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

Recall & Review

beginner

What is the intersection point of two linked lists?

It is the node where two linked lists merge and share the same next nodes onwards.

beginner

Why do two linked lists intersect at a node instead of just having the same value?

Because intersection means they share the exact same node in memory, not just the same value.

intermediate

What is a simple approach to find the intersection point of two linked lists?

Calculate lengths of both lists, advance the longer list by the difference, then move both pointers step-by-step until they meet.

intermediate

What is the time complexity of finding the intersection point using length difference method?

O(m + n), where m and n are the lengths of the two linked lists.

beginner

Can two linked lists intersect if they have different tail nodes?

No, if they intersect, they must share the same tail node onwards.

What does it mean if two linked lists intersect?

A They have the same length

B They have nodes with the same values

C They share the same node in memory from the intersection point onwards

D They start at the same head node

Intersection means the two lists share the exact same node object in memory from that point forward.

Which step is NOT part of the length difference method to find intersection?

A Advance the shorter list by the difference in lengths

B Return the meeting node as intersection

C Move both pointers step-by-step until they meet

D Calculate lengths of both lists

We advance the longer list by the difference, not the shorter one.

If two linked lists do not intersect, what will happen when using the two-pointer method?

A Pointers will loop infinitely

B Pointers will meet at a random node

C Pointers will meet at the head node

D Pointers will eventually both become null without meeting

If no intersection, both pointers reach null at the end without meeting.

What is the space complexity of the length difference method for intersection?

A O(1)

B O(m + n)

C O(m)

D O(n)

Only a few pointers and counters are used, so constant space.

Why is it important to check the tail nodes of both lists when finding intersection?

A Tails determine the length of lists

B If tails differ, lists do not intersect

C Tails contain the intersection value

D Tails are always null

If tails are different nodes, lists cannot intersect.

Explain how to find the intersection point of two linked lists using the length difference method.

Think about aligning the lists so they have equal nodes left to traverse.

Describe why two linked lists must share the same tail node if they intersect.

Consider what happens after the intersection point.

Practice

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

easy

A. The last node of the shorter linked list.

B. The node where the values of both lists are equal but references differ.

C. The first node of the longer linked list.

D. The node where both linked lists share the same reference, meaning they merge.

Solution

1. Step 1: Understand linked list nodes and references

   Each node in a linked list has a reference (pointer) to the next node. Two lists intersect if they share the same node reference at some point.

2. Step 2: Define intersection point

   The intersection point is the node where both lists merge and continue as one, sharing the same reference.

3. Final Answer:

   The node where both linked lists share the same reference, meaning they merge. -> Option D

4. Quick Check:

   Intersection means shared node reference [OK]

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

Common Mistakes:

• Confusing same value with same reference
• Thinking intersection is at the start of lists
• Assuming intersection means equal length

2. Which of the following Go function signatures correctly represents a function to find the intersection point of two singly linked lists?

easy

A. func getIntersectionNode(headA ListNode, headB *ListNode) *ListNode

B. func getIntersectionNode(headA ListNode, headB ListNode) ListNode

C. func getIntersectionNode(headA *ListNode, headB *ListNode) *ListNode

D. func getIntersectionNode(headA *ListNode, headB *ListNode) int

Solution

1. Step 1: Understand Go pointer usage for linked lists

   Linked list nodes are usually handled via pointers to allow traversal and modification. So parameters should be pointers.

2. Step 2: Check return type for intersection node

   The function should return a pointer to the intersecting node or nil if none, so return type is *ListNode.

3. Final Answer:

   func getIntersectionNode(headA *ListNode, headB *ListNode) *ListNode -> Option C

4. Quick Check:

   Use pointers for nodes and return pointer [OK]

Hint: Use pointers for linked list nodes in Go function signatures [OK]

Common Mistakes:

• Passing nodes by value instead of pointer
• Returning wrong type like int instead of node pointer
• Mixing pointer and non-pointer parameters

3. Given the following Go code snippet, what will be printed?

type ListNode struct {
    Val  int
    Next *ListNode
}

func main() {
    common := &ListNode{Val: 8, Next: &ListNode{Val: 10, Next: nil}}
    listA := &ListNode{Val: 3, Next: &ListNode{Val: 7, Next: common}}
    listB := &ListNode{Val: 99, Next: &ListNode{Val: 1, Next: common}}

    intersection := getIntersectionNode(listA, listB)
    if intersection != nil {
        fmt.Println(intersection.Val)
    } else {
        fmt.Println(-1)
    }
}

func getIntersectionNode(headA, headB *ListNode) *ListNode {
    nodes := make(map[*ListNode]bool)
    for headA != nil {
        nodes[headA] = true
        headA = headA.Next
    }
    for headB != nil {
        if nodes[headB] {
            return headB
        }
        headB = headB.Next
    }
    return nil
}

medium

A. 10

B. 8

C. -1

D. 3

Solution

1. Step 1: Understand the linked lists and common node

   listA: 3 -> 7 -> 8 -> 10 listB: 99 -> 1 -> 8 -> 10 The common node starts at value 8.

2. Step 2: Trace getIntersectionNode function

   The function stores all nodes of listA in a map. Then it checks nodes of listB. The first node in listB found in map is the intersection: node with value 8.

