Bit field

A_LCASE = 97
A_UCASE = 65
ALL_26_BITS_SET = 67108863


def is_pangram(sentence):
    letter_flags = 0
    for letter in sentence:
        if 'a' <= letter <= 'z':
            letter_flags |= 1 << ord(letter) - A_LCASE
        elif 'A' <= letter <= 'Z':
            letter_flags |= 1 << ord(letter) - A_UCASE
    return letter_flags == ALL_26_BITS_SET

This solution uses the ASCII value of the letter to set the corresponding bit position. First, some constant values are set.

These values will be used for readability in the body of the is_pangram function. The ASCII value for a is 97. The ASCII value for A is 65. The value for all of the rightmost 26 bits being set is 67108863.

• The for loop loops through the characters of the sentence.
• Each letter is tested for being a through z or A through Z.
• If the lowercased letter is subtracted by a, 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 uppercased 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 OR operator to set the bit. After the loop completes, the function returns True if the letter_flags value is the same value as when all of the rightmost 26 bits are set, which is 67108863.

Although this approach is usually very fast in some other languages, it is comparatively slow in Python.