Insert at End Tail Insert in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Tail Insert in Linked List

What is the printed linked list after inserting 10, 20, and 30 at the end using tail insert?

DSA C

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

#include <stdio.h>
#include <stdlib.h>

Node* tailInsert(Node* head, int val) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->data = val;
    newNode->next = NULL;
    if (head == NULL) {
        return newNode;
    }
    Node* temp = head;
    while (temp->next != NULL) {
        temp = temp->next;
    }
    temp->next = newNode;
    return head;
}

void printList(Node* head) {
    Node* temp = head;
    while (temp != NULL) {
        printf("%d -> ", temp->data);
        temp = temp->next;
    }
    printf("NULL\n");
}

int main() {
    Node* head = NULL;
    head = tailInsert(head, 10);
    head = tailInsert(head, 20);
    head = tailInsert(head, 30);
    printList(head);
    return 0;
}

A 10 -> 20 -> 30 -> NULL

B 30 -> 20 -> 10 -> NULL

C NULL

D 10 -> 20 -> NULL

💡 Hint

Remember that tail insert adds nodes at the end, preserving the order of insertion.

The tailInsert function adds new nodes at the end of the list. After inserting 10, 20, and 30 in that order, the list prints as 10 -> 20 -> 30 -> NULL.

🧠 Conceptual

intermediate

Number of Nodes After Tail Insertions

If you start with an empty linked list and perform tail insertions of 5 nodes, how many nodes will the list contain?

A 0

B 4

C 6

D 5

💡 Hint

Each tail insertion adds exactly one node to the list.

Starting from empty, each tail insertion adds one node. After 5 insertions, the list has 5 nodes.

🔧 Debug

advanced

Identify the Bug in Tail Insert Function

What error will occur when running this tail insert function code snippet?

DSA C

Node* tailInsert(Node* head, int val) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->data = val;
    newNode->next = NULL;
    Node* temp = head;
    if (head == NULL) {
        head = newNode;
    }
    while (temp->next != NULL) {
        temp = temp->next;
    }
    temp->next = newNode;
    return head;
}

A Memory leak due to missing free

B Segmentation fault due to dereferencing NULL pointer

C No error, works correctly

D Compilation error due to missing return

💡 Hint

Check what happens when head is NULL and you try to access temp->next.

When head is NULL, the code sets head = newNode but continues to use temp = head and then temp->next. Since head was NULL before, temp->next access causes segmentation fault.

📝 Syntax

advanced

Correct Syntax for Tail Insert Loop

Which option correctly iterates to the last node in a singly linked list for tail insertion?

A for (; temp->next != NULL; temp = temp->next);

B while (temp != NULL) { temp = temp->next; }

C while (temp->next != NULL) { temp = temp->next; }

D while (temp->next) { temp = temp->next; }

💡 Hint

You want to stop at the last node, not go past it.

Option C correctly stops at the last node where next is NULL. Option C goes past the last node causing temp to be NULL. Option C is syntactically valid but missing braces and less clear. Option C is valid C but less explicit; however, since temp->next is a pointer, it works but option C is clearer and safer.

🚀 Application

expert

Final Linked List After Multiple Tail Inserts and One Deletion

Given an empty linked list, you perform tail insertions of values 1, 2, 3, 4, 5 in order. Then you delete the node with value 3. What is the printed linked list?

DSA C

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

#include <stdio.h>
#include <stdlib.h>

Node* tailInsert(Node* head, int val) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->data = val;
    newNode->next = NULL;
    if (head == NULL) {
        return newNode;
    }
    Node* temp = head;
    while (temp->next != NULL) {
        temp = temp->next;
    }
    temp->next = newNode;
    return head;
}

Node* deleteNode(Node* head, int val) {
    if (head == NULL) return NULL;
    if (head->data == val) {
        Node* temp = head->next;
        free(head);
        return temp;
    }
    Node* curr = head;
    while (curr->next != NULL && curr->next->data != val) {
        curr = curr->next;
    }
    if (curr->next != NULL) {
        Node* temp = curr->next;
        curr->next = temp->next;
        free(temp);
    }
    return head;
}

void printList(Node* head) {
    Node* temp = head;
    while (temp != NULL) {
        printf("%d -> ", temp->data);
        temp = temp->next;
    }
    printf("NULL\n");
}

int main() {
    Node* head = NULL;
    head = tailInsert(head, 1);
    head = tailInsert(head, 2);
    head = tailInsert(head, 3);
    head = tailInsert(head, 4);
    head = tailInsert(head, 5);
    head = deleteNode(head, 3);
    printList(head);
    return 0;
}

A 1 -> 2 -> 4 -> 5 -> NULL

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

C 2 -> 4 -> 5 -> NULL

D 1 -> 2 -> 4 -> NULL

💡 Hint

Deleting node with value 3 removes it from the list, others remain in order.

After inserting 1 to 5, the list is 1 -> 2 -> 3 -> 4 -> 5 -> NULL. Deleting node with value 3 removes it, resulting in 1 -> 2 -> 4 -> 5 -> NULL.

