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

Time Complexity: Merge Two Sorted Linked Lists

O(n + m)

Understanding Time Complexity

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

The question is: How does the number of steps change when the input lists become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode current = dummy;
    while (l1 != null && l2 != null) {
        if (l1.val <= l2.val) {
            current.next = l1;
            l1 = l1.next;
        } else {
            current.next = l2;
            l2 = l2.next;
        }
        current = current.next;
    }
    current.next = (l1 != null) ? l1 : l2;
    return dummy.next;
}

This code merges two sorted linked lists into one sorted linked list by comparing nodes one by one.

Identify Repeating Operations

• Primary operation: The while loop that compares nodes from both lists.
• How many times: It runs until all nodes from both lists are processed, roughly the total number of nodes in both lists combined.

How Execution Grows With Input

As the total number of nodes increases, the number of comparisons and steps grows roughly the same way.

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

Pattern observation: The steps increase directly with the total number of nodes, so doubling the nodes roughly doubles the work.

Final Time Complexity

Time Complexity: O(n + m)

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

Common Mistake

[X] Wrong: "The merge takes quadratic time because of the nested if inside the loop."

[OK] Correct: The if statement is inside a single loop that moves forward through the lists once, so it does not multiply the steps; it just chooses which node to add next.

Interview Connect

Understanding this linear merging process is a key skill for working with linked lists and sorting algorithms, showing you can combine data efficiently.

Self-Check

"What if the input lists were not sorted? How would the time complexity change if we still tried to merge them this way?"

Practice

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

easy

A. To create a new sorted linked list containing all elements from both lists

B. To reverse both linked lists

C. To find the maximum element in both lists

D. To delete duplicate elements from both lists

Solution

1. Step 1: Understand the input lists

   Two linked lists are sorted individually, meaning their elements are in ascending order.

2. Step 2: Merge while maintaining order

   The goal is to combine these two lists into one sorted list that contains all elements from both.

3. Final Answer:

   To create a new sorted linked list containing all elements from both lists -> Option A

4. Quick Check:

   Merging sorted lists means combining all elements in order [OK]

Hint: Merging means combining sorted lists into one sorted list [OK]

Common Mistakes:

• Thinking merge means reversing lists
• Assuming merge removes duplicates
• Confusing merge with finding max element

2. Which of the following Java method signatures correctly represents a function to merge two sorted linked lists of integers?

easy

A. public int merge(ListNode l1, ListNode l2)

B. public void merge(int l1, int l2)

C. public ListNode merge(ListNode l1, ListNode l2)

D. public ListNode merge(int[] l1, int[] l2)

Solution

1. Step 1: Identify correct parameter types

   The function should accept two linked lists, so parameters must be ListNode types, not primitives or arrays.

2. Step 2: Identify correct return type

   The function returns a merged linked list, so return type should be ListNode, not void or int.

3. Final Answer:

   public ListNode merge(ListNode l1, ListNode l2) -> Option C

4. Quick Check:

   Method with ListNode params and return type is correct [OK]

Hint: Method merges two ListNode inputs and returns ListNode [OK]

Common Mistakes:

• Using primitive types instead of ListNode
• Returning void instead of ListNode
• Using arrays instead of linked lists

3. What is the output linked list after merging these two sorted lists?

List 1: 1 -> 3 -> 5 -> null
List 2: 2 -> 4 -> 6 -> null

medium

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

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

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

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

Solution

1. Step 1: Compare first nodes of both lists

   1 (List 1) is less than 2 (List 2), so 1 is first in merged list.

2. Step 2: Continue comparing next nodes

   Next compare 3 and 2, 2 is smaller, add 2, then 3 vs 4, add 3, then 5 vs 4, add 4, then 5 vs 6, add 5, finally add 6.

3. Final Answer:

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

4. Quick Check:

   Merge keeps ascending order combining both lists [OK]

Hint: Pick smaller head node repeatedly to merge sorted lists [OK]

Common Mistakes:

• Appending one list after the other without sorting
• Reversing order of merged list
• Skipping nodes during merge

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

public ListNode merge(ListNode l1, ListNode l2) {
    if (l1 == null) return l2;
    if (l2 == null) return l1;
    if (l1.val < l2.val) {
        l1.next = merge(l1.next, l2);
        return l1;
    } else {
        l2.next = merge(l1, l2.next);
        return l2;
    }
}

medium

A. Missing base case for both lists being null

B. No error; code correctly merges two sorted linked lists

C. Incorrect comparison operator; should be <= instead of <

D. Function does not handle empty lists properly

Solution

1. Step 1: Check base cases

   Code returns the other list if one is null, correctly handling empty lists.

2. Step 2: Check recursive merge logic

   Compares values, recursively merges next nodes, and returns the smaller node, which is correct.

3. Final Answer:

   No error; code correctly merges two sorted linked lists -> Option B

4. Quick Check:

   Recursive merge with base cases is correct [OK]

Hint: Check base cases and recursive calls carefully [OK]

Common Mistakes:

• Forgetting to return merged list
• Incorrect base case handling
• Wrong comparison operator causing order issues

5. You have two sorted linked lists:

List 1: 1 -> 4 -> 5 -> null
List 2: 1 -> 3 -> 4 -> null

After merging, what is the correct merged list?
Also, explain why the merged list contains duplicate values.

hard

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

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

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

D. 1 -> 1 -> 4 -> 5 -> null

Solution

1. Step 1: Merge nodes in ascending order

   Compare heads: both 1, add both 1s. Then 4 vs 3, add 3. Then 4 vs 4, add both 4s. Finally add 5.

2. Step 2: Understand duplicates in merged list

   Duplicates appear because both lists contain the same values independently; merge keeps all elements without removing duplicates.

3. Final Answer:

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

4. Quick Check:

   Merge keeps all elements including duplicates [OK]

Hint: Merge keeps duplicates if present in input lists [OK]

Common Mistakes:

• Removing duplicates during merge
• Appending lists without sorting
• Skipping nodes with same values