For a hash trie, the depth is bounded by the log of the number of elements: O(log(n)). I think O(log(k)) would mean that the bound is based on the size of the

For a hash trie, the depth is bounded by the log of the number of elements: O(log(n)). I think O(log(k)) would mean that the bound is based on the size of the largest key. This may be true of regular tries, but not of hash tries.