Notice that $i <= sqrt(n)$ and $j <= sqrt(n)$.
Now our equality looks like this $p^(i ([sqrt(n)] + 1) + j)^ = q modulo n$.
We can divide by $p^j^$ and get $p^(i[sqrt(n)] + 1)^ = qp^-j^ modulo n$.

Now the application of meet in the middle becomes obvious. We can brute force through the numbers on each side of the equality and find a match.

Using meet in the middle becomes obvious. We can brute force through the numbers on each side of the equality and find a colision.

The algorithm takes O(sqrt(n)) space and O(sqrt(n)) time.
h2. Bidirectional search(interview question)