Task Scheduler - Algorithm

Let $f$ be the maximum frequency and $k$ be the number of tasks with that frequency. The minimum time is: $\max(\text{total tasks}, (f-1) \times (n+1) + k)$. Why? The most frequent task creates $(f-1)$ gaps of size $(n+1)$, plus the final batch of $k$ tasks.

If other tasks fill the gaps, the formula gives total tasks. Otherwise, idle time is needed. Time: $O(n)$. Space: $O(1)$.