Check if elements of Linked List are present in pair
Last Updated :
06 Dec, 2024
Given a singly linked list of integers. The task is to check if each element in the linked list is present in a pair i.e. all elements occur even no. of times.
Examples:
Input: 1 -> 2 -> 3 -> 3 -> 1 -> 2
Output: Yes
Input: 10 -> 20 -> 30 -> 20
Output: No
Method : Using Map
We will traverse the complete linked list from left to right and store all elements in map with their frequency.
After completely traversing the linked list, we will find an element in the map which frequency is not even if any element is found we print "No" otherwise print "Yes".
Below is the implementation of above approach:
C++
// C++ program to check if elements of
// linked lists are present in pair
#include <bits/stdc++.h>
using namespace std;
// A linked list node
struct Node {
int data;
struct Node* next;
Node(int value){
this->data = value;
this->next = NULL;
}
};
// Function to check if elements of
// linked list are present in pair
bool isPair(Node* head)
{
unordered_map<int, int> mp;
while(head != NULL){
mp[head->data]++;
head = head->next;
}
// traversing the map
for(auto i: mp)
if(i.second % 2 == 1) return false;
return true;
}
// Driver program to test above function
int main()
{
/* First constructed linked list is:
10 -> 34 -> 1 -> 10 -> 34 -> 1 */
struct Node* head = new Node(10);
head->next = new Node(34);
head->next->next = new Node(1);
head->next->next->next = new Node(10);
head->next->next->next->next = new Node(34);
head->next->next->next->next->next = new Node(1);
//calling function to check pair elements
if (isPair(head))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
// This code is contributed by Kirti Agarwal(kirtiagarwal23121999)
// Java program to check if elements of
// linked lists are present in pair
import java.util.*;
// A linked list node
class Node {
int data;
Node next;
Node(int value){
this.data = value;
this.next = null;
}
}
class Main
{
// Function to check if elements of
// linked list are present in pair
static boolean isPair(Node head)
{
Map<Integer, Integer> mp = new HashMap<>();
while(head != null){
mp.put(head.data, mp.getOrDefault(head.data, 0) + 1);
head = head.next;
}
// traversing the map
for(Map.Entry<Integer, Integer> entry : mp.entrySet())
if(entry.getValue() % 2 == 1) return false;
return true;
}
// Driver program to test above function
public static void main(String args[])
{
/* First constructed linked list is:
10 -> 34 -> 1 -> 10 -> 34 -> 1 */
Node head = new Node(10);
head.next = new Node(34);
head.next.next = new Node(1);
head.next.next.next = new Node(10);
head.next.next.next.next = new Node(34);
head.next.next.next.next.next = new Node(1);
// calling function to check pair elements
if (isPair(head))
System.out.println("Yes");
else
System.out.println("No");
}
}
# python program for the above approach
# a linked list node
class Node:
def __init__(self, value):
self.data = value
self.next = None
# function to check if elements of
# linked list are preset in pair
def isPair(head):
mp = {}
while(head is not None):
if(mp.get(head.data) is not None):
mp[head.data] += 1
else:
mp[head.data] = 1
head = head.next
# traversing the map
for x, y in mp.items():
if(y % 2 == 1):
return False
return True
# driver program to test above function
# First constructed linked list is:
# 10 -> 34 -> 1 -> 10 -> 34 -> 1
head = Node(10)
head.next = Node(34)
head.next.next = Node(1)
head.next.next.next = Node(10)
head.next.next.next.next = Node(34)
head.next.next.next.next.next = Node(1)
# calling functions to check pair elements
if(isPair(head) is True):
print("Yes")
else:
print("No")
// C# program to check if elements of
// linked lists are present in pair
using System;
using System.Collections.Generic;
// A linked list node
public class Node {
public int data;
public Node next;
public Node(int value)
{
this.data = value;
this.next = null;
}
};
public class GFG
{
// Function to check if elements of
// linked list are present in pair
public static bool isPair(Node head)
{
Dictionary<int, int> mp
= new Dictionary<int, int>();
while (head != null) {
if (mp.ContainsKey(head.data)) {
mp[head.data]++;
}
else {
mp.Add(head.data, 1);
}
head = head.next;
}
// traversing the dictionary
foreach(KeyValuePair<int, int> i in mp)
{
if (i.Value % 2 == 1) {
return false;
}
}
return true;
}
// Driver program to test above function
public static void Main()
{
/* First constructed linked list is:
10 -> 34 -> 1 -> 10 -> 34 -> 1 */
Node head = new Node(10);
head.next = new Node(34);
head.next.next = new Node(1);
head.next.next.next = new Node(10);
head.next.next.next.next = new Node(34);
head.next.next.next.next.next = new Node(1);
// calling function to check pair elements
if (isPair(head)) {
Console.WriteLine("Yes");
}
else {
Console.WriteLine("No");
}
}
}
// This code is contributed by phasing17
// JavaScript Program to check if elements of
// linked lists are present in pair
// a linked list node
class Node{
constructor(value){
this.data = value;
this.next = null;
}
}
// function to check if elements of linked
// list are preseint in pair
function isPair(head){
let mp = new Map();
while(head != null){
if(mp.has(head.data)){
mp.set(head.data, mp.get(head.data) + 1);
}else{
mp.set(head.data, 1);
}
head = head.next;
}
// traversing the map
mp.forEach(function(value, key){
if(value % 2 == 1) return false;
})
return true;
}
// driver program to test above function
/* First constructed linked list is:
10 -> 34 -> 1 -> 10 -> 34 -> 1 */
let head = new Node(10);
head.next = new Node(34);
head.next.next = new Node(1);
head.next.next.next = new Node(10);
head.next.next.next.next = new Node(34);
head.next.next.next.next.next = new Node(1);
// calling function to check pair elements
if(isPair(head))
console.log("Yes");
else
console.log("No")
// This code is contributed by Yash Agarwal(yashagarwal2852002)
Time Complexity: O(N) where N is the number of elements in given Linked List.
Auxiliary Space: O(N) due to map data structure.