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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Intersection Detection with Different Length Lists

What is the output of the following C# code that finds the intersection point of two singly linked lists?

DSA C#

public class ListNode {
    public int val;
    public ListNode next;
    public ListNode(int x) { val = x; next = null; }
}

public class Solution {
    public ListNode GetIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) return null;
        ListNode a = headA, b = headB;
        while (a != b) {
            a = (a == null) ? headB : a.next;
            b = (b == null) ? headA : b.next;
        }
        return a;
    }
}

// Setup linked lists
ListNode common = new ListNode(8);
common.next = new ListNode(10);

ListNode headA = new ListNode(3);
headA.next = new ListNode(7);
headA.next.next = common;

ListNode headB = new ListNode(99);
headB.next = new ListNode(1);
headB.next.next = common;

Solution sol = new Solution();
ListNode intersection = sol.GetIntersectionNode(headA, headB);
System.Console.WriteLine(intersection != null ? intersection.val.ToString() : "null");

A null

B 10

C 8

D 7

💡 Hint

Think about how the pointers switch lists to align lengths.

The code uses two pointers that traverse both lists. When one pointer reaches the end, it switches to the other list's head. This aligns the pointers so they meet at the intersection node, which has value 8.

🧠 Conceptual

intermediate

Understanding Pointer Movement in Intersection Detection

In the two-pointer approach to find the intersection of two linked lists, why do we switch the pointer to the other list's head when it reaches the end?

A To ensure both pointers traverse equal total lengths, aligning them at the intersection point if it exists.

B To reset the pointers so they start over and avoid infinite loops.

C To skip nodes that are not part of the intersection.

D To reverse the linked lists for easier comparison.

💡 Hint

Consider the difference in lengths of the two lists.

Switching pointers to the other list's head makes both pointers traverse the combined length of both lists, so they meet at the intersection point or both reach null if no intersection.

❓ Predict Output

advanced

Output When No Intersection Exists

What will be printed by the following C# code when two linked lists do not intersect?

DSA C#

public class ListNode {
    public int val;
    public ListNode next;
    public ListNode(int x) { val = x; next = null; }
}

public class Solution {
    public ListNode GetIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) return null;
        ListNode a = headA, b = headB;
        while (a != b) {
            a = (a == null) ? headB : a.next;
            b = (b == null) ? headA : b.next;
        }
        return a;
    }
}

ListNode headA = new ListNode(1);
headA.next = new ListNode(2);
headA.next.next = new ListNode(3);

ListNode headB = new ListNode(4);
headB.next = new ListNode(5);
headB.next.next = new ListNode(6);

Solution sol = new Solution();
ListNode intersection = sol.GetIntersectionNode(headA, headB);
System.Console.WriteLine(intersection != null ? intersection.val.ToString() : "null");

A 6

B null

C 1

D 0

💡 Hint

If no intersection exists, what does the code return?

When no intersection exists, both pointers eventually become null at the same time, causing the loop to end and returning null.

🔧 Debug

advanced

Identify the Error in Intersection Detection Code

What error will the following C# code produce when trying to find the intersection point of two linked lists?

DSA C#

public class ListNode {
    public int val;
    public ListNode next;
    public ListNode(int x) { val = x; next = null; }
}

public class Solution {
    public ListNode GetIntersectionNode(ListNode headA, ListNode headB) {
        ListNode a = headA, b = headB;
        while (a != b) {
            a = (a == null) ? headB : a.next;
            b = (b == null) ? headA : b.next;
        }
        return a;
    }
}

A Compilation error

B StackOverflowException

C No error, returns correct intersection

D NullReferenceException

💡 Hint

Check what happens when a or b is null before accessing next.

Using 'a.next ?? headB' causes NullReferenceException when 'a' is null because 'a.next' is accessed without checking if 'a' is null.

🚀 Application

expert

Number of Nodes in Intersection Segment

Given two singly linked lists that intersect, what is the number of nodes in the intersection segment if the first intersecting node has value 15 and the lists share nodes with values [15, 30, 45]?

A 3

B 2

C 1

D 0

💡 Hint

Count all nodes from the intersection node to the end.

The intersection segment includes all nodes starting from the first intersecting node to the end of the lists. Here, nodes with values 15, 30, and 45 are shared, so count is 3.

Practice

1. What does the "Intersection Point of Two Linked Lists" problem ask you to find?

easy

A. The node where two linked lists merge or start sharing nodes

B. The length of the longer linked list

C. The sum of all node values in both lists

D. The last node of the first linked list

Solution

1. Step 1: Understand the problem goal

   The problem asks to find the node where two linked lists merge or start sharing nodes, meaning they have a common reference from that node onward.

2. Step 2: Differentiate from other concepts

   It is not about length, sum, or last node alone, but about the exact node where both lists intersect.

