How do you solve Mark Elements on Array by Performing Queries on LeetCode?

Mark Elements on Array by Performing Queries (LeetCode #3080) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + m log k) time complexity and O(n) space complexity. Key insight: Sorting helps in efficiently finding the smallest elements.

What is the time complexity of Mark Elements on Array by Performing Queries?

The optimal solution for Mark Elements on Array by Performing Queries runs in O(n log n + m log k) time and uses O(n) space. The sorting step takes O(n log n), and for each query, we may need to process up to k elements from the priority queue, leading to a logarithmic factor for each query.

What is the brute force solution for Mark Elements on Array by Performing Queries?

The brute force approach for Mark Elements on Array by Performing Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will process each query one by one, marking the specified index and then finding the smallest unmarked elements. This is straightforward but inefficient as it requires scanning the array multiple times. The optimal Optimal Solution approach improves this to O(n log n + m log k).

What algorithm or data structure is used to solve Mark Elements on Array by Performing Queries?

Mark Elements on Array by Performing Queries uses the following concepts: Array, Hash Table, Sorting, Heap (Priority Queue), Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Mark Elements on Array by Performing Queries?

Clarify the problem requirements and constraints before jumping into coding. Discuss your thought process and approach with the interviewer. Consider edge cases, such as all elements being marked or k being larger than the number of unmarked elements.

What are common mistakes when solving Mark Elements on Array by Performing Queries?

Not properly tracking marked indices, leading to incorrect sums. Failing to handle cases where fewer than k unmarked elements exist.

#3080

Mark Elements on Array by Performing Queries

Medium

ArrayHash TableSortingHeap (Priority Queue)SimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + m log k)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n + m log k)Space O(n)

By sorting the array and using a priority queue, we can efficiently find the smallest unmarked elements. This reduces the need to repeatedly scan the array.

⚙️

Algorithm

3 steps

1Step 1: Create a list of tuples containing each element and its index, then sort this list based on the element values.
2Step 2: Use a priority queue to track the smallest unmarked elements.
3Step 3: For each query, mark the specified index and then mark the next k smallest unmarked elements from the priority queue.

solution.py32 lines

1# Full working Python code
2import heapq
3
4def markElements(nums, queries):
5    n = len(nums)
6    marked = [False] * n
7    results = []
8    total_sum = sum(nums)
9    sorted_nums = sorted((num, i) for i, num in enumerate(nums))
10    min_heap = []
11
12    for num, index in sorted_nums:
13        heapq.heappush(min_heap, (num, index))
14
15    for index, k in queries:
16        if not marked[index]:
17            marked[index] = True
18            total_sum -= nums[index]
19        count = 0
20        while count < k and min_heap:
21            num, idx = heapq.heappop(min_heap)
22            if not marked[idx]:
23                marked[idx] = True
24                total_sum -= num
25                count += 1
26        results.append(total_sum)
27    return results
28
29# Example usage
30nums = [1, 2, 2, 1, 2, 3, 1]
31queries = [[1, 2], [3, 3], [4, 2]]
32print(markElements(nums, queries))

ℹ

Complexity note: The sorting step takes O(n log n), and for each query, we may need to process up to k elements from the priority queue, leading to a logarithmic factor for each query.

1Sorting helps in efficiently finding the smallest elements.
2Using a priority queue allows for quick access to the smallest unmarked elements.

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