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  1. Special String Again
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Special String Again

  • 1337er
     + 0 comments

    I use a deque to keep track of previous counts to find the middle cases. I like this solution because there is a single loop while most other solutions have nested loops.

    from collections import deque

# Complete the substrCount function below.
def substrCount(n, s):
    current_char = None
    total_count = 0
    char_count = 0
    count_deque = deque([0, 0], maxlen=2)

    for i in range(n):
        if s[i] == current_char:
            char_count += 1
        else:
            # check for middle chars
            if i >=3 and count_deque[1] == 1 and s[i-2-char_count] == current_char:
                
                middle_case_count = min([count_deque[0], char_count])
                total_count += middle_case_count
            total_count += sum(range(char_count+1))
            count_deque.append(char_count)
            current_char = s[i]
            char_count = 1
            
    if i >=3 and count_deque[1] == 1 and s[i-1-char_count] == current_char:
        middle_case_count = min([count_deque[0], char_count])
        total_count += middle_case_count
    total_count += sum(range(char_count+1))
    return total_count