Intersection Point of Two Linked Lists in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the 'Intersection Point' in two linked lists?

The intersection point is the node where two linked lists merge and share the same next nodes onward. After this node, both lists have the same sequence of nodes.

beginner

Why can't we just compare node values to find the intersection point?

Because nodes may have the same value but be different objects. Intersection means the exact same node in memory, not just equal values.

intermediate

What is the main idea behind the two-pointer technique to find the intersection?

Use two pointers starting at each list head. Move them forward. When one reaches the end, switch it to the other list's head. They will meet at the intersection or end (null).

intermediate

What is the time complexity of the two-pointer approach to find the intersection point?

The time complexity is O(m + n), where m and n are the lengths of the two linked lists. Each pointer traverses each list at most once.

beginner

What does it mean if the two pointers never meet and both become null?

It means the two linked lists do not intersect. They have no common node.

What does the intersection point of two linked lists represent?

A The last node of the shorter list

B The node with the highest value in both lists

C The first node of the longer list

D The node where both lists share the same next nodes

The intersection point is where both lists merge and share the same nodes onward.

Why do we switch pointers to the other list's head in the two-pointer method?

A To equalize the path lengths so pointers meet at intersection

B To reset the pointers to start over

C To skip nodes with duplicate values

D To reverse the linked lists

Switching pointers balances the difference in list lengths, so they align at the intersection.

If two linked lists do not intersect, what will the two-pointer method return?

A The head of the longer list

B null

C The last node of the first list

D The first node with the same value

If no intersection exists, pointers will both reach null and the method returns null.

What is the space complexity of the two-pointer intersection algorithm?

A O(1)

B O(m + n)

C O(m)

D O(n)

The algorithm uses only two pointers and no extra data structures, so space is constant.

Which of these is NOT true about intersection in linked lists?

A Intersection means nodes are the same object in memory

B After intersection, both lists share nodes

C Nodes with same value but different objects are intersection

D Intersection can be found by pointer comparison

Nodes must be the exact same object, not just have the same value, to be intersection.

Explain how the two-pointer technique finds the intersection point of two linked lists.

Think about balancing the lengths by switching pointers.

Describe what it means for two linked lists to intersect and why node identity matters.

Focus on the difference between value equality and object identity.

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