CF 106258A - The Easy One

CF 106258A - The Easy One

Rating: -
Tags: -
Solve time: 3m 40s
Verified: yes

Solution

Problem Understanding

The task is extremely small and direct: we are given a sequence of responses from multiple people, where each response is either “this is easy” or “this is hard”. The goal is to decide whether the problem can still be considered easy overall.

The rule is simple. If even one person thinks the problem is hard, the final verdict becomes “hard”. Only if everyone agrees that it is easy do we accept it as easy.

So the input is just a list of binary opinions, where one value represents “easy” and the other represents “hard”. The output is a single word that summarizes whether any “hard” opinion exists.

The constraint is tiny, with at most 100 opinions. That immediately rules out any concern about performance. Even a quadratic or nested scan would be fine, but the structure suggests we only need a single pass.

The main edge cases are mostly about minimal inputs and uniform inputs. If there is only one person and they say easy, the answer must be easy. If that same single person says hard, the answer flips immediately. Another subtle case is when the hard opinion appears at the very end or very beginning, which should not affect correctness but can expose incorrect early termination logic if someone implements it carelessly.

Approaches

A brute-force interpretation would treat the input as a collection of values and repeatedly check subsets or simulate a decision process. For example, one might imagine checking every prefix and recomputing whether the prefix is “clean” of hard opinions. This is correct but unnecessarily repetitive, since each check scans the same elements again. In the worst case, that becomes quadratic in the number of people, about $n^2$, which is still trivial for $n \le 100$ but already indicates wasted work.

The key observation is that the decision depends only on existence. We do not care how many people think it is hard, nor where they appear. We only need to know whether at least one “1” exists in the list. That reduces the entire problem to a single scan with a boolean flag.

Once this is seen, the solution becomes a streaming check: as soon as a hard opinion appears, the answer is determined. There is no need to process further elements for correctness.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (rechecking prefixes)	O(n²)	O(1)	Accepted but unnecessary
Single Pass Scan	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Read the number of opinions. This tells us how many values we will scan, but does not affect logic beyond loop bounds.
2. Initialize a flag hard_found as false. This variable represents whether we have already encountered any opinion marking the problem as hard.
3. Iterate through each response in the list. For each value, check if it equals 1, which represents a “hard” vote.
4. If a hard vote is found, set hard_found to true. At this point, the final answer is already determined logically, even if we continue scanning for simplicity of implementation.
5. After processing all values, output “HARD” if hard_found is true, otherwise output “EASY”.

The reasoning behind early detection is that the problem is effectively an OR-reduction over all inputs. Once one true value is seen, the global OR is already true.

Why it works

The decision rule is equivalent to computing a logical OR over all input values, where 1 contributes true and 0 contributes false. The final answer depends only on whether this OR evaluates to true. Since OR is idempotent and order-independent, scanning sequentially and tracking whether any true value appears preserves correctness. No rearrangement or counting is needed, and no later value can override a previously seen hard opinion.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
arr = list(map(int, input().split()))

hard_found = False

for x in arr:
    if x == 1:
        hard_found = True

print("HARD" if hard_found else "EASY")

The solution reads all values in one go and scans them once. The only state we maintain is a boolean flag. There is no need for early exit, though it could be added without changing correctness.

A common implementation mistake is trying to be too clever with counting or segment logic. That introduces unnecessary complexity. Another issue is forgetting that input is space-separated integers on a single line, which can lead to partial reads if handled incorrectly.

Worked Examples

Example 1

Input:

3
0 0 1

We scan the array step by step.

Index	Value	hard_found
1	0	false
2	0	false
3	1	true

After the scan, at least one hard opinion exists, so the output is:

HARD

This trace shows that a single late occurrence is enough to flip the final decision.

Example 2

Input:

1
0

Index	Value	hard_found
1	0	false

No hard value is ever seen, so the result is:

EASY

This confirms the behavior on the minimal input size.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each opinion is checked exactly once in a linear scan
Space	O(1)	Only a single boolean flag is stored

With $n \le 100$, this is far below any practical limits. Even if constraints were larger, the same linear scan remains optimal since every input must be read at least once.

Test Cases

import sys, io

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

    n = int(input())
    arr = list(map(int, input().split()))

    hard_found = False
    for x in arr:
        if x == 1:
            hard_found = True

    return "HARD" if hard_found else "EASY"

# provided samples
assert run("3\n0 0 1\n") == "HARD"
assert run("1\n0\n") == "EASY"

# custom cases
assert run("5\n0 0 0 0 0\n") == "EASY", "all easy"
assert run("4\n1 1 1 1\n") == "HARD", "all hard"
assert run("6\n0 1 0 0 0 0\n") == "HARD", "single early hard"
assert run("2\n0 1\n") == "HARD", "minimal mix"

Test input	Expected output	What it validates
all zeros	EASY	no hard votes
all ones	HARD	uniform hard case
early hard	HARD	early termination correctness
last hard	HARD	late detection correctness

Edge Cases

One edge case is when there is only one participant. The algorithm still works because the loop runs exactly once and directly reflects that single opinion.

Another case is when the hard opinion appears at the very end. A naive implementation that incorrectly assumes early termination or misreads input might miss it, but the linear scan correctly updates the flag when it appears.

A final case is when all values are easy. Here, the flag remains false throughout, and the output correctly remains “EASY”, confirming that absence of evidence is treated as absence of hardness rather than ambiguity.