introduction.md

Introduction

There are many idiomatic ways to solve Isogram. Among them are:

• You can scrub the input with a list comprehension and then compare the len() of the scrubbed letters with the len() of a set of the scrubbed letters.
• You can scrub the input with a couple of calls to replace() and then compare the len() of the scrubbed letters with the len() of a set of the scrubbed letters.
• You can scrub the input with a re.sub() and then compare the len() of the scrubbed letters with the len() of a set of the scrubbed letters.
• You can filter the input with a re.findall() and then compare the len() of the scrubbed letters with the len() of a set of the scrubbed letters.

General guidance

The key to solving Isogram is to determine if any of the letters in the input are repeated. A repeated letter means the input is not an isogram. The letters are "scrubbed" or filtered so that non-alphabetic characters are not considered when determining an isogram. The occurrence of the letter a and the letter A count as a repeated letter, so Alpha would not be an isogram.

The following four approaches compare the length of the scrubbed letters with the length of a setof the scrubbed letters.

Approach: scrub with a list comprehension

def is_isogram(phrase):
    scrubbed = [ltr.lower() for ltr in phrase if ltr.isalpha()]
    return len(set(scrubbed)) == len(scrubbed)

For more information, check the scrub with list comprehension approach

Approach: scrub with replace()

def is_isogram(phrase):
    scrubbed = phrase.replace('-', '').replace(' ', '').lower()
    return len(scrubbed) == len(set(scrubbed))

For more information, check the scrub with replace() approach

Approach: scrub with re.sub()

import re


def is_isogram(phrase):
    scrubbed = re.compile('[^a-zA-Z]').sub('', phrase).lower()
    return len(set(scrubbed)) == len(scrubbed)

For more information, check the scrub with re.sub() approach

Approach: filter with re.findall()

import re


def is_isogram(phrase):
    scrubbed = "".join(re.findall("[a-zA-Z]", phrase)).lower()
    return len(set(scrubbed)) == len(scrubbed)

For more information, check the filter with re.findall() approach

Other approaches

Besides the aforementioned, idiomatic approaches, you could also approach the exercise as follows:

Other approach: Bit field

Another approach can use a bit field to keep track of used letters. For more information, check the bit field approach.

Which approach to use?

All four set approaches are idiomatic. The replace approach is the fastest.

To compare performance of the approaches, check the Performance article.