I knew the diagonal grouping trick (row+col = constant) but fumbled the direction alternation logic.
Group cells by their row+column sum, which identifies each diagonal. Then iterate through the diagonals in order, alternating the direction of traversal (up-right or down-left) based on the diagonal index. Handle the empty matrix case by returning an empty array immediately.
Pro tip: Clarify the expected output format and edge cases (e.g., empty matrix, single row/column) before coding, and discuss time/space complexity upfront to demonstrate thoroughness.
Confirm that diagonals are defined by equal row+column sums, and that traversal alternates direction starting with up-right. Check for empty matrix and single row/column cases.
Use a hash map (dictionary) where keys are diagonal indices (row+col) and values are lists of elements in that diagonal. Alternatively, compute the range of diagonal indices and iterate directly.
Traverse the matrix once, appending each element to the list corresponding to its diagonal index. This takes O(m*n) time.
Iterate through diagonal indices from 0 to m+n-2. For even indices, output the diagonal in reverse order (down-left); for odd indices, output in normal order (up-right).
State that time and space complexity are O(m*n). Walk through a small example to verify correctness, including edge cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a depth-first search (DFS) traversal, passing the current number formed so far (as an integer) down to each child. At each leaf, add the current number to a running total. This avoids string concatenation and handles large numbers efficiently.
Pro tip: Mention that you can also solve it iteratively with a stack to avoid recursion depth issues, and discuss how to handle potential integer overflow by using modulo or big integers if the tree is very deep.
Confirm that each root-to-leaf path forms a number, and that the sum includes all such numbers. Ask about edge cases like empty tree or single node.
Decide between recursive DFS or iterative stack-based DFS. Explain that both work, but recursion is simpler and more readable.
Write a helper function that takes a node and the current number. At each node, update the number as current * 10 + node.val. If leaf, add to sum; else recurse on children.
If the tree is empty, return 0. If a node is a leaf, add its number to the total. Ensure null children are skipped.
Time complexity is O(N) where N is number of nodes, as each node is visited once. Space complexity is O(H) for recursion stack, where H is tree height.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the problem and constraints, then discuss a recursive DFS solution that traverses the nested list while tracking depth. Emphasize that each integer contributes value * depth to the total sum, and analyze time and space complexity.
Pro tip: Mention that you can avoid recursion depth issues by using an iterative stack-based approach, and discuss how to handle very deep nesting or large inputs. This shows awareness of production constraints beyond the basic solution.
Ask clarifying questions about input size, nesting depth limits, and whether the list can be empty. Confirm that depth starts at 1 for the outermost list.
Propose a recursive DFS that passes the current depth. For each element, if it's an integer, add value * depth; if it's a list, recurse with depth + 1.
Trace a small example like [1, [2, [3]]] to demonstrate correctness: 1*1 + 2*2 + 3*3 = 14. This validates the logic and catches edge cases.
State that time complexity is O(N) where N is the total number of elements (including nested lists), and space complexity is O(D) for recursion depth D.
Mention iterative stack-based alternative to avoid recursion limits, and handle edge cases like empty list, single integer, or deeply nested structures.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
No root reference is the interesting constraint here.
Use the parent pointers to find the depth of each node, then align them to the same depth by moving the deeper node up. Finally, move both nodes up simultaneously until they meet; that meeting point is the lowest common ancestor.
Pro tip: Clarify with the interviewer whether the nodes are guaranteed to be in the same tree and whether the tree can be mutated. Also, mention that this approach is O(h) time and O(1) space, which is optimal given the constraints.
Confirm that both nodes are in the same tree and that parent pointers are valid. Ask if the tree can be modified or if extra space is allowed.
Write a helper function to compute the depth of a node by traversing parent pointers up to the root. Compute depths for both given nodes.
If depths differ, move the deeper node up by the difference in depths so that both nodes are at the same level.
Move both nodes up simultaneously until they point to the same node. That node is the lowest common ancestor.
State that the time complexity is O(h) where h is the height of the tree, and space complexity is O(1) since only a few pointers are used.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Part (a) was BFS over stop positions, fine.
Model the problem as a graph where each node is a stop cell (where the ball can rest after hitting a wall or boundary), and edges represent rolling in one of the four directions until stopping. Use BFS to find reachability and shortest path in terms of number of rolls, but track cells traveled separately. For part (c), modify BFS to consider holes as intermediate stopping points and track lexicographically smallest direction string among shortest paths. For part (d), discuss precomputing all-pairs shortest paths or using bidirectional BFS with memoization for repeated queries.
Pro tip: Clarify upfront whether 'cells traveled' counts each cell entered or just the number of rolls, and whether the ball can stop on the target if it's not against a wall. Also, for lexicographic order, define the direction priority (e.g., 'd' < 'l' < 'r' < 'u') and ensure your BFS explores in that order to naturally get the smallest string.
Ask about grid dimensions, whether the ball starts at a given cell, if the target is a stop cell, and how holes affect movement. Confirm direction ordering for lexicographic comparison.
Define nodes as stop cells. Precompute for each open cell and direction the next stop cell and distance traveled. Use BFS to find reachability and minimum rolls, while tracking total cells traveled.
For part (c), treat holes as possible stopping points if the ball passes over them. During BFS, when multiple paths have the same length, choose the one with lexicographically smallest direction string by exploring directions in sorted order.
For static maze, precompute all-pairs shortest paths between all stop cells using BFS from each stop cell, or use bidirectional BFS with caching. Discuss trade-offs between precomputation time and query time.
State time and space complexity for each part. For repeated queries, compare precomputation O(V*(V+E)) vs per-query BFS O(V+E) and suggest when each is appropriate.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.