Why Merge Two Sorted Linked Lists in DSA C?

The Big Idea

What if you could combine two sorted lists perfectly without lifting a pen or losing track?

The Scenario

Imagine you have two lists of names sorted alphabetically on paper. You want to combine them into one big sorted list by comparing each name manually and writing them down one by one.

The Problem

Doing this by hand is slow and easy to mess up. You might lose track of which name comes next or accidentally skip some names. It takes a lot of time and effort to keep the order correct.

The Solution

Using a program to merge two sorted linked lists automatically compares the first items of each list and picks the smaller one step-by-step. It keeps the order perfectly without missing anything and finishes quickly.

Before vs After

✗ Before

while(list1 not empty and list2 not empty) {
  if(list1 head < list2 head) {
    add list1 head to result
    move list1 head forward
  } else {
    add list2 head to result
    move list2 head forward
  }
}

✓ After

struct Node* mergeSortedLists(struct Node* list1, struct Node* list2) {
  if(!list1) return list2;
  if(!list2) return list1;
  if(list1->data < list2->data) {
    list1->next = mergeSortedLists(list1->next, list2);
    return list1;
  } else {
    list2->next = mergeSortedLists(list1, list2->next);
    return list2;
  }
}

What It Enables

This lets you combine sorted data quickly and correctly, making it easy to handle bigger lists or streams of sorted information.

Real Life Example

Think about merging two sorted contact lists from different phones into one combined phonebook without losing the alphabetical order.

Key Takeaways

Manual merging is slow and error-prone.

Automated merging compares and links nodes step-by-step.

Result is a perfectly sorted combined linked list.

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 merge operation

   Merging two sorted linked lists means combining their elements into one list that remains sorted.

2. Step 2: Identify the goal

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

3. Final Answer:

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

4. Quick Check:

   Merge = combine sorted lists [OK]

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

Common Mistakes:

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

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; struct Node* next; };

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

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

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

Solution

1. Step 1: Check struct syntax for linked list node

   In C, a node contains data and a pointer to the next node. The pointer must be declared as 'struct Node* next;'.

2. Step 2: Identify correct declaration

   struct Node { int data; struct Node* next; }; correctly declares 'struct Node' with 'int data' and 'struct Node* next;'. struct Node { int data; Node* next; }; misses 'struct' keyword before Node*, struct Node { int data; struct Node next; }; incorrectly uses 'struct Node next;' (not a pointer), and struct Node { int data; struct Node* prev; }; uses 'prev' which is for doubly linked lists.

3. Final Answer:

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

4. Quick Check:

   Node struct needs pointer to next node [OK]

Hint: Node pointer must use 'struct Node*' in C [OK]

Common Mistakes:

• Omitting 'struct' keyword before Node*
• Using node instead of pointer for next
• Confusing singly linked list with doubly linked list

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

List1: 1 -> 3 -> 5 -> NULL
List2: 2 -> 4 -> 6 -> NULL

Assuming the merge function merges correctly.

medium

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

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

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

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

Solution

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

   Start with heads: 1 (List1) and 2 (List2). 1 < 2, so 1 goes first. Then compare 3 and 2, 2 < 3, so 2 goes next. Continue this way to merge in sorted order.

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

   Merged list: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL.

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL -> Option C

4. Quick Check:

   Merge keeps sorted order [OK]

Hint: Pick smaller node each step to merge sorted lists [OK]

Common Mistakes:

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

4. Consider this buggy merge function snippet in C:

struct Node* merge(struct Node* l1, struct 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;
    }
}

What is the main issue if this function is called with two sorted lists?

medium

A. It reverses the merged list

B. It merges lists correctly without issues

C. It loses nodes when one list is empty

D. It causes infinite recursion when lists have equal elements

Solution

1. Step 1: Analyze base cases

   If one list is empty (!l1 or !l2), it correctly returns the other list.

2. Step 2: Analyze recursive case

   If l1->data < l2->data, it sets l1->next to the merge of the rest and returns l1. Otherwise, sets l2->next to merge(l1, l2->next) and returns l2. However, when l1->data == l2->data, the else branch is taken repeatedly without advancing l1, causing infinite recursion.

3. Final Answer:

   It causes infinite recursion when lists have equal elements -> Option D

4. Quick Check:

   Equal elements cause infinite recursion due to no advancement in l1 [OK]

Hint: Recursive merge must handle equal elements to avoid infinite recursion [OK]

Common Mistakes:

• Thinking equal elements cause infinite recursion
• Believing it loses nodes on empty lists
• Assuming it reverses the list

5. You have two sorted linked lists:
List1: 1 -> 4 -> 7 -> NULL
List2: 2 -> 4 -> 6 -> 8 -> NULL

Write the merged sorted linked list after merging them and removing duplicate values.

hard

A. 1 -> 4 -> 7 -> 2 -> 4 -> 6 -> 8 -> NULL

B. 1 -> 2 -> 4 -> 6 -> 7 -> 8 -> NULL

C. 1 -> 2 -> 4 -> 4 -> 6 -> 7 -> 8 -> NULL

D. 1 -> 2 -> 6 -> 7 -> 8 -> NULL

Solution

1. Step 1: Merge lists in sorted order

   Merge elements: 1, 2, 4, 4, 6, 7, 8.

2. Step 2: Remove duplicate values

   Remove one of the duplicate 4's to get: 1 -> 2 -> 4 -> 6 -> 7 -> 8 -> NULL.

3. Final Answer:

   1 -> 2 -> 4 -> 6 -> 7 -> 8 -> NULL -> Option B

4. Quick Check:

   Merge + remove duplicates = sorted unique list [OK]

Hint: Merge then skip duplicates for unique sorted list [OK]

Common Mistakes:

• Leaving duplicates in merged list
• Appending lists without sorting
• Removing unique elements by mistake