3. Final Answer:

   The node where two linked lists merge or start sharing nodes -> Option A

4. Quick Check:

   Intersection means shared node = A [OK]

Hint: Look for the first common node reference in both lists [OK]

Common Mistakes:

• Confusing intersection with equal values instead of shared nodes
• Thinking intersection means same length
• Looking for sum or last node instead of merge point

2. Which of the following C# code snippets correctly declares a singly linked list node class for this problem?

easy

A. class Node { int val; Node next; Node(int x) { val = x; next = null; } }

B. public class ListNode { int val; ListNode next; ListNode() { val = 0; next = null; } }

C. public struct ListNode { int val; ListNode next; }

D. public class ListNode { public int val; public ListNode next; public ListNode(int x) { val = x; next = null; } }

Solution

1. Step 1: Check for correct access modifiers and constructor

   public class ListNode { public int val; public ListNode next; public ListNode(int x) { val = x; next = null; } } uses public fields and a constructor that initializes value and next pointer properly.

2. Step 2: Identify issues in other options

   class Node { int val; Node next; Node(int x) { val = x; next = null; } } lacks public keyword, so fields are private by default; public struct ListNode { int val; ListNode next; } uses struct which is not typical for linked lists; public class ListNode { int val; ListNode next; ListNode() { val = 0; next = null; } } has no constructor parameter and fields are private.

3. Final Answer:

   public class ListNode { public int val; public ListNode next; public ListNode(int x) { val = x; next = null; } } -> Option D

4. Quick Check:

   Public class with constructor = B [OK]

Hint: Use public fields and constructor with parameter for node value [OK]

Common Mistakes:

• Omitting public keyword causing access issues
• Using struct instead of class for nodes
• Not initializing next pointer to null

3. What will be the output of the following C# code snippet that finds the intersection point of two linked lists? public class ListNode { public int val; public ListNode next; public ListNode(int x) { val = x; next = null; } } public class Solution { public ListNode GetIntersectionNode(ListNode headA, ListNode headB) { ListNode a = headA, b = headB; while (a != b) { a = (a == null) ? headB : a.next; b = (b == null) ? headA : b.next; } return a; } } // Setup lists ListNode common = new ListNode(8); common.next = new ListNode(10); ListNode headA = new ListNode(3); headA.next = new ListNode(7); headA.next.next = common; ListNode headB = new ListNode(99); headB.next = common; Solution sol = new Solution(); ListNode result = sol.GetIntersectionNode(headA, headB); Console.WriteLine(result.val);

medium

A. 99

B. 8

C. 7

D. 10

Solution

1. Step 1: Understand the linked list setup

   List A: 3 -> 7 -> 8 -> 10; List B: 99 -> 8 -> 10. The intersection node is the node with value 8.

2. Step 2: Trace the GetIntersectionNode method

   Two pointers traverse lists; when one reaches end, it switches to the other list head. They meet at intersection node 8.

3. Final Answer:

   8 -> Option B

4. Quick Check:

   Intersection node value = 8 [OK]

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

Common Mistakes:

• Confusing node values instead of node references
• Assuming intersection is last node always
• Not switching pointers to other list head

4. Identify the bug in this C# method intended to find the intersection point of two linked lists: public ListNode GetIntersectionNode(ListNode headA, ListNode headB) { ListNode a = headA, b = headB; while (a != b) { a = a.next; b = b.next; if (a == null && b == null) return null; } return a; }

medium

A. It uses incorrect comparison operator for nodes

B. It returns null too early when both pointers are null

C. It does not switch pointers to the other list when reaching null

D. It does not handle empty lists

Solution

1. Step 1: Analyze pointer movement in the loop

   Both pointers move forward but never switch to the other list head when reaching null, so they may never meet.

2. Step 2: Understand the consequence

   Without switching, if lists differ in length, pointers become null at different times and miss intersection.

3. Final Answer:

   It does not switch pointers to the other list when reaching null -> Option C

4. Quick Check:

   Switch pointers at null to find intersection [OK]

Hint: Switch pointers to other list head when null to align traversal [OK]

Common Mistakes:

• Not switching pointers causes infinite loop or wrong result
• Returning null too early without full traversal
• Confusing node equality with value equality

5. Given two singly linked lists that may or may not intersect, which approach correctly finds the intersection node in O(m+n) time and O(1) space? Options: A) Use two pointers starting at heads; when a pointer reaches end, switch it to the other list's head; continue until they meet. B) Use a hash set to store nodes of first list, then check nodes of second list for presence. C) Calculate lengths of both lists, advance longer list pointer by difference, then move both pointers until node.val match. D) Reverse both lists and compare nodes from start until mismatch. Which is the best approach for this problem?

hard

A. Two pointers switching heads when reaching end

B. Hash set to store nodes of first list

C. Advance longer list pointer by length difference then compare node.val

D. Reverse both lists and compare nodes

Solution

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

   Two pointers switching heads when reaching end uses two pointers switching heads, runs in O(m+n) time and O(1) space.

2. Step 2: Compare with other options

   Hash set to store nodes of first list uses extra space O(m); Advance longer list pointer by length difference then compare node.val is incorrect because it compares values instead of node references; Reverse both lists and compare nodes is destructive and inefficient.

3. Step 3: Choose the best approach

   Two pointers switching heads when reaching end is clean, efficient, and uses constant space without modifying lists.

4. Final Answer:

   Two pointers switching heads when reaching end -> Option A

5. Quick Check:

   Two-pointer switch method = A [OK]

Hint: Use two pointers switching heads for O(1) space and linear time [OK]

Common Mistakes:

• Using extra space unnecessarily
• Modifying lists by reversing
• Comparing node values instead of references