Problem Description

You are given an integer array timeReq and an integer splitTime.

In this problem, you need to eliminate all bacterial strains using white blood cells (WBCs). The scenario works as follows:

Initial Setup:

• You start with only one WBC
• Each bacterial strain i takes timeReq[i] units of time to be eliminated
• The bacterial strains can be eliminated in any order

Rules for WBCs:

1. A single WBC can eliminate only one bacterial strain. After eliminating a strain, the WBC becomes exhausted and cannot perform any other tasks
2. A WBC can split itself into two WBCs, which takes splitTime units of time
3. After splitting, the two WBCs can work in parallel to eliminate different bacteria
4. Only one WBC can work on a single bacterial strain - multiple WBCs cannot attack the same strain together

Goal: Find the minimum time required to eliminate all bacterial strains.

Example Understanding:

• If there's only 1 bacterial strain with timeReq[0] = 5, one WBC can eliminate it directly in 5 time units
• If there are 2 bacterial strains with timeReq = [3, 7] and splitTime = 2, the WBC first splits (taking 2 time units), then both resulting WBCs work in parallel. The total time would be 2 + max(3, 7) = 9 time units
• For multiple strains, you need to strategically decide when to split WBCs to minimize the total time

The key insight is that since WBCs work in parallel after splitting, the total time for any group of bacteria being eliminated by split WBCs is determined by the splitting time plus the maximum elimination time among those bacteria.

Intuition

The key insight is to think about this problem in reverse. Instead of thinking about how to split WBCs to handle multiple bacteria, we can think about merging bacteria together.

Let's understand why this reverse thinking works:

When a WBC splits into two WBCs to handle two bacteria, the total time is splitTime + max(timeReq[i], timeReq[j]) where i and j are the two bacteria being handled. This is because:

• First, we spend splitTime to create two WBCs
• Then both WBCs work in parallel, so we only wait for the slower one to finish

Now, if we think backwards: instead of splitting one WBC into two to handle two bacteria, we can imagine "merging" two bacteria into one "super bacteria" that takes splitTime + max(timeReq[i], timeReq[j]) time to eliminate.

This reverse perspective transforms the problem into: repeatedly merge bacteria until only one remains, and that final bacteria's elimination time is our answer.

But which bacteria should we merge first? Here's where the greedy approach comes in:

Consider if we have bacteria with times [2, 3, 10]. If we merge the bacteria with times 2 and 10 first, we get a new bacteria with time splitTime + 10. Then merging this with the bacteria of time 3 gives us another splitTime + max(splitTime + 10, 3), which involves the large value 10 multiple times in our calculations.

However, if we merge the two smallest times first (2 and 3), we get splitTime + 3. Then merging with 10 gives splitTime + max(splitTime + 3, 10). The large value 10 appears fewer times in our nested calculations.

This suggests we should always merge the two bacteria with the smallest elimination times first. By doing this greedily, we ensure that bacteria with longer elimination times participate in fewer merges, minimizing their impact on the total time.

A min-heap is perfect for this strategy: we repeatedly extract the two smallest values, merge them (calculating the new time as splitTime + larger_value), and insert the result back into the heap. We continue until only one value remains in the heap, which represents the minimum time needed to eliminate all bacteria.

Learn more about Greedy, Math and Heap (Priority Queue) patterns.

Solution Approach

Based on our intuition of merging bacteria instead of splitting WBCs, we implement the solution using a min-heap (priority queue) data structure.

Algorithm Steps:

1. Initialize the Min-Heap: Convert the timeReq array into a min-heap. This allows us to efficiently extract the two smallest values at each step.

   heapify(timeReq)

2. Merge Process: While there's more than one bacteria remaining in the heap:

   • Extract the smallest bacteria (let's call its time min1)
   • Extract the second smallest bacteria (let's call its time min2)
   • Create a new "merged" bacteria with elimination time = min2 + splitTime
   • Push this new bacteria back into the heap

   while len(timeReq) > 1:
       heappop(timeReq)  # Remove the smallest (min1)
       heappush(timeReq, heappop(timeReq) + splitTime)  # Pop min2, add splitTime, push back

3. Return the Result: After all merges, only one element remains in the heap, which represents the minimum time needed to eliminate all bacteria.

   return timeReq[0]

Why This Implementation Works:

• When we merge two bacteria with times min1 and min2 (where min1 ≤ min2), the resulting time is splitTime + max(min1, min2) = splitTime + min2
• We discard min1 (the smaller value) and only keep splitTime + min2 because that's the time it takes for the split WBCs to handle both bacteria in parallel
• By always choosing the two smallest values to merge, we ensure that larger elimination times are involved in fewer merge operations, minimizing the total time

Time Complexity: O(n log n) where n is the number of bacteria

• Initial heapify: O(n)
• We perform n-1 merge operations
• Each merge involves 2 heap pops and 1 heap push, each taking O(log n)
• Total: O(n) + O((n-1) * 3 * log n) = O(n log n)

Space Complexity: O(1) if we can modify the input array in-place, otherwise O(n) for the heap

Example Walkthrough

Let's walk through a concrete example with timeReq = [4, 2, 8, 3] and splitTime = 5.

