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 listclass 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 liststatic 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 Codeint 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
// Java program for the above approachimport 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 |
Python3
# Python program for the above approach# Class to store Node of the listclass Node: def __init__(self, data): self.data = data; self.next = None; # Stores head pointer of the listhead = None;# Function to find the count of the# duplicate integers in the listdef 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 Codeif __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#
// C# program for the above approachusing 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. |
Javascript
<script>// javascript program for the above approach // Class to store Node of the listclass 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> |
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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