-In this way, we spend $O(1)$ time per each pair of a vertex and a character, making the running time $O(mk)$. The major overhead here is that we copy a lot of transitions from $u$ in the first case, while the transitions of the second case form the trie and sum up to $m$ over all vertices. To avoid the copying of $go[u][c]$, we may use a persistent array data structure, using which we initially copy $go[u]$ into $go[v]$ and then only update values for characters in which the transition would differ. This leads to the $O(m \log k)$ algorithm.
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