I spent way too long on edge cases before I even had a working solution.
Track the minimum and maximum positions of values 1..k as you iterate through the permutation. For each k, the values occupy a contiguous segment if and only if maxPos - minPos + 1 == k. Build the binary string by checking this condition at each step.
Pro tip: Mention that this O(N) solution is optimal because you must read the entire permutation, and explicitly state the invariant: the set of positions is contiguous exactly when its size equals the range span. This shows you understand the underlying principle, not just the algorithm.
Confirm that the permutation contains integers 1 through N exactly once, and that for each k from 1 to N, we need to check if the positions of values 1..k form a contiguous block. The output is a string of '1's and '0's of length N.
Realize that a set of k distinct positions is contiguous if and only if the difference between the maximum and minimum positions is exactly k-1. Equivalently, maxPos - minPos + 1 == k.
Iterate through the permutation once, maintaining the minimum and maximum positions seen so far for values 1..k. At each step k, check the condition and append '1' or '0' to the result string.
The algorithm runs in O(N) time and O(N) space (for the position array and output). Edge cases include k=1 (always contiguous) and k=N (always contiguous).
Walk through a small example, such as permutation [2,1,3], to verify the logic: for k=1, positions of {1} is index 1 (0-based), contiguous; for k=2, positions of {1,2} are indices 0 and 1, contiguous; for k=3, all indices, contiguous. Output '111'.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the problem as finding a root that minimizes the number of edges needing reversal to point away from it. Use two DFS passes: first compute the cost for an arbitrary root, then reroot to compute costs for all nodes in O(N) time by tracking how the cost changes when moving the root across an edge.
Pro tip: Mention that the answer is the minimum over all roots of (N-1 - number of edges already pointing away from the root), and emphasize that the rerooting technique generalizes to many tree DP problems.
Clarify that for a fixed root, an edge must be reversed if it points toward the root instead of away. The cost is the count of such edges.
Run a DFS from node 0 (or any node) to count how many edges point away from it. This gives the reversal count for that root.
Use a second DFS to propagate the cost to children. When moving the root from u to v, the cost changes by +1 if edge u->v exists (since it now points toward the new root), else -1.
Track the minimum cost across all nodes during the rerooting pass and return it as the answer.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.