Approach #1: Depth-First Search [Accepted]

Intuition

Two words are similar if they are the same, or there is a path connecting them from edges represented by pairs.

We can check whether this path exists by performing a depth-first search from a word and seeing if we reach the other word. The search is performed on the underlying graph specified by the edges in pairs.

Algorithm

We start by building our graph from the edges in pairs.

The specific algorithm we go for is an iterative depth-first search. The implementation we go for is a typical "visitor pattern": when searching whether there is a path from w1 = words1[i] to w2 = words2[i], stack will contain all the nodes that are queued up for processing, while seen will be all the nodes that have been queued for processing (whether they have been processed or not).

Complexity Analysis

• Time Complexity: $$O(NP)$$ , where $$N$$ is the maximum length of words1 and words2, and $$P$$ is the length of pairs. Each of $$N$$ searches could search the entire graph.

• Space Complexity: $$O(P)$$ , the size of pairs.

Approach #2: Union-Find [Accepted]

Intuition

As in Approach #1, we want to know if there is path connecting two words from edges represented by pairs.

Our problem comes down to finding the connected components of a graph. This is a natural fit for a Disjoint Set Union (DSU) structure.

Algorithm

Draw edges between words if they are similar. For easier interoperability between our DSU template, we will map each word to some integer ix = index[word]. Then, dsu.find(ix) will tell us a unique id representing what component that word is in.

For more information on DSU, please look at Approach #2 in the article here. For brevity, the solutions showcased below do not use union-by-rank.

After putting each word in pairs into our DSU template, we check successive pairs of words w1, w2 = words1[i], words2[i]. We require that w1 == w2, or w1 and w2 are in the same component. This is easily checked using dsu.find.

Complexity Analysis

• Time Complexity: $$O(N \log P + P)$$ , where $$N$$ is the maximum length of words1 and words2, and $$P$$ is the length of pairs. If we used union-by-rank, this complexity improves to $$O(N * \alpha(P) + P) \approx O(N + P)$$ , where $$\alpha$$ is the Inverse-Ackermann function.

• Space Complexity: $$O(P)$$ , the size of pairs.

Analysis written by: @awice.