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

Time Complexity: Intersection Point of Two Linked Lists

O(m + n)

Understanding Time Complexity

We want to find how long it takes to find the point where two linked lists meet.

How does the time needed grow as the lists get longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function getIntersectionNode(headA, headB) {
  let pointerA = headA;
  let pointerB = headB;

  while (pointerA !== pointerB) {
    pointerA = pointerA ? pointerA.next : headB;
    pointerB = pointerB ? pointerB.next : headA;
  }

  return pointerA;
}
    

This code finds the node where two linked lists intersect by moving two pointers through the lists.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves pointerA and pointerB through the lists.
• How many times: Each pointer moves at most twice the length of its list before meeting or ending.

How Execution Grows With Input

As the lists get longer, the pointers travel through each node at most twice.

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

Pattern observation: The number of steps grows roughly in direct proportion to the total length of the lists.

Final Time Complexity

Time Complexity: O(m + n)

This means the time grows linearly with the sum of the lengths of both lists.

Common Mistake

[X] Wrong: "The loop runs forever if lists don't intersect."

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

Interview Connect

Understanding this approach shows you can handle linked list problems efficiently by clever pointer use, a skill valued in many coding challenges.

Self-Check

"What if we used two nested loops to compare every node of one list to every node of the other? How would the time complexity change?"

Practice

1. What does the term Intersection Point of Two Linked Lists mean?

easy

A. The node where two linked lists merge and share the same next nodes

B. The first node of the longer linked list

C. The last node of the shorter linked list

D. The node with the highest value in both lists

Solution

1. Step 1: Understand linked list structure

   Linked lists are chains of nodes where each node points to the next node.

2. Step 2: Define intersection point

   The intersection point is where two lists share the exact same node object, meaning from that node onward, both lists are identical.

3. Final Answer:

   The node where two linked lists merge and share the same next nodes -> Option A

4. Quick Check:

   Intersection means shared node = D [OK]

Hint: Intersection means shared node reference, not just value [OK]

Common Mistakes:

• Thinking intersection means same value, not same node
• Confusing intersection with concatenation
• Assuming intersection is always at head

2. Which of the following JavaScript code snippets correctly initializes two linked lists that intersect at node with value 8?

easy

A. const list1 = { val: 3, next: { val: 8, next: null } };
const list2 = { val: 7, next: { val: 8, next: null } };

B. const intersectNode = { val: 8, next: null };
const list1 = { val: 3, next: intersectNode };
const list2 = { val: 7, next: intersectNode };

C. const list1 = { val: 3, next: null };
const list2 = { val: 7, next: null };

D. const intersectNode = { val: 8, next: null };
const list1 = { val: 3, next: { val: 8, next: null } };
const list2 = { val: 7, next: intersectNode };

Solution

1. Step 1: Check for shared node reference

   const intersectNode = { val: 8, next: null }; const list1 = { val: 3, next: intersectNode }; const list2 = { val: 7, next: intersectNode }; creates one node object for value 8 and both lists point to it, ensuring intersection.

2. Step 2: Compare other options

   const list1 = { val: 3, next: { val: 8, next: null } }; const list2 = { val: 7, next: { val: 8, next: null } }; creates two separate nodes with value 8, no intersection. const list1 = { val: 3, next: null }; const list2 = { val: 7, next: null }; has no intersection node. const intersectNode = { val: 8, next: null }; const list1 = { val: 3, next: { val: 8, next: null } }; const list2 = { val: 7, next: intersectNode }; has list1 pointing to a new node, not the shared one.

3. Final Answer:

   shares the same node object for intersection -> Option B

4. Quick Check:

   Shared node object = A [OK]

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

Common Mistakes:

• Creating separate nodes with same value instead of shared node
• Not linking both lists to the intersecting node
• Confusing node value with node reference

3. Given the following code, what will be the output of console.log(getIntersectionNode(listA, listB)?.val || null);?

function getIntersectionNode(headA, headB) {
  let a = headA, b = headB;
  while (a !== b) {
    a = a ? a.next : headB;
    b = b ? b.next : headA;
  }
  return a;
}

const intersect = { val: 5, next: { val: 6, next: null } };
const listA = { val: 1, next: { val: 2, next: intersect } };
const listB = { val: 9, next: { val: 4, next: intersect } };

medium

A. 5

B. null

C. 6

D. 1

Solution

1. Step 1: Understand the algorithm

   The function uses two pointers that traverse both lists; when they meet, that node is the intersection.

2. Step 2: Identify intersection node

   Both lists share the node with value 5, so the function returns that node.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Intersection node value = 5 [OK]

Hint: Two-pointer method returns intersection node or null [OK]

Common Mistakes:

• Expecting value 6 instead of 5
• Confusing node value with list head value
• Not understanding pointer switching logic

4. The following code attempts to find the intersection node of two linked lists but may cause infinite loop when there is no intersection. What is the bug?

function getIntersectionNode(headA, headB) {
  let a = headA, b = headB;
  while (a !== b) {
    a = a.next || headB;
    b = b.next || headA;
  }
  return a;
}

medium

A. The function should return b instead of a

B. The loop condition should be 'while(a && b && a !== b)'

C. The pointers should not switch lists

D. Using 'a.next || headB' incorrectly jumps to headB when a.next is null

Solution

1. Step 1: Analyze pointer update logic

   Using 'a.next || headB' sets a to headB when a.next is null (while a points to last node), skipping null handling and causing infinite loops when no intersection.

2. Step 2: Correct pointer switching

   The correct approach is to check if 'a' is null, then switch to headB; otherwise, move to a.next.

3. Final Answer:

   Using 'a.next || headB' causes skipping headB when a.next is null -> Option D

4. Quick Check:

   Pointer update must check 'a' null, not 'a.next' [OK]

Hint: Check pointer null before moving to next list head [OK]

Common Mistakes:

• Using 'a.next || headB' instead of 'a ? a.next : headB'
• Not handling null pointers correctly
• Returning wrong pointer at end

5. You have two singly linked lists that intersect. List A length is 7, List B length is 5. How can you find the intersection point efficiently?

hard

A. Reverse both lists and compare from the start

B. Traverse both lists fully and compare node values to find intersection

C. Advance the longer list by difference in lengths, then move both pointers together until they meet

D. Use a hash set to store all values of the shorter list, then check values of the longer list

Solution

1. Step 1: Calculate length difference

   Find lengths of both lists; difference is 2 (7 - 5).

2. Step 2: Advance longer list pointer

   Move pointer of longer list ahead by 2 nodes to align both lists equally from intersection point.

3. Step 3: Move both pointers together

   Traverse both lists simultaneously until pointers are equal, which is the intersection node.

4. Final Answer:

   Advance longer list by difference in lengths, then move both pointers together until they meet -> Option C

5. Quick Check:

   Align pointers by length difference = A [OK]

Hint: Align list starts by length difference before simultaneous traversal [OK]

Common Mistakes:

• Comparing node values instead of node references
• Not aligning pointers before traversal
• Using extra space unnecessarily