Print duplicates from a list of integers
Last Updated :
23 Jul, 2025
Given a Linked list of integers containing duplicate elements, the task is to print all the duplicates in the given Linked List.
Examples:
Input: list = [10, 20, 30, 20, 20, 30, 40, 50, -20, 60, 60, -20, -20]
Output: [20, 30, -20, 60]
Input: list = [5, 7, 5, 1, 7]
Output: [5, 7]
Naive Approach: To basic way to solve the problem is as follows:
- We traverse the whole linked list.
- For each node, we check in the remaining list whether the duplicate node exists or not.
- If it does then store it in an array.
- In the end, print all elements of the array
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Representation of node
struct Node {
int data;
Node* next;
};
// Function to insert a node at the beginning
void insert(Node** head, int item)
{
Node* temp = new Node();
temp->data = item;
temp->next = *head;
*head = temp;
}
// Function to print duplicate nodes in
// the linked list
vector<int> printDuplicatesInList(Node* head)
{
vector<int> dupes;
while (head->next != NULL) {
// Starting from the next node
Node* ptr = head->next;
while (ptr != NULL) {
// If some duplicate node is found
if (head->data == ptr->data) {
dupes.push_back(head->data);
break;
}
ptr = ptr->next;
}
head = head->next;
}
// Return the count of duplicate nodes
return dupes;
}
// Driver code
int main()
{
Node* head = NULL;
insert(&head, 5);
insert(&head, 7);
insert(&head, 5);
insert(&head, 1);
insert(&head, 7);
// Function Call
vector<int> dupesInList = printDuplicatesInList(head);
for (int i = 0; i < dupesInList.size(); i++)
cout << dupesInList[i] << ", ";
cout << endl;
return 0;
}
// Java implementation of the approach
import java.util.*;
class Node {
int data;
Node next;
}
public class PrintDuplicateNodes {
// Function to insert a node at the beginning
static void insert(Node[] head, int item) {
Node temp = new Node();
temp.data = item;
temp.next = head[0];
head[0] = temp;
}
// Function to print duplicate nodes in the linked list
static List<Integer> printDuplicatesInList(Node head) {
List<Integer> dupes = new ArrayList<>();
while (head.next != null) {
// Starting from the next node
Node ptr = head.next;
while (ptr != null) {
// If some duplicate node is found
if (head.data == ptr.data) {
dupes.add(head.data);
break;
}
ptr = ptr.next;
}
head = head.next;
}
// Return the list of duplicate nodes
return dupes;
}
// Driver code
public static void main(String[] args) {
Node[] head = new Node[1];
head[0] = null;
insert(head, 5);
insert(head, 7);
insert(head, 5);
insert(head, 1);
insert(head, 7);
// Function Call
List<Integer> dupesInList = printDuplicatesInList(head[0]);
for (int i = 0; i < dupesInList.size(); i++) {
System.out.print(dupesInList.get(i) + ", ");
}
System.out.println();
}
}
//This code is contributed by chinmaya121221
# Representation of node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Function to insert a node at the beginning
def insert(head, item):
new_node = Node(item)
new_node.next = head
head = new_node
return head
# Function to print duplicate nodes in the linked list
def print_duplicates_in_list(head):
dupes = []
while head is not None:
ptr = head.next
while ptr is not None:
if head.data == ptr.data:
dupes.append(head.data)
break
ptr = ptr.next
head = head.next
return dupes
# Driver code
if __name__ == "__main__":
head = None
head = insert(head, 5)
head = insert(head, 7)
head = insert(head, 5)
head = insert(head, 1)
head = insert(head, 7)
# Function call
dupes_in_list = print_duplicates_in_list(head)
for item in dupes_in_list:
print(item, end=", ")
print()
#Contributed by Aditi Tyagi
// C# implementation of the approach
using System;
using System.Collections.Generic;
// Representation of node
class Node
{
public int data;
public Node next;
}
class Program
{
// Function to insert a node at the beginning
static void Insert(ref Node head, int item)
{
Node temp = new Node
{
data = item,
next = head
};
head = temp;
}
// Function to print duplicate nodes in the linked list
static List<int> PrintDuplicatesInList(Node head)
{
List<int> dupes = new List<int>();
while (head.next != null)
{
// Starting from the next node
Node ptr = head.next;
while (ptr != null)
{
// If some duplicate node is found
if (head.data == ptr.data)
{
dupes.Add(head.data);
