Intersection Point of Two Linked Lists in DSA C# - Build from Scratch

Intersection Point of Two Linked Lists

📖 Scenario: Imagine two train tracks that merge at some point and continue as one track. We want to find the exact point where the two tracks join. In programming, this is like finding the intersection node of two linked lists.

🎯 Goal: You will build a program to find the intersection point of two singly linked lists. If they intersect, print the value at the intersection node; otherwise, print null.

📋 What You'll Learn

Create two linked lists with some shared nodes

Use a helper variable to track lengths of the lists

Implement logic to find the intersection node

Print the value of the intersection node or null if none

💡 Why This Matters

🌍 Real World

Finding the intersection point of linked lists is useful in memory management, detecting shared resources, and merging data streams.

💼 Career

Understanding linked list intersections helps in debugging complex data structures and optimizing algorithms in software development.

1

Create Two Linked Lists with Shared Nodes

Create a class called Node with an int data field and a Node next field. Then create two linked lists: headA and headB. The first list should be 3 -> 6 -> 9 -> 15 -> 30 -> null. The second list should be 10 -> 15 -> 30 -> null, where nodes with values 15 and 30 are shared with the first list.

DSA C#

// Define Node class and create two linked lists headA and headB with shared nodes
// Your code here

Hint

Start by defining the Node class. Then create nodes for the first list and the shared nodes. Finally, link the second list to the shared nodes.

2

Calculate Lengths of Both Lists

Create two integer variables lengthA and lengthB. Write a method GetLength that takes a Node and returns its length. Use this method to set lengthA for headA and lengthB for headB.

DSA C#

public int GetLength(Node head)
{
    // Your code here
}

int lengthA = GetLength(headA);
int lengthB = GetLength(headB);

Hint

Use a loop to count nodes until you reach null. Then return the count.

3

Find the Intersection Node

Write a method GetIntersectionNode that takes headA, headB, lengthA, and lengthB. Move the longer list's pointer ahead by the difference in lengths. Then move both pointers step-by-step until they meet or reach null. Return the intersection Node or null if none.

DSA C#

Node GetIntersectionNode(Node headA, Node headB, int lengthA, int lengthB)
{
    // Your code here
}

Hint

Advance the pointer of the longer list by the difference in lengths. Then move both pointers together until they meet or reach the end.

4

Print the Intersection Node Data or Null

Call GetIntersectionNode with headA, headB, lengthA, and lengthB. Store the result in intersection. If intersection is not null, print its data. Otherwise, print null.

DSA C#

Node intersection = GetIntersectionNode(headA, headB, lengthA, lengthB);
if (intersection != null)
{
    // Your code here
}
else
{
    // Your code here
}

Hint

Check if intersection is not null. If so, print intersection.data. Otherwise, print null.

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