Count duplicates in a given circular linked list
Last Updated :
27 Feb, 2023
Given a circular linked list, the task is to check whether the given list has duplicates or not.
Example:
Input: list = {5, 7, 5, 1, 4, 4}
Output: 2
Explanation: The given list has 2 indices having integers which has already occurred in the list during traversal.
Input: list = {1, 1, 1, 1, 1}
Output: 4
Approach: The given problem has a very similar solution to finding the count of duplicates in a linked list. The idea is to use hashing. Traverse the given circular linked list using the algorithm discussed here. Create a hashmap to store the integers that occurred in the list and for each integer, check if the integer has already occurred. Maintain the count of already occurred integers in a variable.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Class to store Node of the list
class Node {
public:
int data;
Node* next;
Node(int data) {
this->data = data;
this->next = NULL;
}
};
// Function to find the count of the
// duplicate integers in the list
static int checkDuplicate(Node* head)
{
if (head == NULL)
return 0;
// Stores the count of duplicate
// integers
int cnt = 0;
// Stores the integers occurred
set<int> s;
s.insert(head->data);
Node *curr = head->next;
// Loop to traverse the given list
while (curr != head) {
// If integer already occurred
if (s.find(curr->data) != s.end())
cnt++;
// Add current integer into
// the hashmap
s.insert(curr->data);
curr = curr->next;
}
// Return answer
return cnt;
}
// Driver Code
int main()
{
Node *head = new Node(5);
head->next = new Node(7);
head->next->next = new Node(5);
head->next->next->next = new Node(1);
head->next->next->next->next = new Node(4);
head->next->next->next->next->next = new Node(4);
head->next->next->next->next->next->next = head;
cout << checkDuplicate(head) << endl;
return 0;
}
// This code is contributed by Dharanendra L V.
// Java program for the above approach
import java.util.HashSet;
public class GFG {
// Class to store Node of the list
static class Node {
public int data;
public Node next;
public Node(int data) {
this.data = data;
}
}
// Stores head pointer of the list
static Node head;
// Function to find the count of the
// duplicate integers in the list
static int checkDuplicate(Node head) {
if (head == null)
return 0;
// Stores the count of duplicate
// integers
int cnt = 0;
// Stores the integers occurred
HashSet<Integer> s = new HashSet<Integer>();
s.add(head.data);
Node curr = head.next;
// Loop to traverse the given list
while (curr != head) {
// If integer already occurred
if (s.contains(curr.data))
cnt++;
// Add current integer into
// the hashset
s.add(curr.data);
curr = curr.next;
}
// Return answer
return cnt;
}
// Driver Code
public static void main(String[] args) {
head = new Node(5);
head.next = new Node(7);
head.next.next = new Node(5);
head.next.next.next = new Node(1);
head.next.next.next.next = new Node(4);
head.next.next.next.next.next = new Node(4);
head.next.next.next.next.next.next = head;
System.out.println(checkDuplicate(head));
}
}
// this code is contributed by bhardwajji
# Python program for the above approach
# Class to store Node of the list
class Node:
def __init__(self, data):
self.data = data;
self.next = None;
# Stores head pointer of the list
head = None;
# Function to find the count of the
# duplicate integers in the list
def checkDuplicate(head):
if (head == None):
return 0;
# Stores the count of duplicate
# integers
cnt = 0;
# Stores the integers occurred
s = set();
s.add(head.data);
curr = head.next;
# Loop to traverse the given list
while (curr != head):
# If integer already occurredA
if ((curr.data) in s):
cnt+=1;
# Add current integer into
# the hashmap
s.add(curr.data);
curr = curr.next;
# Return answer
return cnt;
# Driver Code
if __name__ == '__main__':
head = Node(5);
head.next = Node(7);
head.next.next = Node(5);
head.next.next.next = Node(1);
head.next.next.next.next = Node(4);
head.next.next.next.next.next = Node(4);
head.next.next.next.next.next.next = head;
print(checkDuplicate(head));
# This code is contributed by umadevi9616
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Class to store Node of the list
class Node {
public int data;
public Node next;
public Node(int data)
{
this.data = data;
}
};
// Stores head pointer of the list
static Node head;
// Function to find the count of the
// duplicate integers in the list
static int checkDuplicate(Node head)
{
if (head == null)
return 0;
// Stores the count of duplicate
// integers
int cnt = 0;
// Stores the integers occurred
HashSet<int> s = new HashSet<int>();
s.Add(head.data);
Node curr = head.next;
// Loop to traverse the given list
while (curr != head) {
// If integer already occurred
if (s.Contains(curr.data))
cnt++;
// Add current integer into
// the hashmap
s.Add(curr.data);
curr = curr.next;
}
// Return answer
return cnt;
}
// Driver Code
public static void Main()
{
head = new Node(5);
head.next = new Node(7);
head.next.next = new Node(5);
head.next.next.next = new Node(1);
head.next.next.next.next = new Node(4);
head.next.next.next.next.next = new Node(4);
head.next.next.next.next.next.next = head;
Console.Write(checkDuplicate(head));
}
}
// This code is contributed by ipg2016107.
<script>
// javascript program for the above approach
// Class to store Node of the list
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
// Stores head pointer of the list
var head;
// Function to find the count of the
// duplicate integers in the list
function checkDuplicate(head) {
if (head == null)
return 0;
// Stores the count of duplicate
// integers
var cnt = 0;
// Stores the integers occurred
var s = new Set();
s.add(head.data);
var curr = head.next;
// Loop to traverse the given list
while (curr != head) {
// If integer already occurred
if (s.has(curr.data))
cnt++;
// Add current integer into
// the hashmap
s.add(curr.data);
curr = curr.next;
}
// Return answer
return cnt;
}
// Driver Code
head = new Node(5);
head.next = new Node(7);
head.next.next = new Node(5);
head.next.next.next = new Node(1);
head.next.next.next.next = new Node(4);
head.next.next.next.next.next = new Node(4);
head.next.next.next.next.next.next = head;
document.write(checkDuplicate(head));
// This code is contributed by gauravrajput1
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)