3 ms·

Note that you could also solve this with AND/OR, with just a little more complexity. To “turn on” the bit for the new leading character as you slide the window

Note that you could also solve this with AND/OR, with just a little more complexity. To “turn on” the bit for the new leading character as you slide the window forward, you of course OR it with the bitmask for the window, and to “turn off” the bit for the old trailing character, you AND its complemented bit-value with the bitmask. Not as elegant as XOR, but still works, and shows that XOR properties aren’t essential to the solution.

rep_lodsb 4y ago
If the same character occurs twice in the window, the second OR will have no effect but the AND will remove it. The trick is that XOR will leave the bit set if there is an odd number of occurrences, and the only way to have the bit count equal the window size is if every char is unique.