How do you solve Find Score of an Array After Marking All Elements on LeetCode?

Find Score of an Array After Marking All Elements (LeetCode #2593) 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: Sorting helps efficiently identify the smallest unmarked element.

What is the time complexity of Find Score of an Array After Marking All Elements?

The optimal solution for Find Score of an Array After Marking All Elements runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the marked elements and indexed array.

What is the brute force solution for Find Score of an Array After Marking All Elements?

The brute force approach for Find Score of an Array After Marking All Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we repeatedly find the smallest unmarked element, add its value to the score, and mark it along with its adjacent 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).

What algorithm or data structure is used to solve Find Score of an Array After Marking All Elements?

Find Score of an Array After Marking All Elements 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 Find Score of an Array After Marking All Elements?

Explain your thought process clearly while coding. Consider edge cases, like arrays with all identical elements. Practice writing clean and efficient code.

What are common mistakes when solving Find Score of an Array After Marking All Elements?

Not properly marking adjacent elements, leading to incorrect scores. Overlooking the need to sort elements before processing.

#2593

Find Score of an Array After Marking All Elements

Medium

ArrayHash TableSortingHeap (Priority Queue)SimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution leverages sorting and a priority queue to efficiently find the smallest unmarked element while keeping track of marked elements. This reduces the time complexity significantly.

⚙️

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: Initialize a marked array and a score variable.
3Step 3: Iterate through the sorted list, marking elements and their neighbors while updating the score.

solution.py18 lines

1import heapq
2
3def findScore(nums):
4    n = len(nums)
5    indexed_nums = [(num, i) for i, num in enumerate(nums)]
6    indexed_nums.sort()
7    marked = [False] * n
8    score = 0
9    for num, i in indexed_nums:
10        if marked[i]:
11            continue
12        score += num
13        marked[i] = True
14        if i > 0:
15            marked[i - 1] = True
16        if i < n - 1:
17            marked[i + 1] = True
18    return score

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the marked elements and indexed array.

1Sorting helps efficiently identify the smallest unmarked element.
2Using a marked array allows quick checks for already processed elements.

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