How do you solve Swap Nodes in Pairs on LeetCode?

Swap Nodes in Pairs (LeetCode #24) can be solved using Brute Force or Optimal Solution (Iterative). The optimal approach is Optimal Solution (Iterative) with O(n) time complexity and O(1) space complexity. Key insight: Recognizing that we can swap nodes directly instead of using values can significantly optimize the solution.

What is the brute force solution for Swap Nodes in Pairs?

The brute force approach for Swap Nodes in Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through the linked list and swapping the values of every two adjacent nodes. This is straightforward but inefficient because it requires multiple passes over the list. The optimal Optimal Solution (Iterative) approach improves this to O(n).

What algorithm or data structure is used to solve Swap Nodes in Pairs?

Swap Nodes in Pairs uses the following concepts: Linked List, Recursion. The recommended patterns to study are: Linked List, Two Pointers, Recursion.

What are the interview tips for Swap Nodes in Pairs?

When you first see this problem, clarify whether you can modify the node pointers and not just the values. Communicate your thought process clearly, especially when discussing edge cases like empty lists or lists with one node. Mention edge cases such as lists with an odd number of nodes or empty lists to show you are thinking critically about the problem.

What are common mistakes when solving Swap Nodes in Pairs?

Students often forget to handle the case where the list has an odd number of nodes, leading to a null pointer exception. Not using a dummy node can complicate the logic for edge cases, especially when the head of the list is involved.

#24

Swap Nodes in Pairs

Medium

Linked ListRecursionLinked ListTwo PointersRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution involves swapping nodes directly in the linked list without using extra space for values. This is efficient because we only traverse the list once, modifying pointers directly.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dummy node to handle edge cases and set it to point to the head.
2Step 2: Use a pointer to traverse the list in pairs.
3Step 3: For each pair, adjust the pointers to swap them.

solution.py20 lines

1# Full working Python code with comments
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def swapPairs(head):
8    dummy = ListNode(0)
9    dummy.next = head
10    prev = dummy
11    while prev.next and prev.next.next:
12        first = prev.next
13        second = first.next
14        # Swapping
15        first.next = second.next
16        second.next = first
17        prev.next = second
18        # Move to the next pair
19        prev = first
20    return dummy.next

ℹ

Complexity note: The time complexity is O(n) because we traverse the list only once. The space complexity is O(1) since we are not using any additional data structures that grow with input size.

1Recognizing that we can swap nodes directly instead of using values can significantly optimize the solution.
2Understanding the role of a dummy node can help manage edge cases more effectively.

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