You perform a fixed number of steps on each character, so we start with O(n * X).

So, now we only have to determine X. For each character, we:
- lookup a character in a dictionary/hashtable, we we determined in a previous Dev102 challenge is either O(1) or O(ln N) depending on the algorithm used.
- push or pop a value from a stack: O(1).

Put them all together, we have N * 1 * 1 or N * ln N * 1. O(N) or prehaps O(N ln N).

Of course, being O(N) isn’t the final word on efficiency. It merely means that the time it taken in given by the function:
T = C*N + K, and your C value (X in the above analysis) is rather high. (O(1) merely means that value is fixed, not that it is low).

I gave the full code on mine at http://www.honestillusion.com

(EDIT: fixed URL link a the bottom)