How do you solve Delete Nodes From Linked List Present in Array on LeetCode?

Delete Nodes From Linked List Present in Array (LeetCode #3217) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a set allows for efficient lookups, significantly improving performance.

What is the time complexity of Delete Nodes From Linked List Present in Array?

The optimal solution for Delete Nodes From Linked List Present in Array runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the linked list once, and the space complexity is O(n) due to the set storing the values from the array.

What is the brute force solution for Delete Nodes From Linked List Present in Array?

The brute force approach for Delete Nodes From Linked List Present in Array is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we iterate through each node in the linked list and check if its value exists in the array. If it does, we remove that node. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Delete Nodes From Linked List Present in Array?

Delete Nodes From Linked List Present in Array uses the following concepts: Array, Hash Table, Linked List. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Delete Nodes From Linked List Present in Array?

Always consider the time and space complexity of your solution. Think about edge cases, such as an empty list or all nodes needing removal. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Delete Nodes From Linked List Present in Array?

Not using a set for lookups, leading to inefficient O(n²) solutions. Failing to handle edge cases, such as removing the head of the linked list.

#3217

Delete Nodes From Linked List Present in Array

Medium

ArrayHash TableLinked ListHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses a set to store the values from the array, allowing for O(1) average time complexity for lookups. We then iterate through the linked list just once, removing nodes as needed.

⚙️

Algorithm

3 steps

1Step 1: Convert the array nums into a set for O(1) lookups.
2Step 2: Initialize a dummy node to simplify edge cases and maintain the head of the modified list.
3Step 3: Iterate through the linked list, checking if each node's value is in the set. If it is, skip it; otherwise, link it to the new list.

solution.py18 lines

1# Full working Python code
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def deleteNodes(nums, head):
8    num_set = set(nums)
9    dummy = ListNode(0)
10    prev = dummy
11    current = head
12    while current:
13        if current.val not in num_set:
14            prev.next = current
15            prev = current
16        current = current.next
17    prev.next = None  # Terminate the list
18    return dummy.next

ℹ

Complexity note: The time complexity is O(n) because we traverse the linked list once, and the space complexity is O(n) due to the set storing the values from the array.

1Using a set allows for efficient lookups, significantly improving performance.
2Iterating through the linked list only once is key to achieving optimal time complexity.

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