Bit field using a for loop

Isogram
Isogram in Python 
A_LCASE = 97
Z_LCASE = 122
A_UCASE = 65
Z_UCASE = 90


def is_isogram(phrase):
    letter_flags = 0

    for ltr in phrase:
        letter = ord(ltr)
        if letter >= A_LCASE and letter <= Z_LCASE:
            if letter_flags & (1 << (letter - A_LCASE)) != 0:
                return False
            else:
                letter_flags |= 1 << (letter - A_LCASE)
        elif letter >= A_UCASE and letter <= Z_UCASE:
            if letter_flags & (1 << (letter - A_UCASE)) != 0:
                return False
            else:
                letter_flags |= 1 << (letter - A_UCASE)
    return True

This solution uses the ASCII value of the letter to set the corresponding bit position.

Some constants are defined for readability in the function. Python doesn't enforce having real constant values, but using all uppercase letters is the naming convention for a Python constant. It indicates that the value is not intended to be changed.

  • The ASCII value for a is 97.

  • The ASCII value for z is 122.

  • The ASCII value for A is 65.

  • The ASCII value for Z is 90.

  • A for loop is used to iterate the characters in the input phrase.

  • The ord() function is used to get the ASCII value of the letter.

  • The if statements look for a character being a through z or A through Z.

  • If the lowercase letter is subtracted by 97, then a will result in 0, because 97 minus 97 equals 0. z would result in 25, because 122 minus 97 equals 25. So a would have 1 shifted left 0 places (so not shifted at all) and z would have 1 shifted left 25 places.

  • If the uppercase letter is subtracted by A, then A will result in 0, because 65 minus 65 equals 0. Z would result in 25, because 90 minus 65 equals 25. So A would have 1 shifted left 0 places (so not shifted at all) and Z would have 1 shifted left 25 places.

In that way, both a lower-cased z and an upper-cased Z can share the same position in the bit field.

So, for a thirty-two bit integer, if the values for a and Z were both set, the bits would look like

      zyxwvutsrqponmlkjihgfedcba
00000010000000000000000000000001

We can use the bitwise AND operator to check if a bit has already been set. If it has been set, we know the letter is duplicated and we can immediately return False. If it has not been set, we can use the bitwise OR operator to set the bit. If the loop completes without finding a duplicate letter (and returning False), the function returns True.

18th Sep 2024 · Found it useful?

Other Approaches to Isogram in Python

Other ways our community solved this exercise
def is_isogram(phrase):
    scrubbed = [ltr.lower() for ltr in phrase if ltr.isalpha()]
    return len(set(scrubbed)) == len(scrubbed)
Scrub with a list comprehension

Use a list comprehension and set to return the answer. 

def is_isogram(phrase):
    scrubbed = phrase.replace('-', '').replace(' ', '').lower()
    return len(scrubbed) == len(set(scrubbed))
Scrub with a series of replace calls

Use a series of replace calls and set to return the answer. 

import re


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

Use a re.sub and set to return the answer. 

import re


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

Use a re.findall and set to return the answer.