# Union and Intersection of two linked lists | Set-3 (Hashing)

Given two Linked Lists, create union and intersection lists that contain union and intersection of the elements present in the given lists. Order of elements in output lists doesn’t matter.

Examples:

```Input:
List1: 10 -> 15 -> 4 -> 20
list2:  8 -> 4 -> 2 -> 10
Output:
Intersection List: 4 -> 10
Union List: 2 -> 8 -> 20 -> 4 -> 15 -> 10
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

We have already discussed Method-1 and Method-2 of this question.
In this post, its Method-3 (Using Hashing) is discussed with a Time Complexity of O(m+n) i.e. better than both methods discussed earlier.

```Implementation:
1- Start traversing both the lists.
a) Store the current element of both lists
with its occurrence in the map.
2- For Union: Store all the elements of the map
in the resultant list.
3- For Intersection: Store all the elements only
with an occurrence of 2 as 2 denotes that
they are present in both the lists.
```

Below is C++ implementation of above steps.

```// C++ program to find union and intersection of
// two unsorted linked lists in O(m+n) time.
#include<bits/stdc++.h>
using namespace std;

struct Node
{
int data;
struct Node* next;
};

/* A utility function to insert a node at the
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));

/* put in the data */
new_node->data = new_data;

/* link the old list off the new node */

/* move the head to point to the new node */
}

/* Utility function to store the elements of both list */
unordered_map<int, int> &eleOcc)
{

// Traverse both lists
while (ptr1 != NULL || ptr2 != NULL)
{
// store element in the map
if (ptr1!=NULL)
{
eleOcc[ptr1->data]++;
ptr1=ptr1->next;
}

// store element in the map
if (ptr2 != NULL)
{
eleOcc[ptr2->data]++;
ptr2=ptr2->next;
}
}
}

struct Node *getUnion(unordered_map<int, int> eleOcc)
{
struct Node *result = NULL;

// Push all the elements into the resultant list
for (auto it=eleOcc.begin(); it!=eleOcc.end(); it++)
push(&result, it->first);

return result;
}

/* Function to get intersection of two linked lists
struct Node *getIntersection(unordered_map<int, int> eleOcc)
{
struct Node *result = NULL;

// Push a node with an element having occurrence
// of 2 as that means the current element is present
// in both the lists
for (auto it=eleOcc.begin(); it!=eleOcc.end(); it++)
if (it->second == 2)
push(&result, it->first);

// return resultant list
return result;
}

/* A utility function to print a linked list*/
void printList(struct Node *node)
{
while (node != NULL)
{
printf ("%d ", node->data);
node = node->next;
}
}

// Prints union and intersection of lists with head1
{
// Store all the elements of both lists in the map
unordered_map<int, int> eleOcc;

Node *intersection_list = getIntersection(eleOcc);
Node *union_list = getUnion(eleOcc);

printf("\nIntersection list is \n");
printList(intersection_list);

printf("\nUnion list is \n");
printList(union_list);
}

/* Drier program to test above function*/
int main()
{

/* create a linked lits 11->10->15->4->20 */

/* create a linked lits 8->4->2->10 */

printf("First list is \n");

printf("\nSecond list is \n");

return 0;
}
```

Output:

```First list is
5 4 3 2 1
Second list is
6 5 3 1
Intersection list is
3 5 1
Union list is
3 4 6 5 2 1
```

We can also handle the case of duplicates by maintaining separate Hash for both the lists.

Time Complexity : O(m + n)
Auxiliary Space : O(m + n)

### Asked in: Amazon, Flipkart, Komli Media, Microsoft, Taxi4Sure, [24]7 Innovation Lab

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