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;Â Â // Class to store Node of the listclass Node {public:Â Â Â Â int data;Â Â Â Â Node* next;Â Â Â Â Â Â Â Â Â Node(int data) {Â Â Â Â Â Â Â Â Â this->data = data;Â Â Â Â Â Â Â Â Â Â this->next = NULL;Â Â Â Â }};Â Â // Function to find the count of the// duplicate integers in the liststatic int checkDuplicate(Node* head){Â Â Â Â if (head == NULL)Â Â Â Â Â Â Â Â return 0;Â
    // Stores the count of duplicate    // integers    int cnt = 0;Â
    // Stores the integers occurred    set<int> s;    s.insert(head->data);    Node *curr = head->next;Â
    // Loop to traverse the given list    while (curr != head) {Â
        // If integer already occurred        if (s.find(curr->data) != s.end())            cnt++;Â
        // Add current integer into        // the hashmap        s.insert(curr->data);        curr = curr->next;    }Â
    // Return answer    return cnt;}Â
// Driver Codeint main(){Â Â Â Â Â Â Â Â Â Node *head = new Node(5);Â Â Â Â head->next = new Node(7);Â Â Â Â Â Â head->next->next = new Node(5);Â Â Â Â head->next->next->next = new Node(1);Â Â Â Â head->next->next->next->next = new Node(4);Â Â Â Â head->next->next->next->next->next = new Node(4);Â Â Â Â head->next->next->next->next->next->next = head;Â
    cout << checkDuplicate(head) << endl;  Â
    return 0;}  // This code is contributed by Dharanendra L V. |
Java
// Java program for the above approachÂ
import java.util.HashSet;Â
public class GFG {Â
    // Class to store Node of the list    static class Node {        public int data;        public Node next;Â
        public Node(int data) {            this.data = data;        }    }Â
    // Stores head pointer of the list    static Node head;Â
    // Function to find the count of the    // duplicate integers in the list    static int checkDuplicate(Node head) {        if (head == null)            return 0;Â
        // Stores the count of duplicate        // integers        int cnt = 0;Â
        // Stores the integers occurred        HashSet<Integer> s = new HashSet<Integer>();        s.add(head.data);        Node curr = head.next;Â
        // Loop to traverse the given list        while (curr != head) {Â
            // If integer already occurred            if (s.contains(curr.data))                cnt++;Â
            // Add current integer into            // the hashset            s.add(curr.data);            curr = curr.next;        }Â
        // Return answer        return cnt;    }Â
    // 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;Â
        System.out.println(checkDuplicate(head));    }}Â
// this code is contributed by bhardwajji |
Python3
# Python program for the above approachÂ
# Class to store Node of the listclass Node: Â Â Â Â def __init__(self, data):Â Â Â Â Â Â Â Â self.data = data; Â Â Â Â Â Â Â Â self.next = None;Â Â Â Â Â # Stores head pointer of the listhead = None;Â
# Function to find the count of the# duplicate integers in the listdef checkDuplicate(head):Â Â Â Â if (head == None):Â Â Â Â Â Â Â Â return 0;Â
    # Stores the count of duplicate    # integers    cnt = 0;Â
    # Stores the integers occurred    s = set();    s.add(head.data);    curr = head.next;Â
    # Loop to traverse the given list    while (curr != head):Â
        # If integer already occurredA        if ((curr.data) in s):            cnt+=1;Â
        # Add current integer into        # the hashmap        s.add(curr.data);        curr = curr.next;         # Return answer    return cnt;Â
# Driver Codeif __name__ == '__main__':Â Â Â Â head =Â Node(5);Â Â Â Â head.next =Â Node(7);Â Â Â Â head.next.next =Â Node(5);Â Â Â Â head.next.next.next =Â Node(1);Â Â Â Â head.next.next.next.next =Â Node(4);Â Â Â Â head.next.next.next.next.next =Â Node(4);Â Â Â Â head.next.next.next.next.next.next = head;Â
    print(checkDuplicate(head));Â
# This code is contributed by umadevi9616 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;Â
class GFG {Â
    // Class to store Node of the list     class Node {        public int data;        public Node next;Â
        public Node(int data)        {            this.data = data;        }    };Â
    // Stores head pointer of the list    static Node head;Â
    // Function to find the count of the    // duplicate integers in the list    static int checkDuplicate(Node head)    {        if (head == null)            return 0;Â
        // Stores the count of duplicate        // integers        int cnt = 0;Â
        // Stores the integers occurred        HashSet<int> s = new HashSet<int>();        s.Add(head.data);        Node curr = head.next;Â
        // Loop to traverse the given list        while (curr != head) {Â
            // If integer already occurred            if (s.Contains(curr.data))                cnt++;Â
            // Add current integer into            // the hashmap            s.Add(curr.data);            curr = curr.next;        }Â
        // Return answer        return cnt;    }Â
    // 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;Â
        Console.Write(checkDuplicate(head));    }}Â
// This code is contributed by ipg2016107. |
Javascript
<script>// javascript program for the above approachÂ
    // Class to store Node of the listclass Node {    constructor(val) {        this.data = val;        this.next = null;    }}Â
    // Stores head pointer of the list    var head;Â
    // Function to find the count of the    // duplicate integers in the list    function checkDuplicate(head) {        if (head == null)            return 0;Â
        // Stores the count of duplicate        // integers        var cnt = 0;Â
        // Stores the integers occurred        var s = new Set();        s.add(head.data);var curr = head.next;Â
        // Loop to traverse the given list        while (curr != head) {Â
            // If integer already occurred            if (s.has(curr.data))                cnt++;Â
            // Add current integer into            // the hashmap            s.add(curr.data);            curr = curr.next;        }Â
        // Return answer        return cnt;    }Â
    // 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;Â
        document.write(checkDuplicate(head));Â
// This code is contributed by gauravrajput1 </script> |
Output:Â
2
Â
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
