Intersection Point of Two Linked Lists in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the intersection node value

Given two singly linked lists that intersect, what is the value of the node where they intersect?

DSA C

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

// List A: 1 -> 2 -> 3 -> 4 -> 5 -> NULL
// List B: 9 -> 4 -> 5 -> NULL (4 and 5 are shared nodes)

// Intersection at node with value 4

int getIntersectionNode(struct Node* headA, struct Node* headB) {
    struct Node *a = headA, *b = headB;
    while (a != b) {
        a = (a == NULL) ? headB : a->next;
        b = (b == NULL) ? headA : b->next;
    }
    return a ? a->data : -1;
}

A 3

B 5

C 4

D -1

💡 Hint

Think about how pointers move through both lists and meet at the intersection.

The function uses two pointers that traverse both lists. When they meet, it is the intersection node. Here, the intersection node has value 4.

❓ Predict Output

intermediate

Output of intersection detection with no intersection

What is the output of the function when two linked lists do not intersect?

DSA C

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

int getIntersectionNode(struct Node* headA, struct Node* headB) {
    struct Node *a = headA, *b = headB;
    while (a != b) {
        a = (a == NULL) ? headB : a->next;
        b = (b == NULL) ? headA : b->next;
    }
    return a ? a->data : -1;
}

// List A: 1 -> 2 -> 3 -> NULL
// List B: 4 -> 5 -> 6 -> NULL
// No intersection

A 0

B 6

C 3

D -1

💡 Hint

If no intersection exists, the pointers will both become NULL at the same time.

When no intersection exists, both pointers eventually become NULL simultaneously, so the function returns -1.

🧠 Conceptual

advanced

Why does the two-pointer technique work for intersection detection?

Why does moving two pointers through both lists and switching heads when reaching NULL guarantee they meet at the intersection node?

A Because both pointers traverse the same total length, so they align at the intersection after switching lists.

B Because the pointers move at different speeds, causing them to meet at the intersection.

C Because the pointers use hashing to detect the intersection node.

D Because the pointers compare node values to find the intersection.

💡 Hint

Think about the total distance each pointer travels.

Each pointer traverses both lists fully, so they cover equal distances. This alignment causes them to meet at the intersection node or both become NULL if no intersection.

🔧 Debug

advanced

Identify the bug in intersection detection code

What is the bug in this code snippet for detecting intersection of two linked lists?

DSA C

int getIntersectionNode(struct Node* headA, struct Node* headB) {
    struct Node *a = headA, *b = headB;
    while (a != NULL && b != NULL) {
        if (a == b) return a->data;
        a = a->next;
        b = b->next;
    }
    return -1;
}

A The loop stops too early; it should continue until both pointers are NULL or equal.

B The pointers should move two steps at a time to detect intersection faster.

C The function should compare node data instead of node addresses.

D The function should return 0 instead of -1 when no intersection.

💡 Hint

Think about what happens if lists have different lengths.

The loop stops when either pointer reaches NULL, missing the chance to switch heads and align pointers for intersection detection.

🚀 Application

expert

Number of nodes in intersection part

Given two intersecting singly linked lists, how many nodes are in the shared intersection part if the first intersection node has value 7 and the lists end at node with value 10?

DSA C

// List A: 3 -> 6 -> 9 -> 7 -> 8 -> 10 -> NULL
// List B: 1 -> 7 -> 8 -> 10 -> NULL
// Intersection starts at node with value 7

// Question: How many nodes are shared from intersection to end?

A 2

B 3

C 4

D 5

💡 Hint

Count nodes starting from the intersection node to the end.

Nodes shared are 7, 8, and 10, totaling 3 nodes.

Practice

1. What does the term intersection point of two linked lists mean?

easy

A. The node where both linked lists have the same value but different references

B. The node where both linked lists share the same reference

C. The first node of the longer linked list

D. The last node of the shorter linked list

Solution

1. Step 1: Understand linked list nodes and references

   Each node in a linked list has a reference (address) to the next node. Two lists intersect if they share the same node reference, not just the same value.

2. Step 2: Define intersection point

   The intersection point is the first node where both lists point to the exact same node object in memory.

3. Final Answer:

   The node where both linked lists share the same reference -> Option B

4. Quick Check:

   Intersection means shared node reference [OK]

Hint: Intersection means same node object, not just same value [OK]

Common Mistakes:

• Confusing same value with same node reference
• Thinking intersection is based on node values
• Assuming intersection is at list ends

2. Which of the following C code snippets correctly declares a node structure for a singly linked list used to find intersection?

easy

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

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

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

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

Solution

1. Step 1: Understand correct struct declaration

   In C, to declare a linked list node, you define a struct with data and a pointer to the same struct type.

