CF 1030A - In Search of an Easy Problem

CF 1030A - In Search of an Easy Problem

Rating: 800
Tags: implementation
Solve time: 3m 6s
Verified: yes

Solution

Problem Understanding

We are given a small group of people, each giving a binary opinion about a single problem. Each response is either 0, meaning the person considers the problem easy, or 1, meaning the person considers it hard. The coordinator’s rule is simple: the problem stays as it is only if everyone considers it easy. If even one person calls it hard, the problem is rejected.

The task is to determine whether any “hard” opinion appears among the responses. If at least one value in the list is 1, the answer becomes “HARD”. Otherwise, if all values are 0, the answer is “EASY”.

The constraint on n is at most 100, which means any linear scan is trivially fast. Even checking all subsets or recomputing multiple passes would be acceptable in terms of performance, but unnecessary. The structure of the problem already suggests that a single pass is sufficient.

There are no tricky hidden edge cases involving ordering or arithmetic. The only meaningful edge case is when n equals 1, since the decision depends entirely on a single value. Another is when all values are 0, which should correctly return “EASY”. A common mistake is incorrectly assuming the presence of 0 implies easiness without verifying that no 1 exists.

A second subtle issue is output formatting. Any capitalization variant is accepted, but the logical condition must still be correct regardless of formatting choice.

Approaches

A brute-force approach would simply scan through the list and explicitly count how many people say “hard”. If the count is greater than zero, we print “HARD”, otherwise “EASY”. This is already optimal in structure, but it can be described as repeatedly checking each element and accumulating a result.

In the worst case, we examine all n values, performing constant work per element. Since n is at most 100, even more complicated approaches would not matter, but the key observation is that we do not need to process relationships between elements, only detect existence of a single value.

The insight is that this is an existence-check problem rather than an aggregation problem. We do not need the total number of 1s, only whether at least one exists. That allows early termination as soon as we find a single 1.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (count all)	O(n)	O(1)	Accepted
Optimal (early stop check)	O(n)	O(1)	Accepted

Algorithm Walkthrough

Optimal approach

1. Read the integer n, which tells us how many opinions we will process. This defines the number of iterations we will perform.
2. Read the list of n integers representing opinions. Each value is either 0 or 1, so we only care about detecting the presence of 1.
3. Initialize a flag variable, for example has_hard = False. This variable tracks whether we have seen at least one hard opinion so far.
4. Iterate through each value in the list. For each value, check whether it equals 1.
5. If a value equals 1, set has_hard = True. At this point, we already know the final answer will be “HARD”, so we can optionally stop early.
6. After processing all values (or stopping early), check the flag. If has_hard is True, output “HARD”. Otherwise output “EASY”.

Why it works

The decision depends only on whether there exists at least one element equal to 1. The flag maintains the invariant that after processing any prefix of the array, it correctly reflects whether a 1 has been seen in that prefix. Since the final array is just the full prefix, the flag correctly represents whether any hard response exists in the input. This guarantees correctness without needing any additional structure or computation.

Python Solution

import sys
input = sys.stdin.readline

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

has_hard = False

for x in arr:
    if x == 1:
        has_hard = True
        break

if has_hard:
    print("HARD")
else:
    print("EASY")

The solution reads the input in linear time and stores the responses in a list. It then scans through the list once, stopping immediately when it encounters a 1. The early break is not required for correctness but reflects the existence-check nature of the problem and avoids unnecessary iterations.

The key implementation detail is the boolean flag. It avoids counting and directly encodes the condition we care about. Another subtle point is handling input splitting correctly, since all responses are given on a single line.

Worked Examples

Example 1

Input:

3
0 0 1

Step	Value	has_hard
Start	-	False
1	0	False
2	0	False
3	1	True

The third value triggers the condition immediately. Once a single 1 is found, the result becomes “HARD”, and further processing is unnecessary.

Output:

HARD

Example 2

Input:

4
0 0 0 0

Step	Value	has_hard
Start	-	False
1	0	False
2	0	False
3	0	False
4	0	False

No value changes the flag, so the array contains no hard opinions.

Output:

EASY

This demonstrates the complementary case where the condition never activates.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each element is checked at most once, with optional early termination
Space	O(1)	Only a boolean flag is used beyond input storage

The constraints cap n at 100, so even a full scan is negligible. The solution easily fits within both time and memory limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys as _sys
    from collections import deque

    data = inp.strip().split()
    n = int(data[0])
    arr = list(map(int, data[1:]))

    has_hard = False
    for x in arr:
        if x == 1:
            has_hard = True
            break

    return "HARD" if has_hard else "EASY"

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

# all easy
assert run("5\n0 0 0 0 0\n") == "EASY"

# single element hard
assert run("1\n1\n") == "HARD"

# single element easy
assert run("1\n0\n") == "EASY"

# alternating values
assert run("6\n0 1 0 1 0 0\n") == "HARD"

Test input	Expected output	What it validates
5 zeros	EASY	all-easy case
1	HARD	minimum hard case
0	EASY	minimum easy case
mixed	HARD	detection anywhere in array

Edge Cases

The most important edge case is when n equals 1. For input 1 0, the loop processes a single element and never sets the flag, so the output is correctly “EASY”. For input 1 1, the flag is set immediately and the output becomes “HARD”. This confirms that the algorithm behaves correctly even when there is no “bulk” input structure.

Another edge case is when all values are 0. For example, 4 0 0 0 0 keeps the flag unchanged throughout the iteration, and the final decision correctly outputs “EASY”. This shows that absence detection is handled properly.

Finally, cases with multiple 1s, such as 5 0 1 0 1 0, test whether the algorithm incorrectly depends on counting. Since the flag is set on the first 1 and never reset, the final result remains “HARD”, independent of how many additional 1s appear.