Intersection Point of Two Linked Lists in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Intersection Point Detection

What will be the output of the following Go code that finds the intersection point of two linked lists?

DSA Go

package main
import "fmt"

type ListNode struct {
    Val  int
    Next *ListNode
}

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

func main() {
    // Create intersecting lists:
    // List A: 1 -> 2 -> 3 -> 4 -> 5
    // List B: 9 -> 4 -> 5 (4 and 5 are shared nodes)
    shared := &ListNode{Val: 4, Next: &ListNode{Val: 5, Next: nil}}
    headA := &ListNode{Val: 1, Next: &ListNode{Val: 2, Next: &ListNode{Val: 3, Next: shared}}}
    headB := &ListNode{Val: 9, Next: shared}

    intersection := getIntersectionNode(headA, headB)
    if intersection != nil {
        fmt.Println(intersection.Val)
    } else {
        fmt.Println("No intersection")
    }
}

A 3

B 4

C 5

D No intersection

💡 Hint

Think about where the two lists start to share nodes.

The two lists intersect starting at the node with value 4, so the function returns that node.

🧠 Conceptual

intermediate

Understanding the Two Pointer Technique

Why does the two pointer technique used in finding the intersection point of two linked lists work?

A Because the pointers use recursion to backtrack and find the intersection.

B Because the pointers move at different speeds, one will catch the other at the intersection.

C Because the pointers compare node values to find the intersection.

D Because both pointers traverse equal total lengths by switching heads, they meet at intersection or end simultaneously.

💡 Hint

Consider what happens when each pointer reaches the end of its list.

Switching heads makes both pointers traverse the combined length of both lists, so they align at the intersection or both reach nil.

🔧 Debug

advanced

Identify the Bug in Intersection Detection

What error will this Go code produce when trying to find the intersection point of two linked lists?

DSA Go

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

A Runtime panic: nil pointer dereference

B Infinite loop

C Returns nil correctly when no intersection

D Compilation error

💡 Hint

Check the order of pointer moves and nil checks inside the loop.

The code moves pointers before checking for nil, causing nil pointer dereference when accessing Next of nil.

❓ Predict Output

advanced

Output When Lists Do Not Intersect

What will be printed by this Go program when two linked lists do not intersect?

DSA Go

package main
import "fmt"

type ListNode struct {
    Val  int
    Next *ListNode
}

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

func main() {
    headA := &ListNode{Val: 1, Next: &ListNode{Val: 2, Next: nil}}
    headB := &ListNode{Val: 3, Next: &ListNode{Val: 4, Next: nil}}

    intersection := getIntersectionNode(headA, headB)
    if intersection != nil {
        fmt.Println(intersection.Val)
    } else {
        fmt.Println("No intersection")
    }
}

A 3

B 1

C No intersection

D Runtime panic

💡 Hint

The lists have no shared nodes.

Since the lists do not share any nodes, the function returns nil and prints 'No intersection'.

🚀 Application

expert

Find Intersection Node Value After Modifications

Given two linked lists intersecting at node with value 7, after modifying the second list by adding nodes before the intersection, what is the value of the intersection node found by the algorithm?

DSA Go

package main
import "fmt"

type ListNode struct {
    Val  int
    Next *ListNode
}

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

func main() {
    shared := &ListNode{Val: 7, Next: &ListNode{Val: 8, Next: nil}}
    headA := &ListNode{Val: 1, Next: &ListNode{Val: 2, Next: shared}}
    headB := &ListNode{Val: 3, Next: &ListNode{Val: 4, Next: &ListNode{Val: 5, Next: shared}}}

    intersection := getIntersectionNode(headA, headB)
    if intersection != nil {
        fmt.Println(intersection.Val)
    } else {
        fmt.Println("No intersection")
    }
}

A 7

B 5

C 3

D No intersection

💡 Hint

The intersection node is shared regardless of nodes added before it.

Adding nodes before the intersection in one list does not change the intersection node value found by the algorithm.

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