3. Final Answer:

   8 -> Option B

4. Quick Check:

   First shared node value = 8 [OK]

Hint: Map nodes of first list, check second list for first match [OK]

Common Mistakes:

• Printing value of last node instead of intersection
• Confusing node value with node reference
• Returning nil when intersection exists

4. The following Go function is intended to find the intersection point of two linked lists but has a bug. What is the bug?

func getIntersectionNode(headA, headB *ListNode) *ListNode {
    lengthA, lengthB := 0, 0
    currA, currB := headA, headB

    for currA != nil {
        lengthA++
        currA = currA.Next
    }
    for currB != nil {
        lengthB++
        currB = currB.Next
    }

    diff := lengthA - lengthB
    if diff < 0 {
        diff = -diff
        for i := 0; i < diff; i++ {
            headA = headA.Next
        }
    } else {
        for i := 0; i < diff; i++ {
            headB = headB.Next
        }
    }

    for headA != nil && headB != nil {
        if headA == headB {
            return headA
        }
        headA = headA.Next
        headB = headB.Next
    }
    return nil
}

medium

A. The function incorrectly advances headA when diff < 0 instead of headB.

B. The function does not handle the case when one list is empty.

C. The pointers headA and headB are advanced incorrectly when diff < 0.

D. The diff calculation should be absolute value before advancing pointers.

Solution

1. Step 1: Analyze pointer advancement logic

   If lengthA < lengthB (diff < 0), the shorter list is A, so we should advance headB to align lengths, but code advances headA incorrectly.

2. Step 2: Identify correct pointer to advance

   When diff < 0, advance headB by |diff| steps, else advance headA. The code does the opposite.

3. Final Answer:

   The function incorrectly advances headA when diff < 0 instead of headB. -> Option A

4. Quick Check:

   Advance pointer of longer list, not shorter [OK]

Hint: Advance pointer of longer list by difference in length [OK]

Common Mistakes:

• Advancing pointer of shorter list instead of longer
• Not taking absolute value of length difference
• Ignoring empty list cases

5. You have two singly linked lists that may or may not intersect. Write a Go function that returns the intersection node using the two-pointer technique without extra space. Which of the following code snippets correctly implements this approach?

hard

A. func getIntersectionNode(headA, headB *ListNode) *ListNode {
if headA == nil || headB == nil {
return nil
}
pA, pB := headA, headB
for pA != pB {
if pA == nil {
pA = headB
} else {
pA = pA.Next
}
if pB == nil {
pB = headA
} else {
pB = pB.Next
}
}
return pA
}

B. func getIntersectionNode(headA, headB *ListNode) *ListNode {
visited := make(map[*ListNode]bool)
for headA != nil {
visited[headA] = true
headA = headA.Next
}
for headB != nil {
if visited[headB] {
return headB
}
headB = headB.Next
}
return nil
}

C. func getIntersectionNode(headA, headB *ListNode) *ListNode {
lengthA, lengthB := 0, 0
currA, currB := headA, headB
for currA != nil {
lengthA++
currA = currA.Next
}
for currB != nil {
lengthB++
currB = currB.Next
}
diff := lengthA - lengthB
if diff < 0 {
diff = -diff
for i := 0; i < diff; i++ {
headA = headA.Next
}
} else {
for i := 0; i < diff; i++ {
headB = headB.Next
}
}
for headA != nil && headB != nil {
if headA == headB {
return headA
}
headA = headA.Next
headB = headB.Next
}
return nil
}

D. func getIntersectionNode(headA, headB *ListNode) *ListNode {
for headA != nil {
for headB != nil {
if headA == headB {
return headA
}
headB = headB.Next
}
headA = headA.Next
}
return nil
}

Solution

1. Step 1: Understand two-pointer technique

   Two pointers start at heads of each list. When a pointer reaches end, it switches to the other list's head. They meet at intersection or nil.

2. Step 2: Analyze code snippets

   func getIntersectionNode(headA, headB *ListNode) *ListNode { if headA == nil || headB == nil { return nil } pA, pB := headA, headB for pA != pB { if pA == nil { pA = headB } else { pA = pA.Next } if pB == nil { pB = headA } else { pB = pB.Next } } return pA } correctly implements this logic with pointer switching and loop until pointers meet. func getIntersectionNode(headA, headB *ListNode) *ListNode { visited := make(map[*ListNode]bool) for headA != nil { visited[headA] = true headA = headA.Next } for headB != nil { if visited[headB] { return headB } headB = headB.Next } return nil } uses extra space (map). func getIntersectionNode(headA, headB *ListNode) *ListNode { lengthA, lengthB := 0, 0 currA, currB := headA, headB for currA != nil { lengthA++ currA = currA.Next } for currB != nil { lengthB++ currB = currB.Next } diff := lengthA - lengthB if diff < 0 { diff = -diff for i := 0; i < diff; i++ { headA = headA.Next } } else { for i := 0; i < diff; i++ { headB = headB.Next } } for headA != nil && headB != nil { if headA == headB { return headA } headA = headA.Next headB = headB.Next } return nil } uses length difference method. func getIntersectionNode(headA, headB *ListNode) *ListNode { for headA != nil { for headB != nil { if headA == headB { return headA } headB = headB.Next } headA = headA.Next } return nil } is inefficient nested loops.

3. Final Answer:

   Option A code snippet correctly implements two-pointer technique without extra space. -> Option A

4. Quick Check:

   Two-pointer switch method = func getIntersectionNode(headA, headB *ListNode) *ListNode { if headA == nil || headB == nil { return nil } pA, pB := headA, headB for pA != pB { if pA == nil { pA = headB } else { pA = pA.Next } if pB == nil { pB = headA } else { pB = pB.Next } } return pA } [OK]

Hint: Switch pointers to other list head when reaching end [OK]

Common Mistakes:

• Using extra space instead of two-pointer method
• Not switching pointers after reaching list end
• Using nested loops causing O(n^2) time