Task Scheduler – Solution & Complexity
Solution Walkthrough
Step 1: Understand the bottleneck
- The hard part is not ordering every task individually; it is respecting the cooldown for the most frequent letter.
- If one task appears much more often than the rest, the schedule needs gaps between its copies, and those gaps must be filled by other tasks or by idle slots.
Step 2: Brute-force baseline
- A direct simulation keeps a max-heap of currently available task counts plus a cooldown queue storing when a task can be used again.
- That works and is a good interview stepping stone, but it still simulates every interval explicitly, including idle time.
Step 3: Optimal counting insight
- Let
maxFreqbe the highest task frequency. Place those copies first asmaxFreq - 1full rows separated byncooldown slots. - If
maxCounttasks tie for that same highest frequency, they all occupy the last column of the frame. The minimum required frame length is(maxFreq - 1) * (n + 1) + maxCount. - The real answer is the larger of that frame length and the total number of tasks, because abundant filler tasks can eliminate all idle intervals.
- Let
Step 4: Final solution (all languages)
The counting formula runs in linear time over the task list and constant extra space because the alphabet size is fixed.
Step 5: Common mistakes and follow-ups
- Forgetting the tie case: if multiple tasks share the same maximum frequency, you need all of them in the last row, not just one.
- Returning the frame length directly even when many filler tasks exist; the answer can never be smaller than
tasks.length. - Follow-up: if asked to print a valid schedule, switch to the heap + cooldown simulation and emit each chosen task or an
idlemarker.