Use BFS from the source node to explore the graph level by level, tracking the parent of each visited node. Once the target is reached, reconstruct the path by backtracking from target to source using the parent pointers. Then state that BFS runs in O(V+E) time and O(V) space.
Pro tip: Mention that BFS is optimal for unweighted graphs and that early termination when the target is dequeued can save time. Also, clarify that the space complexity includes the queue, visited set, and parent map, all O(V).
Explain that BFS is the standard algorithm for shortest paths in unweighted graphs because it explores nodes in order of increasing distance from the source.
Use a queue for BFS, a visited set to avoid revisiting nodes, and a parent map to record the predecessor of each node.
Dequeue a node, check if it's the target (early exit), then enqueue all unvisited neighbors, marking them visited and setting their parent.
If the target was reached, backtrack from target to source using the parent map to build the vertex sequence, then reverse it.
Time complexity is O(V+E) because each vertex and edge is processed once. Space complexity is O(V) for the queue, visited set, and parent map.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one took me a minute to even parse correctly.
Model the problem as finding a minimum-cost path of exactly m vertices in a layered graph where each layer corresponds to a target position. Use dynamic programming with states (vertex, position) to compute the minimum mismatches, then backtrack to reconstruct the path. Optimize by considering graph structure and constraints.
Pro tip: Clarify edge cases upfront: if m=1, just pick the vertex with matching label if exists, else any vertex. Also discuss trade-offs between DP and BFS/A* if the graph is large, showing awareness of scalability.
Ask about graph size, whether it's directed/undirected, if multiple edges exist, and if m can exceed the number of vertices. Confirm that a path can revisit vertices and edges.
Let dp[i][v] = min mismatches for a path of length i ending at vertex v. Initialize dp[1][v] = 0 if label(v)==target[1] else 1. Transition: dp[i][v] = min over predecessors u of dp[i-1][u] + (label(v)!=target[i]).
Iterate i from 2 to m, and for each vertex v, compute dp[i][v] using incoming edges. Store the predecessor u that achieves the minimum to enable path reconstruction.
After filling DP, find vertex v with minimum dp[m][v]. Backtrack using stored predecessors to build the sequence of m vertices.
Time O(m*(V+E)), space O(m*V). Discuss possible optimizations like using only two layers of DP if path reconstruction not needed, or using BFS with priority queue if edge weights are uniform.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.