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Use a stack to handle nested encodings: push the current string and repeat count when encountering '[', and on ']' pop and repeat the substring. Alternatively, use recursion to parse each bracket pair. Clarify constraints and edge cases before coding.
Pro tip: Mention that a stack-based solution runs in O(n) time and space, and that recursion depth could be an issue for deeply nested inputs—showing awareness of trade-offs. Also, test with nested cases like '3[a2[c]]' to demonstrate correctness.
Ask about input constraints (e.g., valid encoding, character set, max nesting depth) and expected output format. Confirm that k is a positive integer and brackets are balanced.
Decide between using a stack (iterative) or recursion. Explain that a stack naturally handles nested structures by storing previous strings and repeat counts.
Iterate through the string: build digits for k, push current string and k onto stack when '[' is found, and on ']' pop and append the repeated substring to the current string.
State that time complexity is O(n) where n is the length of the decoded string, and space complexity is O(n) for the stack and output. Mention that recursion uses call stack space.
Walk through simple cases like '3[a]' and nested cases like '3[a2[c]]' to verify the logic. Discuss edge cases such as empty string or no brackets.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one looked like a plain BFS until you realize the same cell can be visited multiple times with different fuel levels, so your state has to be (row, col, fuel) not just (row, col).
Model the problem as a shortest path on a state graph where each state is (cell, fuel remaining). Use BFS with a deque (0-1 BFS) or Dijkstra, since moving costs 1 step and refueling at gas stations is free. Return the minimum steps when the target is reached, or -1 if unreachable.
Pro tip: Clarify whether refueling at a gas station is optional or mandatory, and whether you can pass through a gas station without refueling. This affects state transitions and can be a common source of bugs.
Ask about grid size, fuel capacity, whether refueling is optional, and if gas stations can be revisited. Confirm movement is 4-directional and blocked cells cannot be entered.
Represent each state as (row, col, fuel_left). The start state is (start_row, start_col, capacity). Use a visited set or 2D array of fuel levels to avoid revisiting states.
Use BFS with a deque (0-1 BFS) because moving costs 1 step and refueling costs 0. For each state, try moving to adjacent cells if fuel_left > 0; if the new cell is a gas station, set fuel_left to capacity at no extra cost.
If start equals target, return 0. If target is blocked or unreachable, return -1. Ensure fuel never goes negative and capacity is respected.
Time complexity is O(m * n * capacity) and space is O(m * n * capacity). Mention that if capacity is large, consider alternative approaches or pruning.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.