Merge Two Sorted Linked Lists in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of merging two sorted linked lists

What is the printed state of the merged linked list after merging these two sorted lists?

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

DSA C

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

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

// After merging, print the list nodes in order separated by ' -> ' and ending with ' -> null'.

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

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

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

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

💡 Hint

Think about how the merge compares the current nodes of both lists and picks the smaller one first.

The merge function compares nodes from both lists and picks the smaller node to continue building the merged list. So the merged list is sorted and contains all nodes from both lists in order.

❓ Predict Output

intermediate

Output after merging lists with duplicates

Given two sorted linked lists:
List 1: 1 -> 2 -> 4 -> null
List 2: 1 -> 3 -> 4 -> null
What is the printed merged list?

DSA C

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

A 1 -> 1 -> 2 -> 3 -> 4 -> 4 -> null

B 1 -> 2 -> 3 -> 4 -> null

C 1 -> 3 -> 4 -> 1 -> 2 -> 4 -> null

D 2 -> 3 -> 4 -> 4 -> null

💡 Hint

Duplicates are allowed and merged in sorted order.

The merge function includes duplicates and keeps the order by always choosing the smaller or equal node first.

🔧 Debug

advanced

Identify the error in this merge function

What error will this merge function cause when merging two sorted linked lists?

DSA C

struct Node* mergeLists(struct Node* l1, struct Node* l2) {
    struct Node* head = NULL;
    if (l1->data < l2->data) {
        head = l1;
        head->next = mergeLists(l1->next, l2);
    } else {
        head = l2;
        head->next = mergeLists(l1, l2->next);
    }
    return head;
}

A Correctly merges the lists

B Infinite recursion causing stack overflow

C Segmentation fault (accessing NULL pointer)

D Compilation error due to missing return

💡 Hint

What happens if one list is empty (NULL)?

The function does not check if l1 or l2 is NULL before accessing l1->data or l2->data, causing a segmentation fault if either list is empty.

❓ Predict Output

advanced

Output of iterative merge of two sorted linked lists

What is the printed merged list after running this iterative merge function on:
List 1: 1 -> 4 -> 7 -> null
List 2: 2 -> 3 -> 6 -> null

DSA C

struct Node* mergeListsIterative(struct Node* l1, struct Node* l2) {
    struct Node dummy;
    struct Node* tail = &dummy;
    dummy.next = NULL;
    while (l1 && l2) {
        if (l1->data < l2->data) {
            tail->next = l1;
            l1 = l1->next;
        } else {
            tail->next = l2;
            l2 = l2->next;
        }
        tail = tail->next;
    }
    tail->next = l1 ? l1 : l2;
    return dummy.next;
}

A 1 -> 2 -> 3 -> 4 -> 7 -> 6 -> null

B 1 -> 4 -> 7 -> 2 -> 3 -> 6 -> null

C 2 -> 3 -> 6 -> 1 -> 4 -> 7 -> null

D 1 -> 2 -> 3 -> 4 -> 6 -> 7 -> null

💡 Hint

The iterative merge picks the smaller node each time and appends it to the result.

The function uses a dummy node and a tail pointer to build the merged list by choosing the smaller node from l1 or l2 until one list is empty, then appends the rest.

🧠 Conceptual

expert

Time complexity of merging two sorted linked lists

What is the time complexity of merging two sorted linked lists with lengths m and n?

A O(m + n)

B O(m * n)

C O(max(m, n)^2)

D O(log(m + n))

💡 Hint

Consider how many nodes are processed in total during the merge.

Each node from both lists is visited exactly once during the merge, so the total time is proportional to the sum of their lengths.

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