I knew the greedy approach from having seen it before: always schedule the most frequent remaining task type, fill the rest of the cooldown window with other types or idle.
Start by identifying the most frequent task type, as it dictates the minimum schedule length. Use the formula max((maxFreq - 1) * (c + 1) + numMaxFreq, totalTasks) to compute the minimum time, then explain how to construct a valid schedule by arranging tasks in blocks separated by idle slots.
Pro tip: Mention that the formula accounts for the case where there are enough other tasks to fill idle slots, making the schedule length equal to the total number of tasks. This shows you understand the edge cases and can reason about the problem's constraints.
Clarify that tasks are unit-time, same type must be separated by at least c idle units, and we can reorder tasks and insert idle slots. The goal is to minimize total time.
Recognize that the most frequent task type determines the lower bound on schedule length. If there are multiple tasks with the same maximum frequency, they can be placed in the last block without needing extra idle slots.
Compute the minimum length as max((maxFreq - 1) * (c + 1) + numMaxFreq, totalTasks). Explain each term: (maxFreq - 1) full blocks of size c+1, plus the final block containing all tasks with max frequency.
Show that the formula gives a lower bound because the most frequent task requires at least (maxFreq - 1) gaps of size c, and the total tasks cannot be less than the number of tasks. Then argue that a schedule achieving this bound exists by placing tasks in blocks and filling idle slots with other tasks.
Describe how to build the schedule: sort tasks by frequency, place the most frequent tasks in each block, then fill remaining slots with other tasks in any order. Time complexity is O(n) to count frequencies and O(n log n) if sorting, space O(n) for the schedule.
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