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

Time Complexity: Merge Two Sorted Linked Lists

O(n + m)

Understanding Time Complexity

When merging two sorted linked lists, we want to know how the time needed grows as the lists get longer.

We ask: How many steps does it take to combine both lists into one sorted list?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def mergeTwoLists(l1, l2):
    dummy = ListNode()
    tail = dummy
    while l1 and l2:
        if l1.val <= l2.val:
            tail.next = l1
            l1 = l1.next
        else:
            tail.next = l2
            l2 = l2.next
        tail = tail.next
    tail.next = l1 or l2
    return dummy.next
    

This code merges two sorted linked lists by repeatedly choosing the smaller head node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

As the total number of nodes increases, the steps to merge grow roughly the same amount.

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

Pattern observation: The work grows linearly with the total number of nodes.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows in direct proportion to the total number of nodes in both lists combined.

Common Mistake

[X] Wrong: "Merging two lists takes O(n * m) time because of nested comparisons."

[OK] Correct: The code only compares one node from each list at a time, moving forward without going back, so it never repeats work for the same nodes.

Interview Connect

Understanding this linear merging process shows you can handle combining sorted data efficiently, a key skill in many coding problems.

Self-Check

"What if the input lists were stored as arrays instead of linked lists? How would the time complexity change?"

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 the order of elements in both lists

C. To find the largest element in both lists

D. To delete duplicate elements from both lists

Solution

1. Step 1: Understand the merging process

   Merging two sorted linked lists means combining them into one list that keeps the sorted order.

2. Step 2: Identify the goal

   The goal is to have a new linked list that contains all elements from both lists in sorted order.

3. Final Answer:

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

4. Quick Check:

   Merging = combine sorted lists [OK]

Hint: Merging keeps order and combines all elements [OK]

Common Mistakes:

• Thinking merging reverses lists
• Only finding largest element
• Removing duplicates instead of merging

2. Which of the following is the correct way to start merging two sorted linked lists in Python?

easy

A. def mergeLists(l1, l2):
current = ListNode(0)
dummy = current.next

B. def mergeLists(l1, l2):
dummy = null
current = dummy

C. def mergeLists(l1, l2):
dummy = ListNode(0)
current = dummy

D. def mergeLists(l1, l2):
dummy = ListNode(0)
current = dummy.next

Solution

1. Step 1: Initialize dummy and current nodes properly

   We create a dummy node with value 0 and set current to dummy to build the merged list.

2. Step 2: Check options for correct syntax

   def mergeLists(l1, l2): dummy = ListNode(0) current = dummy correctly creates dummy and current pointing to dummy. def mergeLists(l1, l2): dummy = ListNode(0) current = dummy.next incorrectly sets current = dummy.next which is initially None.

3. Final Answer:

   def mergeLists(l1, l2): dummy = ListNode(0) current = dummy -> Option C

4. Quick Check:

   Dummy node with value 0 starts merge [OK]

Hint: Start with dummy node value 0 and current pointing to dummy [OK]

Common Mistakes:

• Setting current to dummy.next
• Swapping dummy and current incorrectly
• Setting dummy to null

3. What will be the output linked list after merging these two sorted lists? List 1: 1 -> 3 -> 5 -> null List 2: 2 -> 4 -> 6 -> null Use the merge logic to combine them.

medium

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

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

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

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

Solution

1. Step 1: Compare nodes from both lists in order

   Start with 1 (List 1) and 2 (List 2). 1 is smaller, add 1. Then compare 3 and 2, add 2. Continue comparing and adding smaller nodes.

2. Step 2: Build merged list step-by-step

   Merged list becomes 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null after adding all nodes in sorted order.

3. Final Answer:

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

4. Quick Check:

   Sorted merge combines nodes in ascending order [OK]

Hint: Pick smaller node each step to keep order [OK]

Common Mistakes:

• Appending one list fully before the other
• Mixing node order incorrectly
• Skipping nodes during merge

4. Identify the error in this merge function snippet:

def mergeLists(l1, l2):
    dummy = ListNode(0)
    current = dummy
    while l1 and l2:
        if l1.val < l2.val:
            current.next = l1
            l1 = l1.next
        else:
            current.next = l2
            l2 = l2.next
    current.next = l1 if l1 else l2
    return dummy.next

medium

A. The 'current' pointer is not moved forward inside the loop

B. The dummy node is not initialized correctly

C. The function does not handle empty input lists

D. The return statement should return 'current' instead of 'dummy.next'

Solution

1. Step 1: Check pointer updates inside the loop

   After assigning current.next, the current pointer must move forward to current.next to continue building the list.

2. Step 2: Identify missing current update

   The code misses 'current = current.next' inside the while loop, causing an infinite loop or incorrect list.

3. Final Answer:

   The 'current' pointer is not moved forward inside the loop -> Option A

4. Quick Check:

   Move current forward after linking nodes [OK]

Hint: Always move current pointer after linking a node [OK]

Common Mistakes:

• Forgetting to move current pointer
• Returning wrong node
• Not handling empty lists (though not error here)

5. Given two sorted linked lists where one list is empty and the other is 0 -> 2 -> 4 -> null, what should the merged list be?

medium

A. 0 -> null

B. null

C. 4 -> 2 -> 0 -> null

D. 0 -> 2 -> 4 -> null

Solution

1. Step 1: Understand merging with an empty list

   If one list is empty, the merged list is just the non-empty list as it is already sorted.

2. Step 2: Apply to given lists

   Since the second list is empty and the first is 0 -> 2 -> 4 -> null, the merged list remains 0 -> 2 -> 4 -> null.

3. Final Answer:

   0 -> 2 -> 4 -> null -> Option D

4. Quick Check:

   Merging with empty list returns the other list [OK]

Hint: Empty list merge returns the other list unchanged [OK]

Common Mistakes:

• Returning null when one list is empty
• Reversing the non-empty list
• Returning only the first node