Merge Two Sorted Linked Lists in DSA C++ - Time & Space Complexity

Time Complexity: Merge Two Sorted Linked Lists

O(n + m)

Understanding Time Complexity

We want to know how the time to merge two sorted linked lists grows as the lists get longer.

How does the number of steps change when the input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
    if (!l1) return l2;
    if (!l2) return l1;
    if (l1->val < l2->val) {
        l1->next = mergeTwoLists(l1->next, l2);
        return l1;
    } else {
        l2->next = mergeTwoLists(l1, l2->next);
        return l2;
    }
}
    

This code merges two sorted linked lists by choosing the smaller head each time and recursing on the rest.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Comparing the current nodes of both lists and moving forward.
• How many times: Once for each node in both lists combined, until all nodes are merged.

How Execution Grows With Input

Each step processes one node from either list, so the total steps grow with the total number of nodes.

Input Size (n)	Approx. Operations
10 (total nodes)	About 10 comparisons and moves
100 (total nodes)	About 100 comparisons and moves
1000 (total nodes)	About 1000 comparisons and moves

Pattern observation: The number of operations grows roughly in a straight line with the total number of nodes.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with the total number of nodes in both lists combined.

Common Mistake

[X] Wrong: "The time complexity is O(n * m) because we compare nodes multiple times."

[OK] Correct: Each node is visited only once, so comparisons happen at most once per node, not multiplied.

Interview Connect

Understanding this linear merging process helps you handle many linked list problems efficiently and shows you can analyze recursive code clearly.

Self-Check

"What if we changed the lists to arrays? How would the time complexity of merging change?"

Practice

1. What is the main goal when merging two sorted linked lists into one sorted linked list?

easy

A. Sort each list separately before merging

B. Reverse the order of nodes in both lists

C. Combine nodes from both lists maintaining sorted order

D. Remove duplicate nodes from both lists

Solution

1. Step 1: Understand the merge operation

   Merging means combining two lists into one while keeping the order sorted.

2. Step 2: Maintain sorted order during merge

   We pick the smaller node from the heads of both lists repeatedly to keep the merged list sorted.

3. Final Answer:

   Combine nodes from both lists maintaining sorted order -> Option C

4. Quick Check:

   Merge = Combine sorted lists [OK]

Hint: Always pick the smaller node first to keep order [OK]

Common Mistakes:

• Trying to reverse lists instead of merging
• Sorting lists again after merging
• Ignoring order and just concatenating

2. Which of the following is the correct way to declare a node structure for a singly linked list in C++?

easy

A. struct Node { int data; Node next; };

B. class Node { int data; Node next; };

C. class Node { int data; Node* next; };

D. struct Node { int data; Node* next; };

Solution

1. Step 1: Understand node structure in C++

   A node should have data and a pointer to the next node.

2. Step 2: Check pointer usage

   Using Node* next is correct to point to the next node; Node next would cause infinite size error.

3. Final Answer:

   struct Node { int data; Node* next; }; -> Option D

4. Quick Check:

   Node pointer = Node* next [OK]

Hint: Use pointers for next node in linked list [OK]

Common Mistakes:

• Using object instead of pointer for next node
• Missing semicolon after struct
• Using class without access specifier

3. What will be the output linked list after merging these two sorted lists?
List1: 1 -> 3 -> 5 -> null
List2: 2 -> 4 -> 6 -> null
Using a standard merge algorithm?

medium

A. 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null

B. 1 -> 3 -> 5 -> 2 -> 4 -> 6 -> null

C. 2 -> 4 -> 6 -> 1 -> 3 -> 5 -> null

D. 1 -> 2 -> 4 -> 3 -> 5 -> 6 -> null

Solution

1. Step 1: Compare heads of both lists

   Start with 1 (List1) and 2 (List2). 1 is smaller, so first node is 1.

2. Step 2: Continue picking smaller nodes

   Next compare 3 and 2, pick 2; then 3 and 4, pick 3; then 5 and 4, pick 4; then 5 and 6, pick 5; finally 6.

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null -> Option A

4. Quick Check:

   Sorted merge = 1,2,3,4,5,6 [OK]

Hint: Pick smaller head node repeatedly [OK]

Common Mistakes:

• Appending one list after the other without sorting
• Mixing node order incorrectly
• Skipping nodes during merge

4. Given the following code snippet to merge two sorted linked lists, what is the error?

Node* merge(Node* l1, Node* l2) {
  if (!l1) return l2;
  if (!l2) return l1;
  if (l1->data < l2->data) {
    l1->next = merge(l1->next, l2);
    return l1;
  } else {
    l2->next = merge(l1, l2->next);
    return l2;
  }
}

medium

A. No error, code correctly merges two sorted lists

B. Missing base case for empty lists

C. Incorrect comparison operator used

D. Function does not handle cyclic lists

Solution

1. Step 1: Check base cases

   Code returns the other list if one is empty, which is correct.

2. Step 2: Check recursive merge logic

   It compares data correctly and recursively merges the rest, returning the smaller node.

3. Final Answer:

   No error, code correctly merges two sorted lists -> Option A

4. Quick Check:

   Recursive merge logic correct [OK]

Hint: Check base cases and recursion carefully [OK]

Common Mistakes:

• Forgetting to return merged list
• Wrong base case conditions
• Using wrong pointers in recursion

5. You have two sorted linked lists:
List1: 1 -> 4 -> 5 -> 7 -> null
List2: 2 -> 3 -> 6 -> 8 -> 9 -> null
Write the merged list after merging them. Then, if you want to remove duplicates from the merged list, what will be the final list?
Assume duplicates can appear after merge.

hard

A. Merged: 1 -> 4 -> 5 -> 7 -> 2 -> 3 -> 6 -> 8 -> 9 -> null; After removing duplicates: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

B. Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null; After removing duplicates: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

C. Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null; After removing duplicates: 1 -> 2 -> 3 -> 4 -> 5 -> null

D. Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null; After removing duplicates: 1 -> 4 -> 5 -> 7 -> 2 -> 3 -> 6 -> 8 -> 9 -> null

Solution

1. Step 1: Merge both sorted lists

   Pick nodes in order: 1,2,3,4,5,6,7,8,9 forming a sorted merged list.

2. Step 2: Remove duplicates if any

   Since all values are unique, the list remains the same after removing duplicates.

3. Final Answer:

   Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null; After removing duplicates: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null -> Option B

4. Quick Check:

   Sorted merge with unique values [OK]

Hint: Merge sorted, then remove duplicates if any [OK]

Common Mistakes:

• Concatenating lists without sorting
• Assuming duplicates always exist
• Removing nodes incorrectly