3506. Find Time Required to Eliminate Bacterial Strains 🔒
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
takestimeReq[i]
units of time to be eliminated - The bacterial strains can be eliminated in any order
Rules for WBCs:
- A single WBC can eliminate only one bacterial strain. After eliminating a strain, the WBC becomes exhausted and cannot perform any other tasks
- A WBC can split itself into two WBCs, which takes
splitTime
units of time - After splitting, the two WBCs can work in parallel to eliminate different bacteria
- 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]
andsplitTime = 2
, the WBC first splits (taking 2 time units), then both resulting WBCs work in parallel. The total time would be2 + 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:
-
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)
-
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
- Extract the smallest bacteria (let's call its time
-
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
andmin2
(wheremin1 ≤ min2
), the resulting time issplitTime + max(min1, min2) = splitTime + min2
- We discard
min1
(the smaller value) and only keepsplitTime + 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
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Start EvaluatorExample 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
1from typing import List
2from heapq import heapify, heappop, heappush
3
4class Solution:
5 def minEliminationTime(self, timeReq: List[int], splitTime: int) -> int:
6 # Convert the list into a min-heap for efficient minimum extraction
7 heapify(timeReq)
8
9 # Continue merging until only one element remains
10 while len(timeReq) > 1:
11 # Remove the smallest element (first task to complete)
12 heappop(timeReq)
13
14 # Take the next smallest element and add split time to it
15 # This simulates combining/splitting tasks
16 next_smallest = heappop(timeReq)
17 heappush(timeReq, next_smallest + splitTime)
18
19 # Return the final remaining time
20 return timeReq[0]
21
1class Solution {
2 /**
3 * Calculates the minimum time required to eliminate all tasks.
4 * Strategy: Always merge the two smallest tasks to minimize total time.
5 *
6 * @param timeReq Array containing the time required for each task
7 * @param splitTime Additional time penalty when splitting/merging tasks
8 * @return The minimum total time to complete all tasks
9 */
10 public long minEliminationTime(int[] timeReq, int splitTime) {
11 // Use a min-heap to efficiently get the smallest elements
12 PriorityQueue<Long> minHeap = new PriorityQueue<>();
13
14 // Add all initial time requirements to the heap
15 for (int time : timeReq) {
16 minHeap.offer((long) time);
17 }
18
19 // Keep merging the two smallest tasks until only one remains
20 while (minHeap.size() > 1) {
21 // Remove the smallest task (will be discarded/merged)
22 minHeap.poll();
23
24 // Remove the second smallest task and merge it with split penalty
25 // The merged task takes the time of the second task plus the split time
26 long secondSmallest = minHeap.poll();
27 minHeap.offer(secondSmallest + splitTime);
28 }
29
30 // Return the final remaining task's time
31 return minHeap.poll();
32 }
33}
34
1class Solution {
2public:
3 long long minEliminationTime(vector<int>& timeReq, int splitTime) {
4 // Type alias for long long to simplify code
5 using ll = long long;
6
7 // Min-heap to always process the smallest time requirements first
8 priority_queue<ll, vector<ll>, greater<ll>> minHeap;
9
10 // Initialize heap with all time requirements
11 for (int timeValue : timeReq) {
12 minHeap.push(timeValue);
13 }
14
15 // Keep combining elements until only one remains
16 while (minHeap.size() > 1) {
17 // Remove the smallest element (it gets eliminated)
18 minHeap.pop();
19
20 // Get the second smallest element
21 ll secondSmallest = minHeap.top();
22 minHeap.pop();
23
24 // The second element splits and takes additional time
25 // Push the new time back into the heap
26 minHeap.push(secondSmallest + splitTime);
27 }
28
29 // Return the final remaining time
30 return minHeap.top();
31 }
32};
33
1/**
2 * Calculates the minimum time required to eliminate all tasks
3 * using a greedy approach with priority queue
4 *
5 * @param timeReq - Array of time requirements for each task
6 * @param splitTime - Additional time penalty when combining tasks
7 * @returns The minimum total time to complete all tasks
8 */
9function minEliminationTime(timeReq: number[], splitTime: number): number {
10 // Initialize a min-heap priority queue to store task times
11 const priorityQueue = new MinPriorityQueue();
12
13 // Add all initial task times to the priority queue
14 for (const taskTime of timeReq) {
15 priorityQueue.enqueue(taskTime);
16 }
17
18 // Keep combining the two smallest tasks until only one remains
19 while (priorityQueue.size() > 1) {
20 // Remove the smallest task (will be eliminated)
21 priorityQueue.dequeue()!;
22
23 // Remove the second smallest task and add it back with split time penalty
24 // This represents combining it with the eliminated task
25 const secondSmallest = priorityQueue.dequeue()!;
26 priorityQueue.enqueue(secondSmallest + splitTime);
27 }
28
29 // Return the final remaining task time
30 return priorityQueue.dequeue()!;
31}
32
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)
wherek
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
.
Consider the classic dynamic programming of fibonacci numbers, what is the recurrence relation?
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