The most confusing part of the solution is the decode function's inner loop:
var encodingMap = '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': '🌈', ' ': ' ' ; 83 8 create your own encoding codehs answers
To encode a string, you need to look at one character at a time, change it based on a rule, and add it to a new result string. The most confusing part of the solution is
Using a 5-bit scheme, the word "HELLO" would look like this: (7th letter if A=0): 00111 E (4th letter): 00100 L (11th letter): 01010 L (11th letter): 01010 O (14th letter): 01110 🚀 Extra Challenge: Expanding the Set Copied to clipboard 4
# Conceptual Python approach for 8.3.8 # Map characters (A-Z, space) to 5-bit strings encoding_map = 'A': '00000', 'B': '00001', ... def encode_text(message): # Convert message and map to binary using the dictionary return " ".join([encoding_map.get(c, "") for c in message.upper()]) Use code with caution. Copied to clipboard 4. Advanced/Extra Challenge (6 Bits)
def encoder(text): # Create an empty string to store the result result = ""
def decode_message(binary_string): # Assuming a fixed bit length of 5 (based on our dictionary) bit_length = 5 text_output = ""