Intersection Point of Two Linked Lists in DSA C - Execution Trace

Concept Flow - Intersection Point of Two Linked Lists

Start at head of List1 and List2

↓

Calculate lengths of both lists

↓

Align start pointers by skipping extra nodes in longer list

↓

Traverse both lists together

↓

Check if nodes are same (by reference)

↓

Intersection

↓

End if no intersection

Find lengths, align starts, then move both pointers step-by-step to find the first common node.

Execution Sample

DSA C

Node* getIntersection(Node* head1, Node* head2) {
  int len1 = length(head1);
  int len2 = length(head2);
  while (len1 > len2) { head1 = head1->next; len1--; }
  while (len2 > len1) { head2 = head2->next; len2--; }
  while (head1 && head2) {
    if (head1 == head2) return head1;
    head1 = head1->next;
    head2 = head2->next;
  }
  return NULL;
}

This code finds the intersection node of two linked lists by aligning their starts and moving together.

Execution Table

Step	Operation	List1 Pointer	List2 Pointer	Action	Visual State
1	Calculate length of List1	head1 at Node A	head2 at Node X	Length1 = 5	List1: A->B->C->D->E->NULL List2: X->Y->C->D->E->NULL
2	Calculate length of List2	head1 at Node A	head2 at Node X	Length2 = 5	Same as above
3	Align pointers	head1 at Node A	head2 at Node X	Lengths equal, no skip	Pointers unchanged
4	Compare nodes	Node A	Node X	Not same, move both	Pointers move to B and Y
5	Compare nodes	Node B	Node Y	Not same, move both	Pointers move to C and C
6	Compare nodes	Node C	Node C	Same node found	Intersection at Node C
7	Return intersection	Node C	Node C	Return Node C	End

💡 Intersection found at Node C, pointers matched by reference.

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	After Step 6	Final
head1	Node A	Node A	Node B	Node C	Node C	Node C
head2	Node X	Node X	Node Y	Node C	Node C	Node C
len1	0	5	5	5	5	5
len2	0	5	5	5	5	5

Key Moments - 3 Insights

Why do we skip nodes in the longer list before comparing?

Because the lists may have different lengths, skipping extra nodes aligns the pointers so they have equal nodes left to traverse, ensuring correct comparison (see Step 3 in execution_table).

Why do we compare nodes by reference, not by value?

Intersection means the exact same node in memory, not just same value. So we check if pointers are equal (Step 6), not if data matches.

What if there is no intersection?

Both pointers will reach NULL without matching nodes, and the function returns NULL (not shown here but implied after Step 6).

Visual Quiz - 3 Questions

Look at the execution_table, what is the position of head1 after Step 4?

A Node B

B Node C

C Node A

D Node Y

💡 Hint

Check the 'List1 Pointer' column at Step 4 in execution_table.

At which step do head1 and head2 point to the same node?

A Step 5

B Step 6

C Step 4

D Step 7

💡 Hint

Look for 'Same node found' in the Action column of execution_table.

If List1 was longer by 2 nodes, what would change in the execution_table?

A Steps 4 and 5 would be skipped

B Intersection would be found earlier

C Steps 3 would show skipping 2 nodes in List1

D Lengths would be equal

💡 Hint

Refer to Step 3 where pointer alignment happens by skipping extra nodes.

Concept Snapshot

Intersection Point of Two Linked Lists:
- Find lengths of both lists
- Skip extra nodes in longer list to align
- Move both pointers step-by-step
- Check if pointers are same (by reference)
- Return intersection node or NULL if none

Full Transcript

To find the intersection point of two linked lists, first calculate the lengths of both lists. Then, skip the extra nodes in the longer list so both pointers start equally far from the end. Next, move both pointers forward one node at a time, comparing if they point to the same node by reference. If yes, that node is the intersection. If no intersection exists, both pointers reach NULL and the function returns NULL.

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