Find common elements in three linked lists
Last Updated :
09 Jan, 2023
Given three linked lists, find all common elements among the three linked lists.
Examples:
Input :
10 15 20 25 12
10 12 13 15
10 12 15 24 25 26
Output : 10 12 15
Input :
1 2 3 4 5
1 2 3 4 6 9 8
1 2 4 5 10
Output : 1 2 4
Method 1 : (Simple)
Use three-pointers to iterate the given three linked lists and if any element common print that element.
Time complexity of the above solution will be O(N*N*N)
Method 2 : (Use Merge Sort)
In this method, we first sort the three lists and then we traverse the sorted lists to get the intersection.
Following are the steps to be followed to get intersection of three lists:
- Sort the first Linked List using merge sort. This step takes O(mLogm) time. Refer this post for details of this step.
- Sort the second Linked List using merge sort. This step takes O(nLogn) time. Refer this post for details of this step.
- Sort the third Linked List using merge sort. This step takes O(pLogp) time. Refer this post for details of this step.
- Linearly scan three sorted lists to get the intersection. This step takes O(m + n + p) time. This step can be implemented using the same algorithm as sorted arrays algorithm discussed here.
Time complexity of this method is O(mLogm + nLogn + plogp) which is better than method 1’s time complexity.
Method 3 : (Hashing)
Following are the steps to be followed to get intersection of three lists using hashing:
- Create an empty hash table. Iterate through the first linked list and mark all the element frequency as 1 in the hash table. This step takes O(m) time.
- Iterate through the second linked list and if current element frequency is 1 in hash table mark it as 2. This step takes O(n) time.
- Iterate the third linked list and if the current element frequency is 2 in hash table mark it as 3. This step takes O(p) time.
- Now iterate first linked list again to check the frequency of elements. if an element with frequency three exist in hash table, it will be present in the intersection of three linked lists. This step takes O(m) time.
Time complexity of this method is O(m + n + p) which is better than time complexity of method 1 and 2.
Below is the implementation of the above idea.
C++
// C++ program to find common element
// in three unsorted linked list
#include <bits/stdc++.h>
#define max 1000000
using namespace std;
/* Link list node */
struct Node {
int data;
struct Node* next;
};
/* A utility function to insert a node at the
beginning of a linked list */
void push(struct Node** head_ref, int new_data)
{
struct Node* new_node =
(struct Node *)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
/* print the common element in between
given three linked list*/
void Common(struct Node* head1,
struct Node* head2, struct Node* head3)
{
// Creating empty hash table;
unordered_map<int, int> hash;
struct Node* p = head1;
while (p != NULL) {
// set frequency by 1
hash[p->data] = 1;
p = p->next;
}
struct Node* q = head2;
while (q != NULL) {
// if the element is already exist in the
// linked list set its frequency 2
if (hash.find(q->data) != hash.end())
hash[q->data] = 2;
q = q->next;
}
struct Node* r = head3;
while (r != NULL) {
if (hash.find(r->data) != hash.end() &&
hash[r->data] == 2)
// if the element frequency is 2 it means
// its present in both the first and second
// linked list set its frequency 3
hash[r->data] = 3;
r = r->next;
}
for (auto x : hash) {
// if current frequency is 3 its means
// element is common in all the given
// linked list
if (x.second == 3)
cout << x.first << " ";
}
}
// Driver code
int main()
{
// first list
struct Node* head1 = NULL;
push(&head1, 20);
push(&head1, 5);
push(&head1, 15);
push(&head1, 10);
// second list
struct Node* head2 = NULL;
push(&head2, 10);
push(&head2, 20);
push(&head2, 15);
push(&head2, 8);
// third list
struct Node* head3 = NULL;
push(&head3, 10);
push(&head3, 2);
push(&head3, 15);
push(&head3, 20);
Common(head1, head2, head3);
return 0;
}
// Java program to find common element
// in three unsorted linked list
import java.util.*;
class GFG
{
static int max = 1000000;
/* Link list node */
static class Node
{
int data;
Node next;
};
/* A utility function to insert a node
at the beginning of a linked list */
static Node push(Node head_ref,
int new_data)
{
Node new_node = new Node();
new_node.data = new_data;
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
/* print the common element in between
given three linked list*/
static void Common(Node head1,
Node head2,
Node head3)
{
// Creating empty hash table;
HashMap<Integer,
Integer> hash = new HashMap<Integer,
Integer>();
Node p = head1;
while (p != null)
{
// set frequency by 1
hash. put(p.data, 1);
p = p.next;
}
Node q = head2;
while (q != null)
{
// if the element is already exist in the
// linked list set its frequency 2
if (hash.containsKey(q.data))
hash. put(q.data, 2);
q = q.next;
}
Node r = head3;
while (r != null)
{
if (hash.containsKey(r.data)&&
hash.get(r.data) == 2)
// if the element frequency is 2 it means
// its present in both the first and second
// linked list set its frequency 3
hash. put(r.data, 3);
r = r.next;
}
for (Map.Entry<Integer,
Integer> x : hash.entrySet())
{
// if current frequency is 3 its means
// element is common in all the given
// linked list
if (x.getValue() == 3)
System.out.println(x.getKey() + " ");
}
}
// Driver code
public static void main(String[] args)
{
// first list
Node head1 = null;
head1 = push(head1, 20);
head1 = push(head1, 5);
head1 = push(head1, 15);
head1 = push(head1, 10);
// second list
