This seems a bit overcomplicated to me.
Using a bitmap to track letters you've seen is a great idea. But the parity and counting gives me a headache. How about you start with a zero-size window at the start of the string, then iteratively try to grow it by moving the end forward until it's long enough, and if the character added would be a duplicate, moving the start forward until it's no longer a duplicate? Sort of inchworm-style movement.
I am not smart enough to write Rust, so here it is in Java:
private static int run(String input, int windowSize) {
int windowStart = 0; // start at the start
int windowEnd = 0; // the window is initially zero-size
int lettersSeen = 0; // a bitmap of letters we have seen
while (windowEnd - windowStart = input.length()) throw new IllegalArgumentException("No unique window found");
int letterToAdd = 1
This is not quite optimal in terms of bit operations - i mask the bitmap in the inner loop condition, but you could just test letterToRemove against letterToAdd and break if they're the same. I think that's a bit less readable though.
Note that although this has two nested loops, they don't both independently range over the whole string, so this is not O(n^2). The outer loop moves windowEnd over the whole string, and the inner one moves windowStart over the whole string bit by bit. Each index visits each position in the string at most once.