How do you solve Merge In Between Linked Lists on LeetCode?

Merge In Between Linked Lists (LeetCode #1669) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding linked list manipulation is crucial.

What is the time complexity of Merge In Between Linked Lists?

The optimal solution for Merge In Between Linked Lists runs in O(n) time and uses O(1) space. This complexity is linear because we only traverse the relevant parts of the lists once, making it efficient.

What is the brute force solution for Merge In Between Linked Lists?

The brute force approach for Merge In Between Linked Lists is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will traverse through list1 to find the nodes we need to remove and then create a new list that combines the parts of list1 before and after the removed nodes with list2 in between. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Merge In Between Linked Lists?

Merge In Between Linked Lists uses the following concepts: Linked List. The recommended patterns to study are: Linked List, Two Pointers.

What are the interview tips for Merge In Between Linked Lists?

Explain your thought process clearly. Consider edge cases and discuss them with the interviewer. Practice linked list problems to become comfortable with pointer manipulation.

What are common mistakes when solving Merge In Between Linked Lists?

Not correctly updating the next pointers. Forgetting to handle edge cases like when a = 0 or b is the last node.

#1669

Merge In Between Linked Lists

Medium

Linked ListLinked ListTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

This approach directly modifies the links in list1 to insert list2 in place of the nodes to be removed. It is efficient because it only requires a single pass through the relevant parts of the lists.

⚙️

Algorithm

4 steps

1Step 1: Create a dummy node to simplify edge cases and point it to list1.
2Step 2: Traverse to the (a-1)th node and keep a reference to it.
3Step 3: Traverse to the (b+1)th node to find the node after the section to remove.
4Step 4: Connect the (a-1)th node's next to list2, and the last node of list2 to the (b+1)th node.

solution.py20 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 mergeInBetween(list1, a, b, list2):
8    dummy = ListNode(0)
9    dummy.next = list1
10    prev = dummy
11    for _ in range(a):
12        prev = prev.next
13    curr = prev.next
14    for _ in range(b - a + 1):
15        curr = curr.next
16    prev.next = list2
17    while list2.next:
18        list2 = list2.next
19    list2.next = curr
20    return dummy.next

ℹ

Complexity note: This complexity is linear because we only traverse the relevant parts of the lists once, making it efficient.

1Understanding linked list manipulation is crucial.
2Identifying the nodes to connect is key to solving this problem.

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