Node head2 = null;
head2 = push(head2, 10);
head2 = push(head2, 20);
head2 = push(head2, 15);
head2 = push(head2, 8);
// third list
Node head3 = null;
head3 = push(head3, 10);
head3 = push(head3, 2);
head3 = push(head3, 15);
head3 = push(head3, 20);
Common(head1, head2, head3);
}
}
// This code is contributed by 29AjayKumar
# Python3 program to find common element
# in three unsorted linked list
max = 1000000
# Link list node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# A utility function to insert a node at the
# beginning of a linked list
def push( head_ref, new_data):
new_node = Node(new_data)
new_node.next = head_ref
head_ref = new_node
return head_ref
# Print the common element in between
# given three linked list
def Common(head1, head2, head3):
# Creating empty hash table;
hash = dict()
p = head1
while (p != None):
# Set frequency by 1
hash[p.data] = 1
p = p.next
q = head2
while (q != None):
# If the element is already exist in the
# linked list set its frequency 2
if (q.data in hash):
hash[q.data] = 2
q = q.next
r = head3
while (r != None):
if (r.data in hash) and hash[r.data] == 2:
# If the element frequency is 2 it means
# its present in both the first and second
# linked list set its frequency 3
hash[r.data] = 3
r = r.next
for x in hash.keys():
# If current frequency is 3 its means
# element is common in all the given
# linked list
if (hash[x] == 3):
print(x, end = ' ')
# Driver code
if __name__=='__main__':
# First list
head1 = None
head1 = push(head1, 20)
head1 = push(head1, 5)
head1 = push(head1, 15)
head1 = push(head1, 10)
# Second list
head2 = None
head2 = push(head2, 10)
head2 = push(head2, 20)
head2 = push(head2, 15)
head2 = push(head2, 8)
# Third list
head3 = None
head3 = push(head3, 10)
head3 = push(head3, 2)
head3 = push(head3, 15)
head3 = push(head3, 20)
Common(head1, head2, head3)
# This code is contributed by rutvik_56
// C# program to find common element
// in three unsorted linked list
using System;
using System.Collections.Generic;
class GFG
{
static int max = 1000000;
/* Link list node */
public class Node
{
public int data;
public Node next;
};
/* A utility function to insert a node
at the beginning of a linked list */
static Node push(Node head_ref,
int new_data)
{
Node new_node = new Node();
new_node.data = new_data;
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
/* print the common element in between
given three linked list*/
static void Common(Node head1,
Node head2,
Node head3)
{
// Creating empty hash table;
Dictionary<int,
int> hash = new Dictionary<int,
int>();
Node p = head1;
while (p != null)
{
// set frequency by 1
hash.Add(p.data, 1);
p = p.next;
}
Node q = head2;
while (q != null)
{
// if the element is already exist in the
// linked list set its frequency 2
if (hash.ContainsKey(q.data))
hash[q.data] = 2;
q = q.next;
}
Node r = head3;
while (r != null)
{
if (hash.ContainsKey(r.data)&&
hash[r.data] == 2)
// if the element frequency is 2 it means
// its present in both the first and
// second linked list set its frequency 3
hash[r.data] = 3;
r = r.next;
}
foreach(KeyValuePair<int, int> x in hash)
{
// if current frequency is 3 its means
// element is common in all the given
// linked list
if (x.Value == 3)
Console.Write(x.Key + " ");
}
}
// Driver code
public static void Main(String[] args)
{
// first list
Node head1 = null;
head1 = push(head1, 20);
head1 = push(head1, 5);
head1 = push(head1, 15);
head1 = push(head1, 10);
// second list
Node head2 = null;
head2 = push(head2, 10);
head2 = push(head2, 20);
head2 = push(head2, 15);
head2 = push(head2, 8);
// third list
Node head3 = null;
head3 = push(head3, 10);
head3 = push(head3, 2);
head3 = push(head3, 15);
head3 = push(head3, 20);
Common(head1, head2, head3);
}
}
// This code is contributed by Princi Singh
<script>
// JavaScript program to find common element
// in three unsorted linked list
var max = 1000000;
/* Link list node */
class Node
{
constructor()
{
this.data = 0;
this.next = null;
}
};
/* A utility function to insert a node
at the beginning of a linked list */
function push(head_ref, new_data)
{
var new_node = new Node();
new_node.data = new_data;
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
/* print the common element in between
given three linked list*/
function Common(head1, head2, head3)
{
// Creating empty hash table;
var hash = new Map();
var p = head1;
while (p != null)
{
// set frequency by 1
hash.set(p.data, 1);
p = p.next;
}
var q = head2;
while (q != null)
{
// if the element is already exist in the
// linked list set its frequency 2
if (hash.has(q.data))
hash.set(q.data, 2);
q = q.next;
}
var r = head3;
while (r != null)
{
if (hash.has(r.data)&&
hash.get(r.data) == 2)
// if the element frequency is 2 it means
// its present in both the first and
// second linked list set its frequency 3
hash.set(r.data, 3);
r = r.next;
}
hash.forEach((value, key) => {
// if current frequency is 3 its means
// element is common in all the given
// linked list
if (value == 3)
document.write(key + " ");
});
}
// Driver code
// first list
var head1 = null;
head1 = push(head1, 20);
head1 = push(head1, 5);
head1 = push(head1, 15);
head1 = push(head1, 10);
// second list
var head2 = null;
head2 = push(head2, 10);
head2 = push(head2, 20);
head2 = push(head2, 15);
head2 = push(head2, 8);
// third list
var head3 = null;
head3 = push(head3, 10);
head3 = push(head3, 2);
head3 = push(head3, 15);
head3 = push(head3, 20);
Common(head1, head2, head3);
</script>
Time Complexity : O(m + n + p)
Auxiliary Space : O(m + n + p)
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