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  1. Practice
  2. Algorithms
  3. Implementation
  4. Birthday Chocolate
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Birthday Chocolate

  • vishoo7 + 1 comment

    This is adding from idx to idx + m. It's more efficient to keep a running sum and subtract the beginning of the window and add the next value:

    def solve(n, s, d, m):
    if m > n:
        return 0
    
    i = j = 0
    window_sum = 0
    
    while j < m:
        window_sum += s[j]
        j += 1
        
    count = 1 if window_sum == d else 0
    
    while j < n:
        window_sum -= s[i]
        window_sum += s[j]
        
        if window_sum == d:
            count += 1
        i += 1
        j += 1
        
    return count