Discussion on: Daily Challenge #33 - Did you mean...?

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Here it goes a python one :

I compare by the distances of each char and give less importance to length

fruits = ['cherry', 'pineapple', 'melon', 'strawberry', 'raspberry']

def find_most_similar(word):
    counts = []
    for fruit in fruits:
        count = 0
        if word == fruit:
            return fruit
        else:
            for i in range(min(len(word), len(fruit))):
                count += abs(ord(word[i]) - ord(fruit[i])) + max(0, len(fruit)-len(word))
        counts.append(count)
    min_count = min(counts)
    return fruits[counts.index(min_count)]

Dev Prakash • Aug 12 '19

plzz explain ur code to me ...

Héctor Pascual • Aug 12 '19 • Edited

It's not working 100% but for most basic cases it does, first of all I check if the word passed is inside the fruits list, if not i iterate through each character of the shortest word and I subtract with abs (so that the result is always positive) the value of the characters with the Ord function, then I add the difference of word lengths.

It is kind of a own metric that is not very thinked but it worked to me for these simple tests.

Finally I select as a result the word containing the less value of the metrics I calculated.

After completing the challenge this way I read about levenstein function, it may be a better way to do this.