Insert space for additional async exercises

exercises/101_bit_manipulation2.zig

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+//
+// Another useful application for bit manipulation is setting bits as flags.
+// This is especially useful when processing lists of something and storing
+// the states of the entries, e.g. a list of numbers and for each prime
+// number a flag is set.
+//
+// As an example, let's take the Pangram exercise from Exercism:
+// https://exercism.org/tracks/zig/exercises/pangram
+//
+// A pangram is a sentence using every letter of the alphabet at least once.
+// It is case insensitive, so it doesn't matter if a letter is lower-case
+// or upper-case. The best known English pangram is:
+//
+//           "The quick brown fox jumps over the lazy dog."
+//
+// There are several ways to select the letters that appear in the pangram
+// (and it doesn't matter if they appear once or several times).
+//
+// For example, you could take an array of bool and set the value to 'true'
+// for each letter in the order of the alphabet (a=0; b=1; etc.) found in
+// the sentence. However, this is neither memory efficient nor particularly
+// fast. Instead we choose a simpler approach that is very similar in principle:
+// We define a variable with at least 26 bits (e.g. u32) and set the bit for
+// each letter that is found in the corresponding position.
+//
+// Zig provides functions for this in the standard library, but we prefer to
+// solve it without these extras, after all we want to learn something.
+//
+const std = @import("std");
+const ascii = std.ascii;
+const print = std.debug.print;
+
+pub fn main() !void {
+    // let's check the pangram
+    print("Is this a pangram? {}!\n", .{isPangram("The quick brown fox jumps over the lazy dog.")});
+}
+
+fn isPangram(str: []const u8) bool {
+    // first we check if the string has at least 26 characters
+    if (str.len < 26) return false;
+
+    // we use a 32 bit variable of which we need 26 bits
+    var bits: u32 = 0;
+
+    // loop about all characters in the string
+    for (str) |c| {
+        // if the character is an alphabetical character
+        if (ascii.isAscii(c) and ascii.isAlphabetic(c)) {
+            // then we set the bit at the position
+            //
+            // to do this, we use a little trick:
+            // since the letters in the ASCII table start at 65
+            // and are numbered sequentially, we simply subtract the
+            // first letter (in this case the 'a') from the character
+            // found, and thus get the position of the desired bit
+            bits |= @as(u32, 1) << @truncate(ascii.toLower(c) - 'a');
+        }
+    }
+    // last we return the comparison if all 26 bits are set,
+    // and if so, we know the given string is a pangram
+    //
+    // but what do we have to compare?
+    return bits == 0x..???;
+}