Find the first duplicate element in the linked list

Given a linked list. Find the first element from the left which appears more than once. If all the elements are unique then print -1.

Examples:

Input : 1 2 3 4 3 2 1
Output : 1
In this linked list the element 1 occurs two times
and it is the first element to satisfy the condition.
Hence the answer is 1.

Input : 1 2, 3, 4, 5
Output : -1
All the elements are unique. Hence the answer is -1.

Approach:



  • Count the frequency of all the elements of the linked list using a map.
  • Now, traverse the linked list again to find the first element from the left whose frequency is greater than 1.
  • If no such element exists then print -1.

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// A linked list node
class Node {
public:
    int data;
    Node* next;
};
  
// Given a reference (pointer to pointer)
// to the head of a list and an int,
// appends a new node at the end
void append(Node** head_ref, int new_data)
{
    // allocate node
    Node* new_node = new Node();
  
    Node* last = *head_ref;
  
    // put in the data
    new_node->data = new_data;
  
    // This new node is going to be
    // the last node, so make next of
    // it as NULL
    new_node->next = NULL;
  
    // If the Linked List is empty,
    // then make the new node as head
    if (*head_ref == NULL) {
        *head_ref = new_node;
        return;
    }
  
    // Else traverse till the last node
    while (last->next != NULL)
        last = last->next;
  
    // Change the next of last node
    last->next = new_node;
    return;
}
  
int getFirstDuplicate(Node* node)
{
  
    // Unordered map to store the
    // frequency of elements
    unordered_map<int, int> mp;
    Node* head = node;
  
    // update frequency of all the elements
    while (node != NULL) {
        mp[node->data]++;
        node = node->next;
    }
  
    node = head;
  
    // the first node from the left which
    // appears more than once is the answer
  
    while (node != NULL) {
        if (mp[node->data] > 1)
            return node->data;
        node = node->next;
    }
  
    // all the nodes are unique
    return -1;
}
  
// driver code
int main()
{
    // Start with the empty list
    Node* head = NULL;
  
    // Insert element
    append(&head, 6);
    append(&head, 2);
    append(&head, 1);
    append(&head, 6);
    append(&head, 2);
    append(&head, 1);
  
    cout << getFirstDuplicate(head);
  
    return 0;
}

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Python3

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# Python3 implementation of above approach
  
# Link list node 
class Node :
    def __init__(self):
        self.data = 0
        self.next = None
  
# Given a reference (pointer to pointer)
# to the head of a list and an int,
# appends a node at the end
def append(head_ref, new_data):
  
    # allocate node
    new_node = Node()
  
    last = head_ref
  
    # put in the data
    new_node.data = new_data
  
    # This node is going to be
    # the last node, so make next of
    # it as None
    new_node.next = None
  
    # If the Linked List is empty,
    # then make the node as head
    if (head_ref == None) :
        head_ref = new_node
        return head_ref
      
    # Else traverse till the last node
    while (last.next != None):
        last = last.next
  
    # Change the next of last node
    last.next = new_node
    return head_ref
  
def getFirstDuplicate(node):
  
    # Unordered map to store the
    # frequency of elements
    mp = dict()
    head = node
  
    # update frequency of all the elements
    while (node != None) :
        mp[node.data] = mp.get(node.data, 0) + 1
        node = node.next
      
    node = head
  
    # the first node from the left which
    # appears more than once is the answer
    while (node != None) :
        if (mp[node.data] > 1):
            return node.data
        node = node.next
      
    # all the nodes are unique
    return -1
  
# Driver code
  
# Start with the empty list
head = None
  
# Insert element
head = append(head, 6)
head = append(head, 2)
head = append(head, 1)
head = append(head, 6)
head = append(head, 2)
head = append(head, 1)
  
print(getFirstDuplicate(head))
  
# This code is contributed by Arnab Kundu

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Output:

6

Time Complexity : O(N)

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