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Start by clarifying the game rules and constraints, then propose a data structure like a deque for the snake body and a hash set for O(1) collision checks. Explain how each move updates the snake, checks for collisions and food, and returns the score, emphasizing the O(1) average time per move.
Pro tip: Mention that using a hash set for the snake body allows O(1) collision detection, but also consider the trade-off with memory and the need to handle the tail correctly when it moves. Also, discuss how to handle food advancement efficiently, perhaps with a precomputed list or a queue.
Ask about grid size, initial snake length and position, food placement (predetermined or random), and what happens when the snake fills the grid. Confirm that move(direction) is called each step and returns the score or -1.
Use a deque (or doubly linked list) to represent the snake's body for O(1) additions/removals at both ends, and a hash set to store occupied cells for O(1) collision checks. Optionally, use a 2D array for the grid if needed for food tracking.
For each move, compute the new head position. Check if it's out of bounds or collides with the snake body (excluding the tail if it will move). If food is eaten, grow the snake and update score; otherwise, move the tail. Update the hash set accordingly.
Maintain a list or queue of predetermined food positions. When the head reaches a food position, increment score, remove that food from the list, and do not remove the tail. Return the current score after each move.
Explain that each move is O(1) average time due to hash set operations and deque updates. Discuss edge cases: snake of length 1, moving into the tail's current position (allowed if not growing), and winning condition when snake fills the grid.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Binary search on a peak-finding problem felt unintuitive to me at first.
Use a binary search approach: compare the middle element with its neighbors to decide which half contains a peak, then recurse or iterate in that half. For duplicates, modify the comparison to handle equal neighbors by moving in a consistent direction (e.g., right) to avoid infinite loops. Prove correctness by showing that a peak must exist in the chosen half based on the slope direction.
Pro tip: When explaining the proof, emphasize the invariant that the subarray always contains a peak, and for duplicates, mention that the algorithm still runs in O(log n) by breaking ties consistently.
Confirm that the array is non-empty and discuss edge cases like single element, two elements, and arrays with duplicates. Mention that a peak may be at the boundaries if the boundary element is greater than its only neighbor.
Explain that you compare the middle element with its neighbors. If it's a peak, return it. If the right neighbor is greater, search the right half; otherwise, search the left half. This works because a peak must exist in the direction of the upward slope.
State the invariant: the subarray being searched always contains at least one peak. Show that when you move to the half with the greater neighbor, the invariant is maintained because the slope guarantees a peak in that half. Base case: subarray of size 1 is a peak.
If duplicates are allowed, the simple comparison may fail when neighbors are equal. Modify the algorithm to treat equal neighbors as a downward slope (or consistently move right) to ensure progress. Explain that this still finds a peak in O(log n) time.
State that the algorithm runs in O(log n) time because the search space halves each iteration, and O(1) space because it uses iterative binary search. Summarize the key points and mention that the proof relies on the slope direction.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by explaining the greedy strategy: always schedule the most frequent task next, using a max-heap to track remaining counts and a cooldown queue to enforce the gap. Derive the formula for the minimum time based on the maximum frequency and the number of tasks with that frequency, then describe how to construct the actual schedule by simulating the process. Finally, discuss how to extend to per-task cooldowns by tracking each task's next available time and using a priority queue ordered by frequency.
Pro tip: Mention that the formula approach gives the count but not the schedule; to construct a valid schedule, simulate with a priority queue and a cooldown queue, which also naturally extends to per-task cooldowns.
Clarify that tasks are labeled by characters, each unit runs one task or idle, and the same task must be separated by at least n units. Identify that the goal is to minimize total time and produce a schedule.
Let maxFreq be the highest frequency and maxCount be the number of tasks with that frequency. The minimum time is max((maxFreq-1)*(n+1)+maxCount, totalTasks). Explain why this works.
Simulate time steps: at each step, pick the available task with the highest remaining count, execute it, and put it in a cooldown queue to become available after n units. If no task is available, idle.
Maintain a next-available time for each task type. Use a priority queue of available tasks ordered by remaining count, and a time-ordered queue for tasks in cooldown. At each time, add tasks whose cooldown has expired, then pick the highest-count task.
Discuss time complexity O(totalTasks log k) where k is number of distinct tasks, and space O(k). Compare with the formula approach which is O(k) but doesn't give a schedule.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.