'1 << stuff' = 'bit shift left 1 by stuff many positions'. Say you have a byte, 8 bits, '1' is 00000001 - if we shift it left 3 bits we have 00001000 (8).
'Stuff' there is just any expression (to understand separately) that evaluates to the number of positions to shift; `<<` is the operator for bit-shifting (left, cf. `>>`).
In brief `(s[i - window_size] as u32 - 'a' as u32)` is finding the character that just left the window on the left, represented as a number, starting from 'a' as 0 so that all 26 fit inside 32 bit positions.
Thank you all who kindly took to answering my question. It makes sense now!
The "stuff" assigns an integer value to the char. 'b' - 'a' = 1, for example. This assigns each char a unique integer value. By shifting 1 up that many times, you assign a unique bit position to each char.
For those new to bitshifting, here are the results of the "stuff" operation that @dahfizz describes:
'a' - 'a' = 0, so take 1 and shift it left 0 times:
00000000000000000000000000000001 = 'a'
'b' - 'a' = 1, so take 1 and shift it left 1 time:
00000000000000000000000000000010 = 'b'
'c' - 'a' = 2, so take 1 and shift it left 2 times:
00000000000000000000000000000100 = 'c'
“ fn run(s: &[char], window_size: usize) -> usize { let mut set = 0u32; for i in 0..s.len() { // Turn on bits as they enter the window set ^= 1 << (s[i] as u32 - 'a' as u32);
} “