Treat the grid as a graph and use DFS or BFS to explore each island, counting its area. Keep track of the maximum area found. Alternatively, use Union-Find to merge adjacent land cells and track component sizes.
Pro tip: Clarify whether you can modify the input grid to mark visited cells; if not, use a separate visited set. Also, discuss the trade-offs between DFS (recursive, risk of stack overflow) and BFS (iterative, uses queue) and mention Union-Find as an alternative for very large grids.
Confirm that the grid contains only 0s and 1s, and that an island is a group of connected 1s (4-directionally). The area is the number of cells in the island. We need the maximum area among all islands.
Decide between DFS, BFS, or Union-Find. DFS/BFS are simpler and efficient for most cases; Union-Find is good for dynamic connectivity but overkill here. Mention that DFS can be recursive or iterative.
Iterate through each cell. When encountering a '1' that hasn't been visited, start a traversal (DFS/BFS) to explore the entire island, counting its area. Mark cells as visited to avoid revisiting.
After computing the area of each island, update the maximum area if the current island's area is larger. Return the maximum area after processing all cells.
Time complexity is O(m*n) since each cell is visited once. Space complexity is O(m*n) in the worst case for the visited set or recursion stack. Mention that in-place modification can reduce space to O(1) if allowed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one tripped me up more than it should have.
Combine a dynamic array (for O(1) random access) with a hash map (for O(1) lookup by value). Insert appends to the array and stores the index in the map; delete swaps the target with the last element, updates the map, and pops; getRandom picks a random index from the array.
Pro tip: Explicitly discuss handling duplicates and edge cases (e.g., deleting the last element, empty structure) to show production-level thinking, and mention that the swap-delete trick is the key to O(1) deletion.
Ask whether duplicates are allowed, if the structure needs to support other operations, and confirm that average O(1) is acceptable (amortized for array operations).
Explain that you'll use a dynamic array for O(1) random access and a hash map from value to index (or set of indices) for O(1) lookup.
Insert: append to array, record index in map. getRandom: pick a random index from the array and return the element.
To delete: find the element's index via map, swap it with the last element, update the moved element's index in the map, then pop the last element and remove the entry from the map.
State that all operations are average O(1) (amortized for array append/pop). Discuss edge cases: deleting the last element, duplicates (if allowed), and empty structure.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a two-pointer (fast and slow) technique to find the Nth node from the end in one pass. Move the fast pointer N+1 steps ahead, then advance both pointers until fast reaches the end; the slow pointer will be just before the node to remove. Handle edge cases like removing the head node by using a dummy node.
Pro tip: Always use a dummy node pointing to the head to simplify edge cases, especially when the node to remove is the head. Also, clarify with the interviewer whether N is guaranteed to be valid and whether the list is singly or doubly linked.
Ask about input constraints: Is N always valid? Can N be greater than the list length? Is the list singly linked? Should we return the head? Confirm these to avoid assumptions.
Decide between two-pass (count length then remove) and one-pass (two pointers). For Meta, prefer the one-pass two-pointer approach for efficiency, but mention the two-pass as a simpler alternative.
Create a dummy node pointing to head. Initialize fast and slow pointers at dummy. Move fast N+1 steps. Then move both until fast is null. Remove slow.next by updating slow.next = slow.next.next.
Walk through examples: removing middle node, head node, last node, and N=1. Verify pointer updates and return dummy.next as the new head.
State time complexity O(L) where L is list length, and space O(1). Mention that two-pass is also O(L) time but requires two traversals; one-pass is more efficient.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a sliding window (two-pointer) technique to maintain a window with at most k zeros, expanding the right pointer and shrinking the left when zeros exceed k. Track the maximum window length seen, which represents the longest subarray of 1s after flipping at most k zeros.
Pro tip: Clarify that the problem is equivalent to finding the longest subarray with at most k zeros, and mention that the window size only increases, so you can avoid shrinking it explicitly—this shows deep understanding and can simplify code.
Confirm that the goal is to find the maximum length of a contiguous subarray containing at most k zeros, since flipping those zeros yields all 1s.
Explain that a sliding window with two pointers (left and right) efficiently tracks a valid window with at most k zeros in O(n) time.
Move the right pointer to include new elements, incrementing a zero count when encountering a 0. When zero count exceeds k, move the left pointer until the window is valid again.
After each expansion, update the maximum window length seen so far. The window size is right - left + 1.
After iterating through the array, return the maximum length found, which is the answer.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recursive DFS with a depth parameter passed down.
Clarify the problem definition and constraints, then discuss both recursive and iterative solutions. Emphasize the trade-offs between simplicity and efficiency, and analyze time and space complexity.
Pro tip: Mention that recursion depth could cause stack overflow for deeply nested lists, and propose an iterative BFS/DFS approach as a robust alternative. Also, consider edge cases like empty lists and non-integer elements.
Confirm the definition of depth: the outermost list has depth 1, and each nested list increases depth by 1. Ask about input constraints, such as maximum depth and list size.
Describe a recursive function that traverses the nested list, passing the current depth. For each integer, add depth * integer to the sum; for each list, recurse with depth + 1.
Explain how to use a stack (DFS) or queue (BFS) to avoid recursion limits. Each stack/queue element stores the current list and its depth.
State that both approaches visit each element once, so time complexity is O(n), where n is the total number of elements. Space complexity is O(d) for recursion depth or O(n) for iterative in worst case.
Mention handling empty lists, lists with no integers, and non-integer elements (if allowed). Also consider negative integers and large depth.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Stack-based approach to track unmatched indices, then rebuild the string excluding those.
Clarify that the goal is to remove the minimum number of parentheses to make the string valid, and that multiple valid answers may exist. Then propose a two-pass stack-based solution: first pass removes unmatched closing parentheses, second pass removes unmatched opening parentheses. Walk through an example to demonstrate correctness and discuss time/space complexity.
Pro tip: Mention that you can solve it in one pass using a stack of indices and a boolean array to mark removals, but the two-pass approach is simpler and equally efficient. Also, proactively discuss edge cases like empty string, all opening or all closing parentheses, and already valid strings.
Ask if the string contains only parentheses or other characters, and confirm that we need to remove the minimum number. Discuss whether multiple valid outputs are acceptable.
Propose a stack-based solution: use a stack to track indices of unmatched opening parentheses, and a set to mark indices to remove. Alternatively, use a counter for a two-pass approach.
Write pseudocode or actual code, explaining each step. For example, first pass: remove unmatched closing parentheses; second pass: remove unmatched opening parentheses. Trace through a sample string like 'a)b(c)d'.
State that the solution runs in O(n) time and O(n) space, where n is the length of the string. Mention that the space can be reduced to O(1) if we only need to return the length of the valid string, but O(n) is needed to construct the result.
Mention testing with empty string, '((((', '))))', '()()', and strings with other characters. Confirm that the algorithm handles them correctly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.