Went with a stack pretty quickly, push counts and partial strings as you scan through.
Use a stack to handle nested patterns: push the current string and repeat count when encountering '[', and on ']' pop and expand. Alternatively, use recursion to process each bracket level. Discuss time and space complexity, which are O(n) and O(depth) respectively.
Pro tip: Clarify edge cases upfront (e.g., multi-digit numbers, empty strings) and mention that the stack approach avoids recursion depth limits, showing production awareness.
Restate the decoding rules and confirm nested patterns, multi-digit counts, and valid input assumptions with the interviewer.
Decide between stack-based iteration and recursion, explaining why one is preferred (e.g., stack avoids recursion depth issues).
Trace the algorithm on '3[a2[c]]' to demonstrate correctness and clarify stack operations or recursive calls.
State time complexity O(n) where n is output length, and space complexity O(d) for nesting depth, discussing trade-offs.
Mention handling multi-digit numbers, empty brackets, and invalid inputs, and propose testing strategies.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The base case I handled with DP no problem, just work backwards from the end and at each index take the max of skipping or taking.
Model the problem as a dynamic programming problem where dp[i] represents the maximum profit from index i to the end. For the follow-up with negative values, note that backward jumps can create cycles, so the problem may become NP-hard or require a different approach like graph search with cycle detection.
Pro tip: Clarify with the interviewer whether the array can contain zeros or negative values in the initial problem, as this affects the DP recurrence and the possibility of infinite loops. Also, discuss the time and space complexity trade-offs between top-down memoization and bottom-up DP.
Ask about array size, value ranges (non-negative? zeros allowed?), and whether jumps can go out of bounds. Confirm if the goal is to maximize profit and if you can stop at any point.
Let dp[i] be the maximum profit starting from index i. Then dp[i] = max(skip: dp[i+1], take: arr[i] + dp[i + arr[i] + 1]) with base case dp[n] = 0. Handle out-of-bounds by treating as 0.
Implement bottom-up DP from right to left for O(n) time and O(n) space, or optimize to O(1) space if only the next few states are needed. Discuss potential for greedy approach if values are positive.
Explain that negative values cause backward jumps, potentially creating cycles. This makes the problem equivalent to finding the maximum weight path in a directed graph with possible cycles, which may be NP-hard. Suggest using DFS with memoization and cycle detection, or transforming to a longest path problem in a DAG if cycles can be avoided.
Discuss time and space complexity for both versions. Mention edge cases: all negative values, zeros, large jumps, and cycles. Propose testing strategies.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Treat the grid as a graph and use DFS/BFS to explore each unvisited land cell, marking all connected land cells as visited. Each time you start a traversal from an unvisited land cell, increment the region count. This approach runs in O(m*n) time and space.
Pro tip: Clarify edge cases upfront (empty grid, all water, all land) and discuss trade-offs between DFS (recursion depth) and BFS (queue memory). Mention that you can mutate the grid to save space if allowed.
Confirm grid dimensions, connectivity definition (4-directional), and whether the grid can be modified. Ask about edge cases like empty grid or no land.
Decide between DFS (recursive or iterative) and BFS. Consider recursion depth limits for large grids and memory constraints.
Iterate through each cell; when encountering an unvisited '1', increment count and perform DFS/BFS to mark all connected '1's as visited (e.g., set to '0' or use a visited set).
State time complexity O(m*n) since each cell is visited once, and space complexity O(m*n) worst-case for recursion stack or queue.
Walk through edge cases and consider optimizations like union-find for dynamic connectivity or in-place modification to save space.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.