Practice

1. What does the 'Insert at End' operation do in a singly linked list?

easy

A. Adds a new node at the end (tail) of the list

B. Deletes the last node of the list

C. Adds a new node at the beginning (head) of the list

D. Searches for a node at the end of the list

Solution

1. Step 1: Understand the operation

   Inserting at the end means adding a new node after the current last node.

2. Step 2: Identify the effect on the list

   The new node becomes the new tail, extending the list by one node.

3. Final Answer:

   Adds a new node at the end (tail) of the list -> Option A

4. Quick Check:

   Insert at end = Add node at tail [OK]

Hint: Insert at end means add after last node [OK]

Common Mistakes:

• Confusing insert at end with insert at beginning
• Thinking it deletes a node
• Assuming it searches instead of inserts

2. Which of the following is the correct syntax to create a new node in C for tail insertion?

easy

A. newNode = malloc(sizeof(Node));

B. struct Node newNode = malloc(sizeof(struct Node));

C. Node newNode = malloc(sizeof(Node));

D. struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));

Solution

1. Step 1: Understand node creation in C

   We use malloc to allocate memory and cast it to pointer of struct Node.

2. Step 2: Check syntax correctness

   struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); correctly casts malloc result to (struct Node*) and uses sizeof(struct Node).

3. Final Answer:

   struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); -> Option D

4. Quick Check:

   Correct malloc and cast syntax = struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); [OK]

Hint: Use malloc with (struct Node*) cast for new node [OK]

Common Mistakes:

• Missing cast to (struct Node*)
• Using wrong type without struct keyword
• Assigning malloc to struct variable instead of pointer

3. What will be the output linked list after inserting 4 at the end of the list 1 -> 2 -> 3 -> NULL using tail insertion?

medium

A. 1 -> 2 -> 3 -> NULL

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

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

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

Solution

1. Step 1: Understand tail insertion effect

   Inserting 4 at the end adds it after 3, the current tail.

2. Step 2: Visualize the new list

   The list becomes 1 -> 2 -> 3 -> 4 -> NULL after insertion.

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> NULL -> Option B

4. Quick Check:

   Insert 4 at tail = 1->2->3->4->NULL [OK]

Hint: New node goes after last node, not at front [OK]

Common Mistakes:

• Inserting at head instead of tail
• Placing new node in middle
• Not updating tail pointer

4. Given the following code snippet for tail insertion, what is the error?

void insertAtEnd(struct Node** head, int data) {
  struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
  newNode->data = data;
  newNode->next = NULL;
  if (*head == NULL) {
    *head = newNode;
  } else {
    struct Node* temp = *head;
    while (temp->next != NULL) {
      temp = temp->next;
    }
    temp = newNode;
  }
}

medium

A. temp = newNode; does not link newNode to the list

B. Missing malloc for newNode

C. newNode->next should not be NULL

D. Head pointer is not updated when list is empty

Solution

1. Step 1: Analyze the while loop and insertion

   The code moves temp to last node, but then does temp = newNode; which only changes temp pointer.

2. Step 2: Identify correct linking

   To link newNode, temp->next should be assigned newNode, not temp itself.

3. Final Answer:

   temp = newNode; does not link newNode to the list -> Option A

4. Quick Check:

   Link newNode by temp->next = newNode [OK]

Hint: Assign newNode to temp->next, not temp [OK]

Common Mistakes:

• Assigning newNode to temp pointer instead of temp->next
• Forgetting to set newNode->next to NULL
• Not handling empty list case

5. You have a singly linked list with nodes 10 -> 20 -> 30 -> NULL. You want to insert 40 at the end using tail insertion. Which of the following code snippets correctly performs this insertion and updates the list?

hard

A. struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));\nnewNode->data = 40;\nnewNode->next = NULL;\nhead = newNode;

B. struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));\nnewNode->data = 40;\nnewNode->next = head;\nhead = newNode;

C. struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));\nnewNode->data = 40;\nnewNode->next = NULL;\nstruct Node* temp = head;\nwhile(temp->next != NULL) { temp = temp->next; }\ntemp->next = newNode;

D. struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));\nnewNode->data = 40;\nnewNode->next = NULL;\nstruct Node* temp = head;\nwhile(temp != NULL) { temp = temp->next; }\ntemp->next = newNode;

Solution

1. Step 1: Create new node with data 40 and next NULL

   This prepares the new node to be added at the end.

2. Step 2: Traverse list to last node

   Using while(temp->next != NULL) moves temp to last node (30).

3. Step 3: Link last node's next to newNode

   temp->next = newNode correctly attaches new node at end.

4. Final Answer:

   Traverse to last node with while(temp->next != NULL), then temp->next = newNode -> Option C

5. Quick Check:

   Traverse to tail, then temp->next = newNode [OK]

Hint: Traverse to last node, then link new node at next [OK]

Common Mistakes:

• Setting newNode->next to head (inserts at front)
• Not updating last node's next pointer
• Loop condition allowing temp to become NULL