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You are managing a team of workers responsible for completing a collection of independent tasks. Each task in the tasks array represents the time (in units) required to complete that particular task. Crucially, each task can only be handled by exactly one worker — no task can be shared or split between multiple workers.
Initially, you have exactly one worker available. However, workers possess a unique ability: they can replicate themselves into two independent workers. This replication process takes a fixed amount of time, known as cloneTime. Here's the key insight: if multiple workers initiate replication simultaneously, they all complete their replication in parallel, meaning the total time cost for that round of replication remains exactly cloneTime, regardless of how many workers are replicating.
Each worker must make a critical decision at any point in time: either replicate (splitting into two workers, costing cloneTime units) or execute a task (completing the assigned task and then becoming unavailable permanently). A worker cannot do both — once they start working on a task, they complete it and exit the system.
Your objective is to determine the minimum total time required to complete all tasks, starting from a single worker and making optimal decisions about when to replicate versus when to assign tasks.
Strategic Considerations:
cloneTime units total.tasks = [1]
cloneTime = 11With just one task and one worker, no replication is needed. The single worker directly completes the task in 1 time unit. Total time = 1.
tasks = [1,2]
cloneTime = 57We need two workers for two tasks. The single worker replicates into two workers (costing 5 time units). Then, the two workers execute their tasks in parallel: one takes 1 unit, the other takes 2 units. Since they work simultaneously, the parallel phase takes max(1, 2) = 2 units. Total time = 5 + 2 = 7.
tasks = [1,2,3]
cloneTime = 14Optimal strategy: First, the initial worker replicates (1 time unit). Now we have 2 workers. One worker takes the longest task (3 units), while the other replicates again (1 time unit), giving us 2 more workers. These 2 workers handle the remaining tasks (1 and 2 units) in parallel, taking max(1, 2) = 2 units. The parallel execution gives us: 1 (first replication) + max(3, 1 + max(1, 2)) = 1 + max(3, 3) = 4.
Constraints