recursive function clear solution for Count Morse by Tinus_Trotyl - python coding challenges

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recursive function solution in Clear category for Count Morse by Tinus_Trotyl

D = { "a": ".-",   "b": "-...", "c": "-.-.", "d": "-..",  "e": ".",    "f": "..-.",
      "g": "--.",  "h": "....", "i": "..",   "j": ".---", "k": "-.-",  "l": ".-..",
      "m": "--",   "n": "-.",   "o": "---",  "p": ".--.", "q": "--.-", "r": ".-.",
      "s": "...",  "t": "-",    "u": "..-",  "v": "...-", "w": ".--",  "x": "-..-",
      "y": "-.--", "z": "--..",                                                     }

def count_morse(message: str, letters: str, count=0) -> int:  # set count initial to 0          
    if not letters:
        return count + 1
    for i, letter in enumerate(letters):
        if message.startswith(D[letter]):
            count = count_morse(message[len(D[letter]):], letters[:i] + letters[i + 1:], count)
    return count

April 13, 2023

Comments:

cerankas on Aug. 16, 2023, 1:16 p.m.
<p>Count doesn't need to be passed as argument:</p> <pre class='brush: python'>def count_morse(message: str, letters: str) -> int: count = 0 # local count variable instead of argument if not letters: return count + 1 for i, letter in enumerate(letters): if message.startswith(D[letter]): # increase count instead of assigning, omit count in function call arguments count += count_morse(message[len(D[letter]):], letters[:i] + letters[i + 1:]) return count </pre>

Strings Theory

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