Given a graph-G that has |V| vertices and |E| edges, that is:Connected. Directed. Weighted What is the worst case time complexity to obtain the vertex with the highest number of edges (incoming and outgoing), if G is implemented using adjacency matrix?

Question

BlackTom AI · Accepted Answer

An adjacency matrix for a graph with |V| = n is an n × n matrix where entry A[i][j] indicates the presence/weight of an edge from vertex i to vertex j. For a directed graph, the out-degree of vertex i is the sum of row i, and the in-degree of vertex i is the sum of column i. The total degree (incoming + outgoing) for each vertex requires both sums (or equivalent updates while scanning).

Relevant concepts
- Representation: adjacency matrix uses n × n storage independent of |E|.
- Degree computation: out-degree(i) = sum_j A[i][j], in-degree(i) = sum_j A[j][i].
- Time complexity: count number of constant-time operations needed to compute all degrees.

Step-by-step reasoning and intermediate calculations
1. Let n = |V|. Initialize arrays in[n] and out[n] to 0. 
2. Iterate over every matrix entry: for i from 1 to n, for j from 1 to n, read A[i][j]:
   - If edge exists (or add weight), do out[i] += 1 (or weight) and in[j] += 1 (or weight).
3. Number of entries processed = n rows × n columns = n^2.
4. Each matrix entry requires O(1) work (read + two constant updates). Total work = O(n^2).
5. This work is also a lower bound in the worst case because any algorithm must at least inspect the matrix entries that could encode edges; thus Ω(n^2). Combining gives Θ(n^2).

Why the correct answer is correct
- Using adjacency matrix forces scanning the n×n matrix to accumulate incoming and outgoing counts; that yields Θ(|V|^2) time.

Why other options are incorrect (brief)
- Θ(1), Θ(|V|), Θ(|V| log |V|) are too small — they don’t account for reading the full n×n matrix.
- Θ(|E|) or Θ(|V|+|E|) apply to adjacency-list implementations where work is proportional to edges; for matrix |E| can be as large as n^2, but the matrix forces n^2 work regardless.
- Θ(|E|^2) and Θ(|E| log |E|) are irrelevant to matrix access pattern.

Conclusion
The worst-case time complexity using an adjacency matrix is Θ(|V|^2), which matches the provided answer.

Given a graph-G that has |V| vertices and |E| edges, that is:Connected. Directed. Weighted What is the worst case time complexity to obtain the vertex with the highest number of edges (incoming and outgoing), if G is implemented using adjacency matrix?单项选择题

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