Intersection Point of Two Linked Lists in DSA C++ - Time & Space Complexity

Time Complexity: Intersection Point of Two Linked Lists

O(n + m)

Understanding Time Complexity

We want to know how the time needed to find the intersection point of two linked lists changes as the lists get longer.

How does the number of steps grow when the lists have more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Find intersection point of two singly linked lists
ListNode* getIntersectionNode(ListNode* headA, ListNode* headB) {
    ListNode* a = headA;
    ListNode* b = headB;
    while (a != b) {
        a = (a == nullptr) ? headB : a->next;
        b = (b == nullptr) ? headA : b->next;
    }
    return a;
}
    

This code finds the node where two linked lists meet by traversing both lists with two pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves pointers through the lists.
• How many times: Each pointer moves at most twice the length of the longer list.

How Execution Grows With Input

As the lists get longer, the pointers walk through each node up to two times.

Input Size (n)	Approx. Operations
10	About 20 steps
100	About 200 steps
1000	About 2000 steps

Pattern observation: The steps grow roughly twice as fast as the list length, so the growth is linear.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with the total number of nodes in both lists combined.

Common Mistake

[X] Wrong: "The loop runs infinitely if there is no intersection."

[OK] Correct: The pointers switch lists once, so they will both reach the end (nullptr) and stop, ensuring the loop ends.

Interview Connect

Understanding this time complexity helps you explain how efficient your solution is and shows you can analyze linked list problems clearly.

Self-Check

"What if we used hash sets to store visited nodes? How would the time complexity change?"

Practice

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

easy

A. The last node of each linked list.

B. The node where the values of both linked lists are equal.

C. The first node of the longer linked list.

D. The node where both linked lists share the same reference (memory address).

Solution

1. Step 1: Understand linked list nodes

   Each node in a linked list has a value and a reference (address) to the next node.

2. Step 2: Define intersection point

   The intersection point is where two linked lists share the exact same node reference, not just the same value.

3. Final Answer:

   The node where both linked lists share the same reference (memory address). -> Option D

4. Quick Check:

   Intersection means shared node reference = A [OK]

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

Common Mistakes:

• Confusing same value with same node reference
• Thinking intersection is the last node always
• Assuming intersection means equal length lists

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

easy

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

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

C. struct Node { int data; Node* next(){}; };

D. struct Node { int data; Node next(); };

Solution

1. Step 1: Understand pointer usage in C++ structs

   A linked list node must have a pointer to the next node, so Node* next; is correct.

2. Step 2: Analyze options

   struct Node { int data; Node* next; }; uses Node* next; which is correct. Options A, B, and D misuse pointer or function syntax.

3. Final Answer:

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

4. Quick Check:

   Pointer to next node = C [OK]

Hint: Use pointer (*) for next node in struct [OK]

Common Mistakes:

• Forgetting the * pointer for next node
• Using function syntax instead of pointer
• Declaring next as object instead of pointer

3. Given two linked lists intersecting at node with value 7, what will the following code print?

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

// Lists:
// A: 3 -> 6 -> 9 -> 7 -> 10
// B: 1 -> 7 -> 10

Node* intersection = getIntersectionNode(headA, headB);
if (intersection) std::cout << intersection->data << std::endl;
else std::cout << -1 << std::endl;

medium

A. -1

B. 10

C. 7

D. 3

Solution

1. Step 1: Understand the algorithm

   This code uses two pointers traversing both lists; when one reaches end, it switches to the other list's head. They meet at intersection.

2. Step 2: Trace the pointers

   Both pointers meet at node with value 7, which is the intersection point.

3. Final Answer:

   7 -> Option C

4. Quick Check:

   Intersection node data = 7 [OK]

Hint: Two-pointer switch method finds intersection node [OK]

Common Mistakes:

• Assuming intersection is node with value 10
• Printing -1 when intersection exists
• Confusing node value with node reference

4. Identify the bug in this code snippet that tries to find the intersection point of two linked lists:

Node* getIntersection(Node* headA, Node* headB) {
    while (headA != headB) {
        headA = headA->next;
        headB = headB->next;
    }
    return headA;
}

medium

A. It does not handle different list lengths, causing null pointer errors.

B. It uses incorrect comparison operator for nodes.

C. It returns the wrong node after loop ends.

D. It should use recursion instead of iteration.

Solution

1. Step 1: Analyze pointer movement

   The code moves both pointers forward simultaneously without adjusting for different list lengths.

2. Step 2: Identify consequence

   If lists differ in length, one pointer may reach null before the other, causing a null pointer dereference or infinite loop.

3. Final Answer:

   It does not handle different list lengths, causing null pointer errors. -> Option A

4. Quick Check:

   Unequal lengths cause errors = D [OK]

Hint: Adjust pointers for length difference before comparing [OK]

Common Mistakes:

• Ignoring list length difference
• Dereferencing null pointers
• Assuming simultaneous traversal always works

5. You have two singly linked lists that intersect. List A has length 8, List B has length 5. Which approach correctly finds the intersection point efficiently?

hard

A. Traverse both lists fully and store nodes in hash sets, then find first common node.

B. Calculate length difference, advance longer list by difference, then move both pointers together.

C. Reverse both lists and compare nodes from start until they differ.

D. Compare node values one by one from heads until a matching value is found.

Solution

1. Step 1: Understand length difference method

   Calculate difference (8 - 5 = 3), advance longer list by 3 nodes to align both lists.

2. Step 2: Traverse both lists together

   Move both pointers one step at a time; the first node where they match is the intersection.

3. Final Answer:

   Calculate length difference, advance longer list by difference, then move both pointers together. -> Option B

4. Quick Check:

   Align by length difference then traverse = B [OK]

Hint: Advance longer list by length difference before simultaneous traversal [OK]

Common Mistakes:

• Comparing node values instead of references
• Reversing lists unnecessarily
• Using extra space when not needed