Intersection of Two Linked Lists in O(m + n) Time Complexity using C++

Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null.

For example, the following two linked lists begin to intersect at node c1:

The test cases are generated such that there are no cycles anywhere in the entire linked structure.

Note that the linked lists must retain their original structure after the function returns.

Custom Judge:

The inputs to the judge are given as follows (your program is not given these inputs):

intersectVal - The value of the node where the intersection occurs. This is 0 if there is no intersected node.
listA - The first linked list.
listB - The second linked list.
skipA - The number of nodes to skip ahead in listA (starting from the head) to get to the intersected node.
skipB - The number of nodes to skip ahead in listB (starting from the head) to get to the intersected node.

The judge will then create the linked structure based on these inputs and pass the two heads, headA and headB to your program. If you correctly return the intersected node, then your solution will be accepted.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
Output: Intersected at '8'
Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect).
From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Intersected at '2'
Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect).
From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: No intersection
Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

Constraints:

The number of nodes of listA is in the m.
The number of nodes of listB is in the n.
1 <= m, n <= 3 * 10⁴
1 <= Node.val <= 10⁵
0 <= skipA < m
0 <= skipB < n
intersectVal is 0 if listA and listB do not intersect.
intersectVal == listA[skipA] == listB[skipB] if listA and listB intersect.

Follow up: Could you write a solution that runs in O(m + n) time and use only O(1) memory?

Understanding the Problem

The core challenge of this problem is to identify the node at which two singly linked lists intersect. This problem is significant in various applications such as finding common paths in network routing, merging data streams, and more. A common pitfall is assuming that the lists intersect at the same position from the head, which is not necessarily true.

Approach

To solve this problem, we need to consider the following steps:

Calculate the lengths of both linked lists.
Align the start of the longer list with the shorter list by skipping the extra nodes.
Traverse both lists simultaneously until we find the intersection node or reach the end of the lists.

Naive Solution

A naive solution would involve using nested loops to compare each node of the first list with each node of the second list. This approach has a time complexity of O(m * n), which is not optimal for large lists.

Optimized Solution

An optimized solution involves using two pointers to traverse the lists. By aligning the pointers to start at the same distance from the end of the lists, we can find the intersection in O(m + n) time with O(1) space complexity.

Algorithm

Here is a step-by-step breakdown of the optimized algorithm:

Initialize two pointers, one for each list.
Calculate the lengths of both lists.
Align the pointers by skipping the extra nodes in the longer list.
Traverse both lists simultaneously until the pointers meet at the intersection node or reach the end of the lists.

Code Implementation

// Definition for singly-linked list.
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(nullptr) {}
};

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        // If either head is null, there is no intersection
        if (!headA || !headB) return nullptr;

        // Initialize two pointers
        ListNode *ptrA = headA;
        ListNode *ptrB = headB;

        // Traverse both lists
        while (ptrA != ptrB) {
            // Move to the next node or switch to the other list's head
            ptrA = ptrA ? ptrA->next : headB;
            ptrB = ptrB ? ptrB->next : headA;
        }

        // Either both pointers are null (no intersection) or they meet at the intersection node
        return ptrA;
    }
};

Complexity Analysis

The time complexity of this approach is O(m + n) because we traverse each list at most twice. The space complexity is O(1) as we only use two additional pointers.

Edge Cases

Consider the following edge cases:

One or both lists are empty.
The lists do not intersect.
The intersection occurs at the head of one or both lists.

Our algorithm handles these cases by checking for null pointers and ensuring the pointers traverse the entire length of both lists.

Testing

To test the solution comprehensively, consider the following test cases:

Lists with no intersection.
Lists with intersection at various positions.
Lists of different lengths.

Use a variety of test cases to ensure the solution is robust and handles all edge cases effectively.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller, manageable parts.
Think about the properties of the data structures involved.
Consider edge cases and how to handle them.
Practice solving similar problems to improve your problem-solving skills.

Conclusion

In this blog post, we discussed how to find the intersection of two linked lists efficiently. We explored the problem definition, understood the core challenges, and presented an optimized solution with a detailed explanation. By practicing such problems, you can improve your problem-solving skills and become proficient in algorithm design.

Additional Resources

For further reading and practice, consider the following resources: