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May 30, 2021 14:47
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AlgorithmFriday_Week8
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| # First define the list of chars for all digits | |
| num_dict = {'2': ['a', 'b', 'c'], | |
| '3': ['d', 'e', 'f'], | |
| '4': ['g', 'h', 'i'], | |
| '5': ['j', 'k', 'l'], | |
| '6': ['m', 'n', 'o'], | |
| '7': ['p', 'q', 'r', 's'], | |
| '8': ['t', 'u', 'v'], | |
| '9': ['w', 'x', 'y', 'z'], | |
| # just incase 0 and 1 is part of input, we can safely skip | |
| '0': [''], '1': [''],} | |
| # using a recursive approach | |
| def all_combinations(num): | |
| if num==None: return [] | |
| try: | |
| num = str(num) | |
| if len(num)==0: return [] | |
| if len(num)==1: return num_map[num] | |
| # function that returns comb of only 2 digits | |
| def comb_2(prev, next): return [x+y for x in prev for y in next] | |
| prev = num_dict[num[0]] | |
| for i in num[1:]: | |
| # print(prev) | |
| prev = comb_2(prev, num_dict[i]) | |
| return prev | |
| except: | |
| return [] | |
| all_combinations(23) |
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Hello @Julisam, congratulations 🎉 your solution has been selected as one of the winning solutions in Week 8 of #AlgorithmFridays.
Your solution is clean, readable and passed the tests. It's nice to see an iterative approach for solving this problem. What do you think the time complexity of your solution is?
Out of the many winning solutions, only 3 will be selected for the $20 prize in a raffle draw. The raffle draw will hold today, Friday June 4 at 3.00pm WAT (7.00 am PST)
If you are interested in being a part of the raffle draw, please send me a DM on Twitter @meekg33k so I can share the event invite with you.
NB: Only solutions of participants who indicated interest in the raffle draw will be considered.
Thanks once again for participating and see you later today for Week 9 of #AlgorithmFridays.