Overview

This is a problem about counting the number of “energetic cheering messages” that contain \(K\) or more exclamation marks ! among \(N\) messages.

Analysis

The key insights for this problem are as follows:

1. Each message can be evaluated independently: There’s no need to consider relationships between messages; we can simply examine them one by one in order.

2. The evaluation condition is simple: For each message, we just need to count the occurrences of ! and check whether the count is \(K\) or more.

Thinking Through a Concrete Example

For example, suppose \(N = 3\), \(K = 2\) with the following messages: - Ganbare!! → 2 ! characters → \(2 \geq 2\), so it’s “energetic” ✓ - Hello → 0 ! characters → \(0 \geq 2\) is false, so it’s “not energetic” ✗ - Go!!!Go! → 4 ! characters → \(4 \geq 2\), so it’s “energetic” ✓

In this case, the answer is \(2\).

Why a Naive Approach Is Sufficient

In this problem, each message has a maximum length of \(10^3\), and there are at most \(10^4\) messages. So even if we examine every character of every message, it takes at most \(10^3 \times 10^4 = 10^7\) operations. This is well within the range that can be processed quickly enough.

Algorithm

1. Initialize a variable count to \(0\) to record the number of “energetic cheering messages”
2. For each of the \(N\) messages \(S_i\), repeat the following:
   • Count the number of ! characters in string \(S_i\)
   • If the count of ! is \(K\) or more, increment count by \(1\)
3. Output the final value of count

In Python, you can easily count the number of ! characters in a string \(S\) using the string method S.count('!').

Complexity

• Time complexity: \(O(N \times L)\)
  • \(L\) is the maximum length of each message (at most \(10^3\))
  • This is because the count() method scans all characters for each message
• Space complexity: \(O(L)\)
  • Since we read and process each message one at a time, we only need space to store one message

Implementation Notes

• By utilizing Python’s count() method, you can concisely obtain the occurrence count of a specific character

• If you read and process messages one at a time, there’s no need to store all messages in an array, which improves memory efficiency

• Note that the condition is “\(K\) or more”, so use >= (not >)

  Source Code

N, K = map(int, input().split())
count = 0
for _ in range(N):
    S = input()
    if S.count('!') >= K:
        count += 1
print(count)

This editorial was generated by claude4.5opus.