break;
}
ptr = ptr.next;
}
head = head.next;
}
// Return the count of duplicate nodes
return dupes;
}
// Driver code
static void Main(string[] args)
{
Node head = null;
Insert(ref head, 5);
Insert(ref head, 7);
Insert(ref head, 5);
Insert(ref head, 1);
Insert(ref head, 7);
// Function Call
List<int> dupesInList = PrintDuplicatesInList(head);
foreach (int item in dupesInList)
{
Console.Write(item + ", ");
}
Console.WriteLine();
}
}
// this code is contributed by uttamdp_10
// Define a class for the linked list node
class Node {
constructor(data) {
this.data = data;
this.next = null;
}
}
// Function to insert a node at the beginning of the linked list
function insert(head, item) {
const temp = new Node(item);
temp.next = head[0];
head[0] = temp;
}
// Function to print duplicate nodes in the linked list
function printDuplicatesInList(head) {
const dupes = [];
while (head.next !== null) {
// Starting from the next node
let ptr = head.next;
while (ptr !== null) {
// If some duplicate node is found
if (head.data === ptr.data) {
dupes.push(head.data);
break;
}
ptr = ptr.next;
}
head = head.next;
}
// Return the list of duplicate nodes
return dupes;
}
// Driver code
const head = [null];
insert(head, 5);
insert(head, 7);
insert(head, 5);
insert(head, 1);
insert(head, 7);
// Function Call
const dupesInList = printDuplicatesInList(head[0]);
for (let i = 0; i < dupesInList.length; i++) {
process.stdout.write(dupesInList[i] + ", ");
}
console.log();
Time Complexity: O(N2), where N is the number of elements in the list.
Auxiliary Space: O(N), an additional vector is used to store the duplicate values.
Printing duplicates from a list of integers using Hashing:
- We traverse the whole linked list.
- Create a hashtable
- For each node
- If it is not present in the hashtable, insert it with frequency 1
- If it is already present, update its frequency
- At the end, print all elements of the hashtable with frequency more than 1
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Representation of node
struct Node {
int data;
Node* next;
};
// Function to insert a node at the beginning
void insert(Node** head, int item)
{
Node* temp = new Node();
temp->data = item;
temp->next = *head;
*head = temp;
}
// Function to print duplicate nodes in
// the linked list
map<int, int> printDuplicatesInList(Node* head)
{
// Create a hash table insert head
map<int, int> M;
// Traverse through remaining nodes
int count = 0;
for (Node* curr = head; curr != NULL;
curr = curr->next) {
// If the current element
// is not found then insert
// current element with
// frequency 1
if (M.find(curr->data) == M.end())
M[curr->data] = 1;
// Else update the frequency
else
M[curr->data]++;
}
// Return the map with frequency
return M;
}
// Driver code
int main()
{
Node* head = NULL;
insert(&head, 5);
insert(&head, 7);
insert(&head, 5);
insert(&head, 1);
insert(&head, 7);
// Function Call
map<int, int> dupesInList = printDuplicatesInList(head);
// Traverse the map to print the
// frequency
for (auto& it : dupesInList) {
if (it.second > 1)
cout << it.first << ", ";
}
cout << endl;
return 0;
}
// Java implementation of the approach
import java.util.*;
class Node {
int data;
Node next;
}
public class PrintDuplicateNodes {
// Function to insert a node at the beginning
static void insert(Node[] head, int item) {
Node temp = new Node();
temp.data = item;
temp.next = head[0];
head[0] = temp;
}
// Function to print duplicate nodes in the linked list
static Map<Integer, Integer> printDuplicatesInList(Node head) {
Map<Integer, Integer> M = new HashMap<>();
// Traverse through remaining nodes
Node curr = head;
while (curr != null) {
// If the current element is not found then insert
// the current element with frequency 1
if (!M.containsKey(curr.data)) {
M.put(curr.data, 1);
}
// Else update the frequency
else {
M.put(curr.data, M.get(curr.data) + 1);
}
curr = curr.next;
}
// Return the map with frequency
return M;
}
// Driver code
public static void main(String[] args) {
Node[] head = new Node[1];
head[0] = null;
insert(head, 5);
insert(head, 7);
insert(head, 5);
insert(head, 1);
insert(head, 7);
// Function Call
Map<Integer, Integer> dupesInList = printDuplicatesInList(head[0]);
// Traverse the map to print the frequency
for (Map.Entry<Integer, Integer> entry : dupesInList.entrySet()) {
