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Clarify the problem and edge cases, then propose a linear scan comparing each adjacent pair after normalizing case. Discuss time and space complexity, and optionally mention alternative approaches like using regular expressions or built-in functions.
Pro tip: Demonstrate awareness of Unicode and locale-specific case folding, and suggest using case-insensitive comparison without altering the original string to avoid unnecessary memory usage.
Confirm that case-insensitive comparison means 'A' and 'a' are identical, and discuss handling of empty strings, single-character strings, and non-alphabetic characters.
Describe a single-pass approach: iterate through the string from index 0 to n-2, compare each character with the next using case-insensitive equality, and increment a counter when they differ.
State that the time complexity is O(n) where n is the string length, and space complexity is O(1) if no extra data structures are used.
Mention how to perform case-insensitive comparison (e.g., using lower() or casefold() in Python, or Character.toLowerCase() in Java) and note that creating a lowercased copy uses O(n) space, so prefer on-the-fly comparison.
Optionally mention alternative approaches like using zip and sum in Python, and walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify that the allocator manages a contiguous array of cells, where allocations can only start at indices that are multiples of 8. Use a balanced binary search tree (e.g., sortedcontainers.SortedList) to track free blocks, ensuring O(log n) allocation and deallocation. For each allocation, find the leftmost free block with start index divisible by 8, assign an ID, and split the block; for erase, free all cells of that ID and merge adjacent free blocks.
Pro tip: Mention that in practice, memory allocators often use size-class-based free lists to reduce fragmentation and improve speed, but for this problem, a balanced BST is sufficient. Also, discuss how to handle edge cases like allocation requests larger than any free block.
Confirm that the memory is a contiguous array of cells, allocations must start at indices divisible by 8, and alloc returns the leftmost valid free block. Ask about the expected size of memory and number of operations to choose the right data structure.
Use a balanced BST (e.g., SortedList) to store free blocks as (start, size) sorted by start. Maintain a dictionary mapping ID to its allocated blocks (list of (start, size)). This allows efficient finding of the leftmost valid free block and merging on erase.
Iterate through free blocks in order of start index. For each block, compute the smallest start >= block.start that is divisible by 8. If that start + x <= block.end, allocate there, update the free block (split or remove), and record the allocation. If no block fits, return -1 or raise an error.
Retrieve all blocks allocated to the given ID from the dictionary. For each block, add it back to the free blocks list and merge with adjacent free blocks. Remove the ID from the dictionary.
Discuss time complexity: O(log n) for finding and updating free blocks using a balanced BST, plus O(k) for merging adjacent blocks on erase. Handle edge cases: allocation larger than any free block, erasing non-existent ID, and memory exhaustion.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify that the problem requires an online algorithm with O(log n) or O(1) update time. Propose maintaining a balanced BST or Fenwick tree to track segment boundaries, updating the count by checking neighbors when a point's color changes. Analyze time and space complexity, and discuss trade-offs for different data structures.
Pro tip: Emphasize that the key is to only check the immediate neighbors of the updated point; this local update ensures efficiency. Also, mention that using a hash map for colors and a sorted set for positions can simplify implementation.
Restate the problem: points on a number line, each can be colored; after each update (assign or change color), return the number of maximal contiguous same-color segments. Clarify that updates are online and we need efficient queries.
For each update, we need to know the colors of the immediate left and right neighbors (if they exist) to determine how the segment count changes. This suggests maintaining a data structure that supports fast neighbor queries.
Use a balanced binary search tree (e.g., TreeSet in Java) or a Fenwick tree over compressed coordinates to store colored points and quickly find predecessors/successors. Use a hash map to store the color of each point.
When updating a point's color, compute the change in segment count by considering the old and new colors relative to neighbors. Adjust the count accordingly: if the point was isolated, adding it may merge segments; if it was part of a segment, changing color may split or merge.
Each update requires O(log n) time for neighbor queries and O(1) for count adjustment. Discuss edge cases: first/last point, no neighbors, same color as both neighbors, etc. Also mention space complexity O(n).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.