Classic monotonic stack problem once you recognize it, but I didn't recognize it fast enough.
Use a monotonic decreasing stack to track indices of days with unresolved higher measurements. Iterate through the array, and for each day, pop indices from the stack while the current measurement is strictly higher, recording the distance. Push the current index onto the stack. After the iteration, remaining indices have no higher measurement, so their answers remain 0.
Pro tip: Emphasize that the stack stores indices, not values, to compute distances easily. Also, explicitly state that the algorithm runs in O(n) time because each index is pushed and popped at most once, and space is O(n) for the stack and output array.
Clarify that for each day, we need the number of days until a strictly higher measurement. If none, return 0. The solution must be O(n) time.
Use a stack to keep track of indices of days that are waiting for a higher measurement. The stack will maintain a decreasing order of measurements from bottom to top.
For each day i, while the stack is not empty and the current measurement is greater than the measurement at the index on top of the stack, pop the index j and set result[j] = i - j. Then push i onto the stack.
After the loop, any indices left in the stack have no higher measurement to the right, so their result remains 0 (initialized).
Justify O(n) time: each index is pushed and popped at most once. Space is O(n) for the stack and output array. Correctness: the stack invariant ensures that when a higher measurement is found, it is the next strictly higher one for the popped indices.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the distance definition and edge cases first, then propose a DFS-based solution that tracks ancestor distances using a hash map to count pairs in O(n) time. Explain how to handle k=0 and empty tree, and analyze time and space complexity.
Pro tip: Explicitly state that distance counts intermediate nodes, so distance = number of edges - 1; this shows attention to detail and avoids off-by-one errors. Also, mention that for k=0, no valid pairs exist because nodes must be distinct and distance counts intermediate nodes.
Confirm the distance definition (intermediate nodes) and edge cases like k=0, empty tree, and single node. Ask if k can be negative or if nodes are guaranteed unique.
Use DFS with a hash map to store distances from the current node to nodes in its subtree. At each node, combine distance counts from left and right subtrees to find pairs with total distance k.
For k=0, return empty list since distinct nodes cannot have distance 0 (intermediate nodes count). For empty tree, return empty list. For k<0, return empty list.
Time complexity is O(n) because each node is visited once and hash map operations are O(1) on average. Space complexity is O(n) for the hash map and recursion stack.
Compare with O(n^2) brute-force approach (computing all pairs). Mention that the hash map approach is optimal for this problem, but if memory is constrained, a two-pass approach could be considered.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.