if (entry.getValue() > 1) {
System.out.print(entry.getKey() + ", ");
}
}
System.out.println();
}
}
//This code is contributed by chinmaya121221
class Node:
def __init__(self, data):
self.data = data
self.next = None
def insert(head, item):
temp = Node(item)
temp.next = head[0]
head[0] = temp
def print_duplicates_in_list(head):
M = {}
# Traverse through the linked list
curr = head
while curr is not None:
# If the current element is not found, insert it with frequency 1
if curr.data not in M:
M[curr.data] = 1
# Else update the frequency
else:
M[curr.data] += 1
curr = curr.next
# Return a dictionary with frequencies
return M
# Driver code
head = [None]
insert(head, 5)
insert(head, 7)
insert(head, 5)
insert(head, 1)
insert(head, 7)
# Function Call
dupes_in_list = print_duplicates_in_list(head[0])
# Traverse the dictionary to print elements with frequency > 1
for key, value in dupes_in_list.items():
if value > 1:
print(key, end=", ")
print()
// C# implementation of the approach
using System;
using System.Collections.Generic;
// Representation of node
class Node
{
public int data;
public Node next;
}
class Program
{
// Function to insert a node at the beginning
static void Insert(ref Node head, int item)
{
Node temp = new Node
{
data = item,
next = head
};
head = temp;
}
// Function to print duplicate nodes in the linked list
static Dictionary<int, int> PrintDuplicatesInList(Node head)
{
// Create a dictionary to insert head
Dictionary<int, int> M = new Dictionary<int, int>();
// Traverse through remaining nodes
for (Node curr = head; curr != null; curr = curr.next)
{
// If the current element is not found then insert
// current element with frequency 1
if (!M.ContainsKey(curr.data))
M[curr.data] = 1;
// Else update the frequency
else
M[curr.data]++;
}
// Return the dictionary with frequency
return M;
}
// Driver code
static void Main(string[] args)
{
Node head = null;
Insert(ref head, 5);
Insert(ref head, 7);
Insert(ref head, 5);
Insert(ref head, 1);
Insert(ref head, 7);
// Function Call
Dictionary<int, int> dupesInList = PrintDuplicatesInList(head);
// Traverse the dictionary to print the frequency
foreach (var kvp in dupesInList)
{
if (kvp.Value > 1)
Console.Write(kvp.Key + ", ");
}
Console.WriteLine();
}
}
// this code is contributed by uttamdp_10
// Define a class for the linked list node
class Node {
constructor(data) {
this.data = data;
this.next = null;
}
}
// Function to insert a node at the beginning of the linked list
function insert(head, item) {
// Create a new node with the given data
const temp = new Node(item);
// Set the next pointer of the new node to the current head
temp.next = head[0];
// Update the head to the new node
head[0] = temp;
}
// Function to find duplicate nodes in the linked list and their frequencies
function printDuplicatesInList(head) {
// Create a Map to store the frequency of each element
const frequencyMap = new Map();
let curr = head;
// Traverse through the linked list
while (curr !== null) {
// If the current element is not found in the Map, add it with a frequency of 1
if (!frequencyMap.has(curr.data)) {
frequencyMap.set(curr.data, 1);
}
// If the element already exists in the Map, update its frequency
else {
frequencyMap.set(curr.data, frequencyMap.get(curr.data) + 1);
}
curr = curr.next;
}
// Create an array to store the duplicate keys (elements with frequency > 1)
const duplicateKeys = [];
for (const [key, value] of frequencyMap.entries()) {
if (value > 1) {
duplicateKeys.push(key);
}
}
return duplicateKeys;
}
// Create the head of the linked list
const head = [null];
// Insert elements at the beginning of the linked list
insert(head, 5);
insert(head, 7);
insert(head, 5);
insert(head, 1);
insert(head, 7);
// Find and print duplicate keys in the linked list
const duplicateKeys = printDuplicatesInList(head[0]);
// Print the duplicate keys
for (let i = 0; i < duplicateKeys.length; i++) {
process.stdout.write(duplicateKeys[i] + ", ");
}
console.log();
Time Complexity: O(N*logN), for traversing the list O(N) time will consume, and to find the element in msp, O(Log N) time will be taken, So the time complexity of the above approach is O(N*logN).
Auxiliary Space: O(N), map is used to store the elements.
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