For each vocabulary word (V_i), we can consider changing each letter to every other letter. This gives us (3*L) possibilities for words differing by exactly one letter. We can store a hash of each of these "(1)-mistaken" strings in a map.
To avoid an (\mathcal{O}(L)) time factor for hashing each possibility, we can use a rolling hash. For example, to hash an array (A_{1..L}), we can define (h(A_1, ..., A_L) := (A_1 * p^1 + ... + A_L* p^L) \text{ mod } P), where (p) and (P) are prime numbers. We can then precompute modded prefix sums in (\mathcal{O}(L)). Given any (h(A_1, ..., A_L)), we will be able to to replace a value (A_i) with (A_i') without rehashing, by computing ((h(A_1, ..., A_L) - h(A_1, ..., A_{i}) + h(A_1, ..., A_{i - 1}) + A_i' * p^i) \text{ mod } P) in (O(1)) time (where the second and third terms are prefix sums).
For each hash (h), we can compute two tables of counts:
- (\text{H}[h][j]) storing the number of (1)-mistaken strings with hash (h), mistaken at index (j)
- (\text{Hsum}[h] = \sum_j H[h][j]) storing the total number of (1)-mistaken strings with hash (h)
For each query (W_i), we can first compute (h(W_i)) in (\mathcal{O}(L)). Then, we can go through all (3*L) possibilities of strings derived from swapping (W_i) at each index (j_1) (deriving a new hash (h') from (h(W_i)) in (\mathcal{O}(1)) time as described above). We can see if this swap matches any previously (1)-mistaken strings (mistake at some other index (j_2)). Each such replacement will contribute ((\text{Hsum}[h'] - \text{H}[h'][j_1])/2) to the answer. That is, we need to exclude hashes where (j_1) equals a previously-mistaken index, and divide by (2) to avoid double counting ((j_1, j_2)) and ((j_2, j_1)).
The overall time and space complexity is (\mathcal{O}{(3L(N + Q))}).