You are given an integer array timeReq and an integer splitTime.
In the microscopic world of the human body, the immune system faces an extraordinary challenge: combatting a rapidly multiplying bacterial colony that threatens the body's survival.
Initially, only one white blood cell (WBC) is deployed to eliminate the bacteria. However, the lone WBC quickly realizes it cannot keep up with the bacterial growth rate.
The WBC devises a clever strategy to fight the bacteria:
ith bacterial strain takes timeReq[i] units of time to be eliminated.splitTime units of time. Once split, the two WBCs can work in parallel on eliminating the bacteria.You must determine the minimum time required to eliminate all the bacterial strains.
Note that the bacterial strains can be eliminated in any order.
Example 1:
Input: timeReq = [10,4,5], splitTime = 2
Output: 12
Explanation:
The elimination process goes as follows:
t = 2 + 10 = 12. The other WBC splits again, using 2 units of time.t = 2 + 2 + 4 and t = 2 + 2 + 5.Example 2:
Input: timeReq = [10,4], splitTime = 5
Output:15
Explanation:
The elimination process goes as follows:
t = 5 + 10 and t = 5 + 4.
Constraints:
2 <= timeReq.length <= 1051 <= timeReq[i] <= 1091 <= splitTime <= 109First, consider the case where there is only one type of bacteria. In this case, there is no need to split the white blood cell (WBC); it can directly eliminate the bacteria, and the time cost is timeSeq[0].
If there are two types of bacteria, the WBC needs to split into two, and each WBC eliminates one type of bacteria. The time cost is splitTime + max(timeSeq[0], timeSeq[1]).
If there are more than two types of bacteria, at each step, we need to consider splitting the WBCs into multiple cells, which is difficult to handle with a forward-thinking approach.
Instead, we can adopt a reverse-thinking approach: instead of splitting the WBCs, we merge the bacteria. We select any two types of bacteria i and j to merge into a new type of bacteria. The time cost for this merge is splitTime + max(timeSeq[i], timeSeq[j]).
To minimize the involvement of bacteria with long elimination times in the merging process, we can greedily select the two bacteria with the smallest elimination times for merging at each step. Therefore, we can maintain a min-heap, repeatedly extracting the two bacteria with the smallest elimination times and merging them until only one type of bacteria remains. The elimination time of this final bacteria is the answer.
The time complexity is O(n times log n), and the space complexity is O(n), where n is the number of bacteria.
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