组成字符串ku的最大次数

The provided code defines a function solution that takes a string and calculates how many times the substring "ku" can be formed from the characters in that string. It does so by counting the occurrences of the characters 'k' and 'u', and since "ku" consists of one 'k' and one 'u', the result is determined by the lesser of the two counts. Here's how the code works in detail:

1. Input Conversion: The input string is converted to lowercase to ensure that the character counting is case-insensitive.

2. Counting Characters: It counts how many times 'k' and 'u' appear in the string using the count method.

3. Calculate Result: The result is the minimum of the two counts, which indicates how many complete "ku" pairs can be formed.

4. Testing the Function: In the if __name__ == '__main__': block, a few test cases are provided to validate the functionality of the solution method.

Here’s a concise breakdown of the solution along with the test cases included:

def solution(s: str) -> int:
    # Convert the string to lowercase to ignore case
    s = s.lower()
    
    # Count occurrences of 'k' and 'u'
    count_k = s.count('k')
    count_u = s.count('u')
    
    # Return the minimum count of 'k' and 'u'
    return min(count_k, count_u)

if __name__ == '__main__':
    # Test cases
    print(solution("AUBTMKAxfuu") == 1)  # Should return True
    print(solution("KKuuUuUuKKKKkkkkKK") == 6)  # Should return True
    print(solution("abcdefgh") == 0)  # Should return True

Explanation of Test Cases:

1. "AUBTMKAxfuu": Contains 1 'k' and 2 'u's, allowing for 1 "ku".
2. "KKuuUuUuKKKKkkkkKK": Contains 10 'k's and 6 'u's, allowing for 6 "ku"s.
3. "abcdefgh": Contains no 'k' or 'u', hence 0 "ku"s.

The code is efficient and works based on simple counting, which is effective for this problem. The overall complexity is O(n), where n is the length of the string, due to the counting operations. Let me know if you need further details or explanations!