HHMathewChan
Posted on June 25, 2022
Question
- Given a name, return the letter with the highest
indexin alphabetical order,- with its corresponding index,
- in the form of a string.
- You are prohibited to use
max()nor is reassigning a value to the alphabet list allowed.
Example
alphabet = ["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"]
alphabet_index(alphabet, "Flavio") ➞ "22v"
alphabet_index(alphabet, "Andrey") ➞ "25y"
alphabet_index(alphabet, "Oscar") ➞ "19s"
My attempt
- algorithm
>find the letter with the highest index in alphabetical order
initialise highest_index to 0
initialise highest_index_character to empty string
>>Find the highest index of the character and the relevant character in the name
initialise highest_index to 0
initialise highest_index_character to empty string
for each character in the name:
if character's index in alphabet is higher than in highest_index
set highest_index to character's index
set highest_index_character to character
return string of highest_index and highest_index_character
- code
alphabet = ["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 alphabet_index(alphabet: list, name: str):
# The name are in upper case, but alphabet is in lower case
name = name.lower()
# initialise highest_index to 0
highest_index = 0
# initialise highest_index_character to empty string
highest_index_character = str()
# for each character in the name:
for character in name:
# if character's index in alphabet is higher than in highest_index
if alphabet.index(character) + 1 > highest_index:
# set highest_index to character's index
highest_index = alphabet.index(character) + 1
# set highest_index_character to character
highest_index_character = character
# return string of highest_index and highest_index_character
return str(str(highest_index) + highest_index_character)
Other solution
- don't find any particular impressive code. If there is shorter way to do this, please let me know.
My reflection
- This takes me long time to write out the algorithm in appropriate formate, need to practice more
Credit
- challenge find on edabit
💖 💪 🙅 🚩
HHMathewChan
Posted on June 25, 2022
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