[dzaima]

A rough sketch for a more direct way to extend the XOR trick to finding more than two differences:

For e.g. 3 differences: instead of a binary xor (i.e. binary-digit-wise sum mod 2), do a binary-digit-wise sum mod 3 (negating one input); a 0 (mod 3) sum result for a given bit means that the bit is the same in all entries, and 1 (mod 3) or 2 (mod 3) mean that you can partition on the bit, resulting in partitions with sizes `sum` and `input_different_element_count - sum`; then you repeat this recursively until they reach containing just 1 difference. (rounding the modulo up to the next power of two instead of odd modulos for the summing is perfectly fine, the final infinite-precision sum is in the range of [0; diffcount] anyway)

Extends trivially to more than 3 differences, and collapses to the basic trick for 2 differences. The accumulator size is O(log(diffcount) * element_size), but the recursive partitioning takes O(n) space or O(diffcount * n) time (plus some logarithm something maybe). Tradeoffs are probably reasonably possible, but the basic hashset approach can reduce its O(n) space requirement at the cost of taking >O(n) time too by partitioning on a hash.