FTR: This can be done with "only" two loops (opposed to the "naive" 3 in the article) -- without any extra data structures. It's sufficient to keep track of how many characters have been unique so far. For each "new" character, check whether it's different from all of them. If so, increase count. Otherwise, reset count to the distance to the match.

  def main():
    count = 1
    for i in range(1, len(SIGNAL)):
      for j in range(0, count):
        if SIGNAL[i - j - 1] == SIGNAL[i]:
          count = j
          break

      count = count + 1
  
      if count == 4:
        print(i + 1)
        break

If we consider the word length a constant this should be O(n).