# Intersection point of two linked lists

### Problem Statement

Given two singly linked lists, which may or may not intersect each other. Write a function to return the intersection point of the two linked lists. If they don’t intersect, return NULL.

### Problem Statement Understanding

What do we mean by the intersection of two linked lists?

By intersection, we mean that the two linked lists merge into one linked list at some node. This means they will have some nodes common from the end.

Example: The above 2 linked lists intersect at the node having value 9.

In this problem, we have to find this intersection point of the two given linked list.

Now that we have understood what we have to do, think of how we can approach this problem?

### Approach 1 (Using hash-set)

One thing we can do is to traverse one of the linked lists and for each node check if it is present in the other linked list. And we return the first matching node as our intersection point.

Now, how to check for the presence of a node in the linked list?

One approach would be that to check for each node of one of the linked lists, we traverse through the other one completely.

But this method would take a lot of time. So, can we do better to check the presence of a node?

We can use a hash-set to store all nodes of the first linked list and check if any node of the second linked list is there or not. By doing so, we reduce the total time taken with the expense of extra space.

Now that you have an idea of what we are going to do, before moving ahead, try to implement it yourself.

### Algorithm

• Declare a hash-set.
• Insert all the nodes of one of the linked lists into the hash-set.
• Now, traverse the other linked list, and for each node check if it is present in the hash-set or not.
• If the node is present, then that is the required intersection point.
• If no node of the second linked list is found in the hash-set, then the two linked lists do not intersect.

### Code Implementation

```#include
using namespace std;

struct Node {
int val;
Node* next;

Node(int value){
val = value;
next = NULL;
}
};

Node* new_node = new Node(new_val);
}

while(i){
cout<val<<" ";
i = i->next;
}
cout<<"\n";
}

Node* get_intersection(Node* l1, Node* l2){
unordered_set list1;
for(Node* i=l1;i!=NULL;i=i->next){
list1.insert(i);
}
for(Node* i=l2;i!=NULL;i=i->next){
if(list1.find(i)!=list1.end()){
// we found the intersection
return i;
}
}
// if no intersection found
return NULL;
}

int main(){
Node* l1 = NULL;

push_front(&l1, 4);
push_front(&l1, 3);
push_front(&l1, 2);
push_front(&l1, 1);

Node* l2 = NULL;

push_front(&l2, 5);
push_front(&l2, 7);

// l1-> 1 2 3 4
// l2-> 7 5

Node* l3 = NULL;
push_front(&l3, 2);
push_front(&l3, 7);
push_front(&l3, 9);

Node *i = l1, *j = l2;
while(i && i->next) i = i->next;
while(j && j->next) j = j->next;

i->next = l3;
j->next = l3;

cout<<"l1: "; printList(l1);
// 1 2 3 4 9 7 2
cout<<"l2: "; printList(l2);
// 7 5 9 7 2

Node *intersection = get_intersection(l1,l2);

if(intersection != NULL){
cout<<"l1 and l2 intersect at : "<val;
}
else {
cout<<"l1 and l2 doesn't intersect at eny point";
}
}
```

### Output

l1: 1 2 3 4 9 7 2
l2: 7 5 9 7 2

l1 and l2 intersect at : 9

Time Complexity: O(n+m), as in the worst case when there is no intersection point, we need to traverse both linked lists.
Space Complexity: O(n), as we are storing all the nodes of the first linked list in a hash-set.
Where n and m are the numbers of nodes in the first and second linked lists, respectively.

Can we do some improvements?

### Approach 2 (By finding the lengths of two linked lists)

Yes, we can do some improvements if we know the lengths of the linked lists. Let’s understand this with an example:

Example 1) Let n and m be the lengths of the 2 linked lists (n = 7, m = 5).
2) Let d be the difference in their lengths (d = 2).
3) Let us look at the distance of the intersection of two linked lists from the beginning. For l1 it is 5 and for l2 it is 3.
4) Now if we move d=2 steps in the larger linked list then we will be at the same distance from the intersection point. Now, if we move together in both the lists, we will meet at the intersection point.

In this approach, we don’t use any extra space.

Now that you have an idea of what we are going to do, before moving ahead, try to implement it yourself.

### Algorithm

• Find the lengths of the two linked lists n and m.
• Find their absolute difference d.
• Now traverse both the linked lists together till their nodes become equal.
• Nodes will become equal only if they are the same node or they both are NULL.
• Once they become equal, we can return the equal node as our intersection point.

### Dry Run ### Code Implementation

```#include
using namespace std;

struct Node {
int val;
Node* next;

Node(int value){
val = value;
next = NULL;
}
};

Node* new_node = new Node(new_val);
}

while(i){
cout<val<<" ";
i = i->next;
}
cout<<"\n";
}

int len = 0;
len++;
}
return len;
}

Node* get_intersection(Node* l1, Node* l2){

int n = get_length(l1);
int m = get_length(l2);

int d = abs(n-m);

if(n>m){
while(d--) l1 = l1->next;
} else {
while(d--) l2 = l2->next;
}

// now l1 will become equal to l2
// only if they intersect or
// when they both become NULL
while(l1!=l2){
l1 = l1->next;
l2 = l2->next;
}
return l1;
}

int main(){
Node* l1 = NULL;

push_front(&l1, 4);
push_front(&l1, 3);
push_front(&l1, 2);
push_front(&l1, 1);

Node* l2 = NULL;

push_front(&l2, 5);
push_front(&l2, 7);

// l1-> 1 2 3 4
// l2-> 7 5

Node* l3 = NULL;
push_front(&l3, 2);
push_front(&l3, 7);
push_front(&l3, 9);

Node *i = l1, *j = l2;
while(i && i->next) i = i->next;
while(j && j->next) j = j->next;

i->next = l3;
j->next = l3;

cout<<"l1: "; printList(l1);
// 1 2 3 4 9 7 2
cout<<"l2: "; printList(l2);
// 7 5 9 7 2

Node *intersection = get_intersection(l1,l2);

if(intersection != NULL){
cout<<"l1 and l2 intersect at : "<val;
}
else {
cout<<"l1 and l2 doesn't intersect at eny point";
}
}
```

Time Complexity: O(n+m), as we need to traverse both linked lists to find their lengths.

In this blog, we have seen how to find the intersection point of two linked lists by two different methods. Problems like these are good for strengthening your concepts in LinkedList. I would highly recommend you to practice more such problems from Linked List.