How do you solve Maximum Total from Optimal Activation Order on LeetCode?

Maximum Total from Optimal Activation Order (LeetCode #3645) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Group elements by limit values.

What is the time complexity of Maximum Total from Optimal Activation Order?

The optimal solution for Maximum Total from Optimal Activation Order runs in O(n log n) time and uses O(n) space. Sorting takes O(n log n) and the heap operations are O(log n) for each element.

What is the brute force solution for Maximum Total from Optimal Activation Order?

The brute force approach for Maximum Total from Optimal Activation Order is Brute Force, with O(n!) time complexity and O(n) space complexity. Try every possible activation order to find the maximum total. This method is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Total from Optimal Activation Order?

Maximum Total from Optimal Activation Order uses the following concepts: Array, Two Pointers, Greedy, Sorting, Heap (Priority Queue). The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Maximum Total from Optimal Activation Order?

Explain your thought process clearly. Discuss edge cases like all limits being the same. Practice similar greedy problems to build intuition.

What are common mistakes when solving Maximum Total from Optimal Activation Order?

Not respecting the limit constraints during activation. Forgetting to remove elements from the heap after activation.

#3645

Maximum Total from Optimal Activation Order

Medium

ArrayTwo PointersGreedySortingHeap (Priority Queue)GreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n log n)
Space	O(n)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

Group elements by their limits and activate the highest values first. This ensures we maximize the total while respecting the limit constraints.

⚙️

Algorithm

3 steps

1Step 1: Pair values with their limits and sort by limit.
2Step 2: Use a max-heap to keep track of the highest values that can be activated.
3Step 3: Iterate through the sorted pairs, adding values to the heap and activating the top values based on the current active count.

solution.py11 lines

1import heapq
2
3def maxActivationValue(value, limit):
4    items = sorted(zip(value, limit), key=lambda x: x[1])
5    max_total, active_count, max_heap = 0, 0, []
6    for v, l in items:
7        if active_count < l:
8            heapq.heappush(max_heap, -v)
9            active_count += 1
10        max_total += -heapq.heappop(max_heap) if max_heap else 0
11    return max_total

ℹ

Complexity note: Sorting takes O(n log n) and the heap operations are O(log n) for each element.

1Group elements by limit values.
2Use a max-heap to efficiently manage the highest values.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.