2. Step 2: Check each option

   typedef struct Node { int data; struct Node* next; } Node; correctly uses typedef and struct Node with a pointer to struct Node. struct Node { int data; Node* next; }; misses typedef and uses Node* without struct keyword. typedef struct { int data; struct Node* next; } Node; misses struct name. struct Node { int data; struct Node next; }; incorrectly uses struct Node next (not a pointer).

3. Final Answer:

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

4. Quick Check:

   Correct struct with pointer to same struct [OK]

Hint: Use 'struct Node* next;' inside struct for pointer to next node [OK]

Common Mistakes:

• Forgetting 'struct' keyword before pointer type
• Using struct variable instead of pointer for next
• Missing typedef or struct name

3. Given the following code snippet to find the intersection point of two linked lists, what will be the output?

Node* getIntersectionNode(Node* headA, Node* headB) {
    Node *a = headA, *b = headB;
    while (a != b) {
        a = (a == NULL) ? headB : a->next;
        b = (b == NULL) ? headA : b->next;
    }
    return a;
}

// Lists:
// List A: 1 -> 2 -> 3 \
//                      6 -> 7 -> NULL
// List B:       4 -> 5 /

// Intersection at node with value 6
Node* result = getIntersectionNode(headA, headB);
printf("%d", result->data);

medium

A. 3

B. 5

C. 6

D. NULL

Solution

1. Step 1: Understand the two-pointer approach

   Two pointers traverse each list; when reaching end, switch to other list head. They meet at intersection or NULL if none.

2. Step 2: Trace pointers on given lists

   Lists intersect at node with value 6. Pointers will meet there after switching lists.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Two-pointer meet at intersection node value 6 [OK]

Hint: Two pointers switch lists to align lengths and meet at intersection [OK]

Common Mistakes:

• Assuming pointers meet at first node
• Not switching pointers to other list head
• Confusing node value with pointer address

4. Identify the bug in the following code snippet intended to find the intersection point of two linked lists:

Node* getIntersectionNode(Node* headA, Node* headB) {
    Node *a = headA, *b = headB;
    while (a != b) {
        if (a == NULL || b == NULL) {
            a = headB;
            b = headA;
        } else {
            a = a->next;
            b = b->next;
        }
    }
    return a;
}

medium

A. The loop may run infinitely if no intersection exists

B. Pointers should not switch lists when NULL

C. The function does not handle empty lists

D. The comparison should be 'a->data == b->data' instead of 'a != b'

Solution

1. Step 1: Analyze loop termination condition

   The loop ends when pointers a and b are equal (same node or NULL).

2. Step 2: Check pointer switching logic

   When a or b is NULL, they switch to other list head. But if no intersection, pointers cycle indefinitely without becoming equal.

3. Final Answer:

   The loop may run infinitely if no intersection exists -> Option A

4. Quick Check:

   Loop lacks proper NULL check, causing infinite loop [OK]

Hint: Add explicit NULL check to break loop if no intersection [OK]

Common Mistakes:

• Assuming pointers always meet
• Comparing node data instead of node pointers
• Not handling empty list inputs

5. You have two singly linked lists that may or may not intersect. Write a function to find the intersection point efficiently. Which approach below is best and why? A) Use two pointers starting at heads, switch to other list when reaching end, until they meet. B) Use nested loops to compare every node of list A with every node of list B. C) Store all nodes of list A in a hash set, then check nodes of list B for presence. D) Reverse both lists and compare nodes from the start until they differ.

hard

A. Two-pointer switching approach; O(n) time, O(1) space

B. Nested loops; O(n*m) time, O(1) space

C. Hash set approach; O(n+m) time, O(n) space

D. Reverse lists; modifies input and O(n) time

Solution

1. Step 1: Evaluate time and space complexity of each approach

   Two-pointer switching approach; O(n) time, O(1) space uses two pointers switching lists, runs in linear time and constant space. Nested loops; O(n*m) time, O(1) space is nested loops, very slow. Hash set approach; O(n+m) time, O(n) space uses extra space for hash set. Reverse lists; modifies input and O(n) time modifies lists, which is not ideal.

2. Step 2: Choose best approach for efficiency and safety

   Two-pointer switching approach; O(n) time, O(1) space is best because it is efficient (O(n)) and does not use extra space or modify lists.

3. Final Answer:

   Two-pointer switching approach; O(n) time, O(1) space -> Option A

4. Quick Check:

   Two-pointer method is optimal for intersection detection [OK]

Hint: Two-pointer switch method is fastest and uses no extra space [OK]

Common Mistakes:

• Choosing nested loops causing slow performance
• Using extra memory unnecessarily
• Modifying input lists causing side effects