I recognized this as an Eulerian path/trail problem pretty quickly, which was lucky.
Recognize this as the Route Inspection Problem (Chinese Postman Problem) for undirected graphs, where the minimum number of strokes equals the number of odd-degree vertices divided by 2 (or 1 if all degrees are even). Explain the reasoning: each stroke can cover at most two odd-degree vertices as endpoints, and by pairing odd vertices with shortest paths, you can achieve this lower bound. Then outline an algorithm to compute the answer.
Pro tip: Mention that this is a classic problem with a known polynomial-time solution, and that the answer is simply max(1, odd_count/2). This shows you can map new problems to known results, a key skill at OpenAI.
State that this is the Chinese Postman Problem (Route Inspection) for undirected graphs, where we want to cover all edges with minimum trails.
Explain that in any graph, the number of vertices with odd degree is even. Each stroke (trail) has two endpoints, which must be odd-degree vertices (or the same vertex if the trail is a cycle).
Argue that each stroke can cover at most two odd-degree vertices as endpoints, so at least odd_count/2 strokes are needed. If odd_count=0, at least 1 stroke is needed (unless the graph has no edges).
Describe how to pair up odd-degree vertices and add shortest paths between each pair to make all degrees even, resulting in an Eulerian graph that can be drawn in one stroke. Removing the added paths splits it into odd_count/2 strokes.
Conclude that the minimum number of strokes is max(1, odd_count/2). Mention that computing odd_count is O(V+E), and if needed, the pairing can be done in O(V^3) using Floyd-Warshall and matching.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.