CF 102697133 - Question-able Difficulty

Rating: -
Tags: -
Solve time: 2m 17s
Verified: yes

Solution

Problem Understanding

The task is deliberately simple. We receive one string, representing a line of text, and must decide whether that line should be considered a question. The definition is purely syntactic: the final character of the string must be ?. If it is, we print YES; otherwise we print NO. The official problem specifies a single input string and a one-second, 256 MB limit.

The key distinction is between a question mark appearing somewhere in the text and a question mark being the final character. For example, Is this? is a question, while Is this? really is not, because the final character is y.

There is no meaningful large numerical constraint here because the input consists of one string. The running time is naturally proportional to the amount of input that must be read. Even if the string were very long, an algorithm that inspects every character would still be linear in the input size and easily fits the stated time limit for ordinary input sizes. More importantly, we can avoid inspecting the whole string after reading it, because the answer depends on exactly one character.

The first edge case is a string containing only a question mark.

The correct output is YES. A careless implementation that expects the question mark to have another character before it could reject this valid case.

The second edge case is a question mark that appears before the end.

hello? world

The correct output is NO. Searching the string for the presence of ? would incorrectly return YES, because the question mark is not the final character.

The third edge case involves trailing whitespace.

hello?

The correct result is NO if that trailing space is actually part of the input string, because the final character is a space rather than ?. This is why the implementation should preserve the meaningful contents of the input line rather than blindly searching for a question mark. In Python, rstrip('\n') removes the line ending while preserving other trailing characters.

Approaches

A straightforward brute-force interpretation is to scan every character and determine which one is the final character. This is correct because after processing the complete string, the last character encountered is exactly the character that controls the answer. For a string of length L, this performs L character inspections in the worst case.

There is no genuine performance failure here. The input itself contains L characters, so reading it already costs O(L) time. A full scan is therefore asymptotically optimal if we count input processing. The only improvement available after the string has been read is to avoid the scan and directly inspect its final character.

The useful observation is that the problem contains no information about the middle of the string that affects the answer. Once the input line has been read, only s[-1] matters. We can consequently replace the full scan with one direct character comparison.

The brute-force scan works because it eventually discovers the final character, but it performs work on characters that cannot affect the result. The observation that the answer depends only on the final position lets us reduce the post-input computation to O(1).

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(L)	O(1)	Accepted
Optimal	O(L) including input, O(1) after input	O(1)	Accepted

Here L is the length of the input string. The distinction between the two approaches is mostly educational rather than practical for this problem, since input reading already takes linear time.

Algorithm Walkthrough

Read the entire input line and remove only its newline character. We keep other characters unchanged because the actual final character determines the answer.
Access the final character using s[-1]. Python's negative indexing gives the character at the last position directly, so there is no need to scan the string.
Compare that character with ?. If they are equal, print YES; otherwise, print NO.

Why it works

The algorithm checks exactly the property used in the definition of a question: whether the final character is ?. If s[-1] == '?', the string satisfies the definition and the algorithm prints YES. If s[-1] != '?', the definition is not satisfied and the algorithm prints NO. No other character can change this decision, so inspecting the final character is sufficient.

Python Solution

import sys
input = sys.stdin.readline

s = input().rstrip('\n')

print("YES" if s[-1] == '?' else "NO")

The input is read with input(), which includes the newline when one is present. We remove exactly that newline with rstrip('\n'). Using strip() would be less precise because it also removes spaces and other whitespace from the ends of the string, potentially changing what the final character actually is.

The expression s[-1] accesses the final character directly. This avoids an unnecessary loop over the string.

The comparison is performed before choosing the output. A question mark produces YES, and every other final character produces NO.

The problem guarantees a valid input string, so accessing s[-1] is safe. There is no integer arithmetic, so integer overflow is irrelevant, and there are no indexing calculations involving an explicit length that could introduce an off-by-one error.

Worked Examples

Sample 1

The input is:

Welcome to the CodeRams contest?

The relevant state is:

String	Final character	Condition	Output
`Welcome to the CodeRams contest?`	`?`	`? == ?`	`YES`

The entire sentence does not need to be analyzed. The final character is a question mark, so the condition is satisfied.

Sample 2

The input is:

We hope you like the problems!

The relevant state is:

String	Final character	Condition	Output
`We hope you like the problems!`	`!`	`! == ?` is false	`NO`

The exclamation mark at the end immediately determines the answer. Even if a question mark appeared somewhere earlier in the sentence, it would not matter.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(L)	Reading the input string takes O(L); the final-character check is O(1).
Space	O(L)	Python stores the input string, while the algorithm uses O(1) additional space.

The algorithm is comfortably within the one-second and 256 MB limits given by the official problem. The input-reading cost dominates the computation, since checking the final character itself takes constant time.

Test Cases

import sys
import io

def solve():
    input = sys.stdin.readline
    s = input().rstrip('\n')
    print("YES" if s[-1] == '?' else "NO")

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 samples
assert run("Welcome to the CodeRams contest?\n") == "YES\n", "sample 1"
assert run("We hope you like the problems!\n") == "NO\n", "sample 2"
assert run("Good luck today\n") == "NO\n", "sample 3"

# Minimum-size meaningful input
assert run("?\n") == "YES\n", "single question mark"

# All characters are ordinary text
assert run("hello\n") == "NO\n", "ordinary word"

# Question mark appears internally but is not final
assert run("hello? world\n") == "NO\n", "question mark not at end"

# Boundary case with a trailing space
assert run("hello? \n") == "NO\n", "trailing space changes final character"

Test input	Expected output	What it validates
`?`	`YES`	Minimum-size meaningful input and direct final-character access
`hello`	`NO`	Ordinary text without a question mark
`hello? world`	`NO`	Prevents confusing the presence of `?` with `?` at the end
`hello?`	`NO`	Checks that trailing characters are not discarded accidentally

Edge Cases

For the single-character input ?, the string has length one and s[-1] is that same question mark. The comparison succeeds and the output is YES. A solution that assumes the string must contain multiple characters would fail here.

For hello? world, the question mark is present but is followed by a space and more text. The algorithm never searches for a question mark anywhere in the string. It reads only s[-1], which is d, so it correctly prints NO.

For an input such as hello? , the final character is a space. The implementation uses rstrip('\n') rather than strip(), so that space remains part of the string. Consequently s[-1] is the space and the answer is NO.

The central invariant is simple: at the moment the answer is produced, the character being compared is exactly the final character of the input line. Since the problem's definition depends exclusively on that character, every possible valid string is classified correctly.