Step 1: Initialize the min-heap

• Convert the array into a min-heap: [2, 3, 8, 4]
• The heap property ensures we can always access the smallest element

Step 2: First merge operation

• Extract smallest: 2 (heap becomes [3, 4, 8])
• Extract second smallest: 3 (heap becomes [4, 8])
• Create merged bacteria: 3 + 5 = 8
• Push back to heap: [4, 8, 8]
• Interpretation: We split one WBC to handle bacteria requiring 2 and 3 time units. Total time for this group is 5 (split) + max(2,3) = 8

Step 3: Second merge operation

• Extract smallest: 4 (heap becomes [8, 8])
• Extract second smallest: 8 (heap becomes [8])
• Create merged bacteria: 8 + 5 = 13
• Push back to heap: [8, 13]
• Interpretation: We split another WBC to handle the bacteria requiring 4 units and the group requiring 8 units. Time is 5 + max(4,8) = 13

Step 4: Final merge operation

• Extract smallest: 8 (heap becomes [13])
• Extract second smallest: 13 (heap becomes [])
• Create merged bacteria: 13 + 5 = 18
• Push back to heap: [18]
• Interpretation: The initial WBC splits to handle the group requiring 8 units and the group requiring 13 units. Total time is 5 + max(8,13) = 18

Result: 18 time units

The WBC splitting tree looks like:

         Initial WBC (splits at t=0)
        /                           \
   (splits at t=5)              (completes at t=8)
    /           \                    |
(splits at t=10)  (done at t=9)   Bacteria[8]
  /        \           |
(t=12)   (t=13)   Bacteria[4]
  |         |
Bacteria[2] Bacteria[3]

The critical path is 5 + 5 + max(2,3) + 5 = 18 units, which matches our heap result.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The initial heapify operation takes O(n) time. The while loop runs n-1 times (reducing the heap size from n to 1). In each iteration, we perform:

• One heappop operation: O(log k) where k is the current heap size
• Another heappop operation: O(log k)
• One heappush operation: O(log k)

Since the heap size decreases from n to 2, the total time for all heap operations in the loop is: O(log n + log(n-1) + ... + log 2) × 3 = O(n × log n)

Therefore, the overall time complexity is O(n) + O(n × log n) = O(n × log n).

Space Complexity: O(n)

The heapify operation modifies the input list in-place and converts it into a min-heap structure. The heap uses the same space as the input list, which is O(n). No additional data structures are created that scale with the input size, only a constant amount of extra space is used for variables during the heap operations.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Modifying Input Array Without Permission

The solution modifies the input array timeReq directly by converting it to a heap. This can be problematic if:

• The caller expects the original array to remain unchanged
• The same array is used elsewhere in the program
• The problem explicitly states not to modify the input

Solution: Create a copy of the input array before performing heap operations:

def minEliminationTime(self, timeReq: List[int], splitTime: int) -> int:
    # Create a copy to avoid modifying the original array
    heap = timeReq.copy()
    heapify(heap)

    while len(heap) > 1:
        heappop(heap)
        next_smallest = heappop(heap)
        heappush(heap, next_smallest + splitTime)

    return heap[0]

2. Empty Array or Single Element Edge Cases

The code doesn't explicitly handle edge cases like:

• Empty array (timeReq = [])
• Single element array (timeReq = [5])

While the current implementation handles a single element correctly (the while loop won't execute), an empty array would cause an IndexError when accessing timeReq[0].

Solution: Add explicit edge case handling:

def minEliminationTime(self, timeReq: List[int], splitTime: int) -> int:
    if not timeReq:
        return 0  # No bacteria to eliminate

    if len(timeReq) == 1:
        return timeReq[0]  # Only one bacteria, no splitting needed

    heap = timeReq.copy()
    heapify(heap)

    while len(heap) > 1:
        heappop(heap)
        next_smallest = heappop(heap)
        heappush(heap, next_smallest + splitTime)

    return heap[0]

3. Integer Overflow in Languages with Fixed Integer Size

In languages like C++ or Java with fixed-size integers, repeatedly adding splitTime could cause integer overflow for large inputs or large splitTime values. Python handles arbitrary precision integers automatically, but when porting this solution to other languages, this becomes a concern.

Solution for other languages: Use appropriate data types (like long long in C++ or Long in Java) and consider adding overflow checks if necessary.

4. Misunderstanding the Merge Logic

A common mistake is to add splitTime to both extracted elements or to the sum of both elements, rather than just to the maximum (which is min2 after extraction):

# INCORRECT: Adding splitTime to both
wrong_result = heappop(heap) + heappop(heap) + splitTime

# INCORRECT: Adding splitTime to the smaller one
min1 = heappop(heap)
min2 = heappop(heap)
wrong_result = min1 + splitTime  # Should be min2 + splitTime

Solution: Always remember that the merge time is splitTime + max(min1, min2), which simplifies to splitTime + min2 when min1 ≤ min2.