My first instinct was to just iterate the whole array every day and that's obviously too slow.
Start by clarifying the problem: we need to support batch replacements (swap all occurrences of one value with another) and efficiently compute the total sum after each day. Propose a data structure that maintains a mapping from values to their current sum and count, and a global sum, allowing O(1) updates per replacement. Then discuss edge cases and trade-offs.
Pro tip: Mention that using a union-find (disjoint set) with path compression can handle replacements in near-constant time, but a simpler hash map approach is often sufficient and easier to implement. Also, highlight that you would validate the input and consider the impact of large batches.
Ask about the size of the array, number of days, range of values, and whether replacements are cumulative. Confirm that we need the total sum after each day's batch.
Propose maintaining a hash map from value to its total sum and count, and a global sum. For each replacement (x, y), update the map and global sum in O(1) time.
Process each replacement in the batch sequentially, updating the data structure. If multiple replacements affect the same value, ensure they are applied in order.
Explain that each replacement is O(1), so total time is O(n + m) where n is array size and m is total replacements. Discuss alternative approaches like union-find and their trade-offs.
List edge cases: replacing a value with itself, replacing a value not present, multiple replacements in one day, large values causing overflow, and empty array.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Chebyshev distance threw me off for longer than I should admit.
First, compute the initial inconvenience using multi-source BFS with Chebyshev distance (8-directional). Then, consider flipping each 0 to 1 and efficiently update the inconvenience, using binary search on the answer to check feasibility with at most one flip.
Pro tip: Mention that Chebyshev distance can be transformed to Manhattan distance via coordinate rotation, which may simplify BFS or allow using standard techniques. Also, emphasize the trade-off between time and space when choosing between binary search and direct computation.
Clarify that Chebyshev distance is the maximum of absolute differences in x and y, equivalent to 8-directional moves. The inconvenience is the maximum over all empty cells of the minimum distance to any center.
Use multi-source BFS from all 1s to compute the distance to the nearest center for every cell. This gives the initial inconvenience as the maximum distance among 0s.
For each 0, flipping it to 1 can only reduce distances for cells that are closer to this new center. The new inconvenience is the maximum over all 0s (except the flipped one) of the minimum of their original distance and the distance to the new center.
Binary search on the answer D. For a given D, check if there exists a 0 such that flipping it makes all 0s have distance ≤ D. This can be checked by finding the intersection of Chebyshev balls of radius D around all 0s that have original distance > D.
Discuss time and space complexity: BFS is O(mn), binary search adds O(log(maxDist)) factor, and each check is O(mn). Mention alternative approaches like computing the two farthest 0s and using their intersection.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.