Read the number of candidate strings. We process them in input order because the tie-breaking rule depends on that order.
- For each string, create a Boolean array of length 26, initially all false. Each position represents whether one lowercase letter has appeared in the current string.
- Scan the string character by character. If the character is a space, ignore it because spaces are not letters. Otherwise convert the character to an index from 0 through 25.
- If that letter has not been seen before, mark it as seen and increment the current distinct-letter count. Repeated occurrences do nothing.
- Compare the current count with the best count found so far. Replace the stored answer only when the current count is strictly greater. Equality must leave the old answer untouched because the old candidate appeared earlier.
- After all strings have been processed, print the stored candidate.
Why it works
For every processed string, the Boolean array contains exactly the set of lowercase letters that have appeared in that string. Each letter contributes one to the count exactly when its first occurrence is encountered, so the final count is exactly the number of distinct letters. The global best is updated only for a strictly larger count, which means it always represents the earliest candidate among all strings attaining the maximum count. Thus the printed string satisfies both the primary condition and the tie-breaking condition.
Python Solution
Pythonimport sysinput = sys.stdin.readline
def solve(): n = int(input())
best_string = "" best_count = -1
for _ in range(n): s = input().rstrip("\n")
seen = [False] * 26 count = 0
for ch in s: if ch == ' ': continue
idx = ord(ch) - ord('a') if not seen[idx]: seen[idx] = True count += 1
if count > best_count: best_count = count best_string = s
print(best_string)
if __name__ == "__main__": solve()
The input is read with readline so spaces inside each candidate string are preserved. We remove only the trailing newline with rstrip("\n"). Using strip() would also remove leading and trailing spaces, which would change the original string and is unnecessary.
The seen array is recreated for every candidate because distinctness is measured independently for each string. The expression ord(ch) - ord('a') maps a to 0 and z to 25, exactly matching the 26-element array.
The comparison uses > rather than >=. This is the implementation of the tie-breaking rule, and changing it is a common source of wrong answers.
There is no integer-overflow concern in Python. The count can never exceed 26 anyway, because there are only 26 lowercase letters.
Worked Examples
For Sample 1, the important state is the number of different letters discovered in each candidate.
| Candidate | Distinct letters | Best count | Best string |
|---|---|---|---|
coderams club |
10 | 10 | coderams club |
neon lights |
8 | 10 | coderams club |
kraftwerk |
7 | 10 | coderams club |
this is a long string of text |
14 | 14 | this is a long string of text |
abababababababababcabababab |
3 | 14 | this is a long string of text |
short |
5 | 14 | this is a long string of text |
The fourth string becomes the answer because it introduces 14 different letters. The long string of repeated a, b, and c demonstrates why occurrences must not be counted individually.
For Sample 3, the tie behavior is the central part of the trace.
| Candidate | Distinct letters | Best count | Best string |
|---|---|---|---|
abcde |
5 | 5 | abcde |
abc |
3 | 5 | abcde |
edcba |
5 | 5 | abcde |
When edcba reaches the same count of 5, the answer is not replaced. The first candidate with that maximum count remains the result, exactly as required.
The second official sample has the same full-alphabet idea in a slightly different form. The pangram-like first string contains all 26 lowercase letters, while the second string contains only 25, so the first string wins.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(∑L i ) | Every character of every candidate is inspected once. |
| Space | O(1) | The seen array always contains exactly 26 entries, independent of input size. |
The algorithm performs a constant amount of work per input character and uses only a fixed-size alphabet array. With the stated one second and 256 MB limits, this is the appropriate linear-time solution.
Test Cases
The statement does not publish an explicit maximum string-length bound, so the maximum-size test below uses a large generated candidate rather than claiming a particular official maximum length.
Python# helper: run solution on input string, return output stringimport sysimport io
def solve(): input = sys.stdin.readline
n = int(input()) best_string = "" best_count = -1
for _ in range(n): s = input().rstrip("\n")
seen = [False] * 26 count = 0
for ch in s: if ch == ' ': continue
idx = ord(ch) - ord('a') if not seen[idx]: seen[idx] = True count += 1
if count > best_count: best_count = count best_string = s
print(best_string)
def run(inp: str) -> str: old_stdin = sys.stdin
| Test input | Expected output | What it validates |
|---|---|---|
1 followed by a |
a |
Minimum number of candidates |
aaaaaaaa, bbbb, abc |
abc |
Repeated letters must be counted once |
abc, abc , a b c |
abc |
Spaces are ignored and ties preserve the first candidate |
| A repeated 26-letter alphabet versus 25 distinct letters | The 26-letter candidate | Large input and complete alphabet coverage |
Edge Cases
A candidate containing only spaces has no distinct letters. For example, with input 2 followed by and a, the first candidate receives count 0 and the second receives count 1, so the answer is a. The algorithm skips every space and correctly leaves the first count at zero.
A string with many repeated copies of one letter must not be rewarded for its length. For input 3, followed by aaaaaaaa, bbbb, and abc, the counts are 1, 1, and 3, so the output is abc. The Boolean array records a only once regardless of how many times it occurs.
Spaces between letters are also harmless. For input 2, followed by a b c and abc, both candidates contain exactly three distinct letters. The first candidate remains the answer because the update condition is strictly greater, not greater than or equal.
Finally, two different strings can contain exactly the same set of letters in a different order. With 3, abcde, abc, and edcba, the first and third candidates both have five distinct letters. The third candidate never replaces the first because its count is equal rather than greater, giving the required output abcde.