The greedy insight is to think about the 'cost' of giving a task to worker 1 versus worker 2, specifically the difference reward1[i] minus reward2[i].
Model the problem as maximizing total reward with a cardinality constraint: assign each task to the worker giving higher reward, but if the count of tasks assigned to Worker 1 exceeds k, switch the k tasks with the smallest penalty (difference between rewards) to Worker 2. Alternatively, use a greedy approach with a priority queue or sort by penalty. Explain the algorithm, prove correctness, and analyze time complexity.
Pro tip: Emphasize that the greedy choice is optimal because the penalty for switching a task is independent of other assignments, and mention that this can be solved in O(n log n) time, which is efficient for large n.
Clarify that each task has two rewards: one for Worker 1 and one for Worker 2. Worker 1 must get exactly k tasks, Worker 2 gets the rest. Goal: maximize sum of rewards.
Assign each task to the worker who gives the higher reward. Count how many tasks are assigned to Worker 1.
If Worker 1 has more than k tasks, compute the penalty (reward1 - reward2) for each task assigned to Worker 1. Switch the tasks with the smallest penalties to Worker 2 until Worker 1 has exactly k tasks.
If initial assignment gives Worker 1 fewer than k tasks, switch tasks from Worker 2 to Worker 1, choosing those with the largest gain (reward1 - reward2).
Sorting tasks by penalty/gain takes O(n log n) time, and the rest is O(n). Space is O(n) for storing tasks.
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Model the problem as finding a root that minimizes the number of edges directed away from it. For each node, compute the number of edges that need reversal if it were the root, using two DFS passes to accumulate counts efficiently. Then return the minimum count over all nodes.
Pro tip: Start by explaining the brute-force O(n^2) approach and then optimize to O(n) using rerooting DP. This shows you can think iteratively and care about efficiency, which is crucial for large-scale systems at Uber.
Clarify that we need to choose a root r and reverse edges so that all nodes can reach r. The goal is to minimize reversals over all r.
For each node as root, perform a DFS to count edges that point away from the root (i.e., need reversal). This takes O(n^2) time.
First, root the tree arbitrarily (e.g., at node 0) and compute the number of reversals needed for that root. Then, use a second DFS to compute the count for all other nodes by adjusting based on the edge between parent and child.
Write code to perform the two DFS passes, track the minimum reversals, and return the optimal root. Analyze time and space complexity as O(n).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start from the top-right corner of the matrix and move left whenever you encounter a 1, or down whenever you encounter a 0. This staircase traversal finds the leftmost column with a 1 in O(m + n) time, which is faster than O(m*n).
Pro tip: Explicitly state the time and space complexity and compare it to the brute-force approach. Mention that the matrix is sorted, which allows the staircase optimization, and handle edge cases like no 1s present.
Confirm that the matrix is binary, each row is sorted (0s then 1s), and we need the leftmost column index containing at least one 1. Ask about edge cases: what if no 1 exists? What if multiple 1s in a column?
Mention that checking every cell takes O(m*n) time, which is too slow. This sets the stage for a better approach.
Start at the top-right cell. If it's 1, record the column and move left; if it's 0, move down. Repeat until out of bounds. This works because rows are sorted.
The algorithm moves at most m steps down and n steps left, so O(m+n) time and O(1) space. Handle cases where no 1 is found by returning -1.
Walk through a small example to verify correctness, such as a 3x4 matrix, and check edge cases like all zeros or all ones.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.