Bit field

Pangram
Pangram in C#
public static class Pangram
{
    public static bool IsPangram(string input) {
        int phrasemask = 0;
        foreach (char letter in input)
        {
            // a-z
            if (letter > 96 && letter < 123)
                phrasemask |= 1 << (letter - 'a');
            // A - Z
            else if (letter > 64 && letter < 91)
                phrasemask |= 1 << (letter - 'A');
        }
        //26 binary 1s
        return phrasemask == 67108863;        
    }
}

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

The string loops through its characters and looks for a character being a through z or A through Z. The ASCII value for a is 97, and for z is 122. The ASCII value for A is 65, and for Z is 90.

  • If the lower-cased 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 upper-cased 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 an unsigned 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 if the phrasemask value is the same value as when all 26 bits are set, which is 67108863.

This approach is fast, but it may be considered to be more idiomatic of the C language than C#.

22nd May 2024 · Found it useful?