#2593

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Find Score of an Array After Marking All Elements

The problem involves marking elements in an array and calculating the score based on adjacent elements.

array hash table sorting heap priority queue simulation

Problem Statement

You are given an array nums consisting of positive integers. Starting with a score of 0, the algorithm proceeds by selecting the smallest unmarked element, marking it along with its adjacent elements, and adding its value to the score. This process continues until all elements are marked.

Your task is to return the total score after performing the above marking algorithm. The constraints on the array length and values require an efficient solution, particularly in how we handle marking and adjacent elements to avoid redundant calculations.

Examples

Example 1

Input: nums = [2,1,3,4,5,2]

Output: 7

We mark the elements as follows:

1 is the smallest unmarked element, so we mark it and its two adjacent elements: [2,1,3,4,5,2].
2 is the smallest unmarked element, so we mark it and its left adjacent element: [2,1,3,4,5,2].
4 is the only remaining unmarked element, so we mark it: [2,1,3,4,5,2]. Our score is 1 + 2 + 4 = 7.

Example 2

Input: nums = [2,3,5,1,3,2]

Output: 5

We mark the elements as follows:

1 is the smallest unmarked element, so we mark it and its two adjacent elements: [2,3,5,1,3,2].
2 is the smallest unmarked element, since there are two of them, we choose the left-most one, so we mark the one at index 0 and its right adjacent element: [2,3,5,1,3,2].
2 is the only remaining unmarked element, so we mark it: [2,3,5,1,3,2]. Our score is 1 + 2 + 2 = 5.

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 106

Solution Approach

Array Scanning

To solve the problem, we iterate over the array to identify the smallest unmarked element. During each iteration, we mark the element and its adjacent ones while accumulating the score. This array scanning is the core approach, ensuring that elements are processed in order.

Hash Table Lookup

A hash table can be used to efficiently track marked elements. Instead of repeatedly scanning the array for marked elements, we can use a hash set to quickly check if an element has been marked, ensuring that the process remains efficient.

Simulating Marking

Simulate the marking process by updating the state of adjacent elements as they are processed. When marking an element, its left and right neighbors are also marked (if applicable), and this simulation continues until all elements are processed.

Complexity Analysis

Metric	Value
Time	O(N)
Space	O(1)

The time complexity is O(N) due to the single scan through the array and the constant-time hash table lookups. The space complexity is O(1) because we only need a constant amount of space to track the marking process, regardless of the input size.

What Interviewers Usually Probe

The candidate can efficiently simulate the marking process and handle adjacent element marking without redundancies.
The candidate demonstrates a clear understanding of how hash lookups optimize the process of marking elements.
The candidate can explain how the array scanning mechanism works within the algorithm.

Common Pitfalls or Variants

Common pitfalls

Not correctly managing the marking process for adjacent elements, leading to missed or incorrectly marked elements.
Using inefficient data structures or algorithms that cause the solution to exceed the time complexity limits.
Failing to correctly simulate the marking process, causing incorrect score calculations.

Follow-up variants

Consider optimizing the algorithm to handle larger arrays with even stricter time limits.
Change the rule for marking adjacent elements, such as marking only one neighbor instead of two.
Alter the marking process to skip certain elements based on additional conditions, like element frequency.

FAQ

What is the primary pattern used in the 'Find Score of an Array After Marking All Elements' problem?

The primary pattern is array scanning combined with hash lookups to simulate marking elements and calculate the score efficiently.

How do I handle marking adjacent elements in this problem?

When marking an element, also mark its left and right neighbors, ensuring that no element is processed more than once.

Why is a hash table used in this problem?

A hash table helps track marked elements efficiently, allowing for constant-time lookups to check if an element has already been processed.

What is the time complexity of the solution?

The time complexity is O(N), where N is the number of elements in the array, due to the single pass required to process the elements.

Can this problem be solved in less than O(N) time?

No, given that every element needs to be processed at least once, achieving a time complexity better than O(N) is not feasible for this problem.

terminal

Solution

Solution 1: Priority Queue (Min Heap)

We use a priority queue to maintain the unmarked elements in the array, and each item in the queue is a tuple $(x, i)$, where $x$ and $i$ represent the element value and index of the array respectively. An array $vis$ is used to record whether the element in the array is marked.

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class Solution:
    def findScore(self, nums: List[int]) -> int:
        n = len(nums)
        vis = [False] * n
        q = [(x, i) for i, x in enumerate(nums)]
        heapify(q)
        ans = 0
        while q:
            x, i = heappop(q)
            ans += x
            vis[i] = True
            for j in (i - 1, i + 1):
                if 0 <= j < n:
                    vis[j] = True
            while q and vis[q[0][1]]:
                heappop(q)
        return ans

Solution 2: Sorting

We can create an index array $idx$ where $idx[i]=i$, and then we sort the index array $idx$ according to the element values in the array $nums$. If the element values are the same, then sort them according to the index values.

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class Solution:
    def findScore(self, nums: List[int]) -> int:
        n = len(nums)
        vis = [False] * n
        q = [(x, i) for i, x in enumerate(nums)]
        heapify(q)
        ans = 0
        while q:
            x, i = heappop(q)
            ans += x
            vis[i] = True
            for j in (i - 1, i + 1):
                if 0 <= j < n:
                    vis[j] = True
            while q and vis[q[0][1]]:
                heappop(q)
        return ans

Continue Practicing

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