Intersection Point of Two Linked Lists in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the struct for a linked list node.

DSA Go

type ListNode struct {
    Val  int
    [1] *ListNode
}

Drag options to blanks, or click blank then click option'

ANext

BPrev

CHead

DChild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Prev' which is for doubly linked lists.

Using 'Head' which is not a node field.

Using 'Child' which is unrelated here.

The field 'Next' points to the next node in a singly linked list.

2fill in blank

medium

Complete the code to advance the pointer in the linked list.

DSA Go

for headA != nil && headB != nil {
    headA = headA.[1]
    headB = headB.Next
}

Drag options to blanks, or click blank then click option'

AHead

BPrev

CNext

DChild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Prev' which points backward.

Using 'Head' which is not a node pointer.

Using 'Child' which is unrelated.

To move forward in the list, use the 'Next' pointer.

3fill in blank

hard

Fix the error in the condition to check if two nodes are the same.

DSA Go

if headA [1] headB {
    return headA
}

Drag options to blanks, or click blank then click option'

A===

B=

C!=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' which assigns instead of compares.

Using '!=' which checks for inequality.

Using '===' which is not valid in Go.

Use '==' to compare if two pointers point to the same node.

4fill in blank

hard

Fill both blanks to correctly reset pointers when they reach the end.

DSA Go

if headA == nil {
    headA = [1]
}
if headB == nil {
    headB = [2]
}

Drag options to blanks, or click blank then click option'

AheadB

BheadA

CheadAOriginal

DheadBOriginal

Attempts:

3 left

💡 Hint

Common Mistakes

Resetting pointers to themselves causing infinite loops.

Using wrong variables for reset.

When a pointer reaches the end, reset it to the start of the other list to find intersection.

5fill in blank

hard

Fill all three blanks to complete the function that finds the intersection node.

DSA Go

func getIntersectionNode(headA, headB *ListNode) *ListNode {
    a, b := headA, headB
    for a != b {
        if a == nil {
            a = [1]
        } else {
            a = a.Next
        }
        if b == nil {
            b = [2]
        } else {
            b = b.Next
        }
    }
    return [3]
}

Drag options to blanks, or click blank then click option'

AheadB

BheadA

Ca

Db

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 'b' instead of 'a'.

Resetting pointers incorrectly causing infinite loops.

Not returning the intersection node.

Reset pointer 'a' to headB and 'b' to headA when they reach nil. Return 'a' which is the intersection or nil.

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