The greedy logic is straightforward once you see it: always pick the largest unvisited neighbor at each step, and start each new component from the largest unvisited node.
Model the problem as a greedy DFS where at each step you visit the largest possible unvisited node, ensuring that the remaining unvisited nodes can still be reached in a valid DFS order. Use a priority queue or sorted adjacency lists to always pick the largest neighbor, and handle disconnected components by restarting at the largest unvisited node. Prove correctness by showing that any deviation from the greedy choice would lead to a lexicographically smaller sequence.
Pro tip: Clarify that the lexicographically largest sequence is not simply sorting all nodes descending; the DFS constraint forces a specific interleaving. Mention that you can simulate the process with a stack and a set of unvisited nodes, always pushing the largest available neighbor, and that the starting node for each component should be the largest unvisited node.
Restate the problem: we need the lexicographically largest DFS visitation sequence over all possible choices of start nodes, neighbor orders, and component restarts. Clarify that the graph is undirected and may be disconnected.
At each step, choose the largest unvisited node that can be visited next while still allowing a valid DFS traversal of the remaining graph. This often means picking the largest neighbor of the current node, but if none, restart at the largest unvisited node.
Use a max-heap or sorted adjacency lists to always explore the largest neighbor first. Maintain a visited set and a stack for DFS. When the stack is empty, pick the largest unvisited node as the new start and continue.
Argue by exchange: if at some point you could visit a larger node than the one chosen by the greedy algorithm, swapping would yield a lexicographically larger sequence, contradicting the greedy choice. Also show that restarting at the largest unvisited node is optimal for disconnected components.
Time complexity is O((n+m) log n) with a priority queue, or O(n+m) if adjacency lists are pre-sorted. Discuss edge cases: empty graph, single node, complete graph, and graphs with multiple components.
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