DailyCodingProblem/Facebook221018 at master · Paroag/DailyCodingProblem · GitHub

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#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Tue Oct 15 15:13:20 2019

@author: oscarbarbier

Good morning! Here's your coding interview problem for today.

This problem was asked by Facebook.

Given the mapping a = 1, b = 2, ... z = 26, and an encoded message,
    count the number of ways it can be decoded.

For example, the message '111' would give 3, since it could be decoded as 'aaa', 'ka', and 'ak'.

You can assume that the messages are decodable. For example, '001' is not allowed.
"""

Solution Oscar

def count_decode (chaine) :
    if len(chaine) == 1 or not chaine :
        return 1
    elif int(chaine[0]) > 2 :
        return(count_decode(chaine[1:]))
    elif chaine[0] == "2" :
        if int(chaine[1]) > 6 :
            return(count_decode(chaine[1:]))
        elif int(chaine[1]) == 0 :
            return(count_decode(chaine[2:]))
        else :
            return(count_decode(chaine[1:]) + count_decode(chaine[2:]))
    elif chaine[0] == "1" :
        if int(chaine[1]) == 0 :
            return(count_decode(chaine[2:]))
        else :
            return(count_decode(chaine[1:]) + count_decode(chaine[2:]))


Solution Charles

 from functools import reduce
    def product(a,b):
    return a * b

def fibo(n):
    def go(a,b,n):
        if n == 0:
            return a
        else:
            return go(b,a+b,n-1)
    return go(1,1,n)

def fb(txt):
    l, s, ind = [], 0, 0
    for c in txt:
        print(c)
        if c in {'1', '2'}:
            s += 1
        else:
            if c == '0':
                s -= 1
            else:
                if s > 0 and txt[ind - 1] == '1':
                    s += 1
                if s > 0 and txt[ind - 1] == '2' and c in {'3', '4', '5', '6'}:
                    s +=1
                if s == 0:
                    s += 1
            l.append(s)
            s = 0
        ind += 1
        if ind == len(txt) and c in {'1', '2'}:
            l.append(s)
    print(l)
    fibs = [fibo(x) for x in l]
    print(fibs)
    return reduce(product, fibs)