Check whether the sum of element of a node is equal to the given key value

Last Updated : 20 Sep, 2022

Given an integer k and a linked list, every node of which consists of a pair of integer variable first and second to hold the data, and a pointer pointing to the next node in the list. The task is to find whether the sum of data variables of any of the nodes is equal to k. If yes then print Yes else print No.

Examples:

Input: (1, 2) -> (2, 3) -> (3, 4) -> (4, 5) -> NULL, k = 5
Output: Yes
For the second node, the sum of data variables is 2 + 3 = 5.

Input: (1, 2) -> (2, 3) -> (3, 4) -> (4, 5) -> NULL, k = 15
Output: No

Approach: Traverse the whole linked list until the sum of elements of a node is equal to the key value. When sum of element of a node is equal to key value then print Yes. If there is no such node whose sum of element is equal to the key value then print No.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Represents node of the linked list
struct Node {
    int first;
    int second;
    Node* next;
};
 
// Insertion in linked list
void insert(Node** head, int f, int s)
{
    Node* ptr = *head;
    Node* temp = new Node();
    temp->first = f;
    temp->second = s;
    temp->next = NULL;
 
    if (*head == NULL)
        *head = temp;
    else {
        while (ptr->next != NULL)
            ptr = ptr->next;
 
        ptr->next = temp;
    }
}
 
// Function that returns true
// if the sum of element in a node = k
bool checkK(Node* head, int k)
{
 
    // Check every node of the linked list
    while (head != NULL) {
 
        // If sum of the data of the current node = k
        if ((head->first + head->second) == k)
            return true;
 
        // Get to next node in the list
        head = head->next;
    }
 
    // No matching node found
    return false;
}
// Driver code
int main()
{
    Node* head = NULL;
    insert(&head, 1, 2);
    insert(&head, 2, 3);
    insert(&head, 3, 4);
    insert(&head, 4, 5);
    int k = 5;
 
    if (checkK(head, k))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
class GfG {
 
    // Represents node of the linked list
    static class Node {
        int first;
        int second;
        Node next;
    }
    static Node head = null;
 
    // Insertion in linked list
    static void insert(int f, int s)
    {
        Node ptr = head;
        Node temp = new Node();
        temp.first = f;
        temp.second = s;
        temp.next = null;
 
        if (head == null)
            head = temp;
        else {
            while (ptr.next != null)
                ptr = ptr.next;
 
            ptr.next = temp;
        }
    }
 
    // Function that returns true
    // if the sum of element in a node = k
    static boolean checkK(Node head, int k)
    {
 
        // Check every node of the linked list
        while (head != null) {
 
            // If sum of the data of the current node = k
            if ((head.first + head.second) == k)
                return true;
 
            // Get to next node in the list
            head = head.next;
        }
 
        // No matching node found
        return false;
    }
    // Driver code
    public static void main(String[] args)
    {
        // Node* head = NULL;
        insert(1, 2);
        insert(2, 3);
        insert(3, 4);
        insert(4, 5);
        int k = 5;
 
        if (checkK(head, k) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by Prerna Saini

Python3

# Python3 implementation of the approach
  
# Represents node of the linked list
class Node:
     
    def __init__(self):
         
        self.first = 0
        self.second = 0
        self.next = None
 
# Insertion in linked list
def insert(head, f, s):
 
    ptr = head
    temp = Node()
    temp.first = f
    temp.second = s
    temp.next = None
  
    if (head == None):
        head = temp
    else:
        while (ptr.next != None):
            ptr = ptr.next
  
        ptr.next = temp
         
    return head
     
# Function that returns true
# if the sum of element in a node = k
def checkK(head, k):
 
    # Check every node of the linked list
    while (head != None):
  
        # If sum of the data of the current node = k
        if ((head.first + head.second) == k):
            return True
  
        # Get to next node in the list
        head = head.next
  
    # No matching node found
    return False
 
# Driver code
if __name__=='__main__':
     
    head = None
    head = insert(head, 1, 2)
    head = insert(head, 2, 3)
    head = insert(head, 3, 4)
    head = insert(head, 4, 5)
    k = 5
  
    if (checkK(head, k)):
        print("Yes")
    else:
        print("No")
  
# This code is contributed by rutvik_56

C#

using System;
 
// c# implementation of the approach
public class GfG {
 
    // Represents node of the linked list
    public class Node {
        public int first;
        public int second;
        public Node next;
    }
    public static Node head = null;
 
    // Insertion in linked list
    public static void insert(int f, int s)
    {
        Node ptr = head;
        Node temp = new Node();
        temp.first = f;
        temp.second = s;
        temp.next = null;
 
        if (head == null) {
            head = temp;
        }
        else {
            while (ptr.next != null) {
                ptr = ptr.next;
            }
 
            ptr.next = temp;
        }
    }
 
    // Function that returns true
    // if the sum of element in a node = k
    public static bool checkK(Node head, int k)
    {
 
        // Check every node of the linked list
        while (head != null) {
 
            // If sum of the data of the current node = k
            if ((head.first + head.second) == k) {
                return true;
            }
 
            // Get to next node in the list
            head = head.next;
        }
 
        // No matching node found
        return false;
    }
    // Driver code
    public static void Main(string[] args)
    {
        // Node* head = NULL;
        insert(1, 2);
        insert(2, 3);
        insert(3, 4);
        insert(4, 5);
        int k = 5;
 
        if (checkK(head, k) == true) {
            Console.WriteLine("Yes");
        }
        else {
            Console.WriteLine("No");
        }
    }
}

Javascript

<script>
 
// Javascript implementation of the approach
 
// Structure of a node of linked list
class Node {
        constructor() {
                this.first = 0;
                this.second = 0;
                this.next = null;
             }
        }
         
// Insertion in linked list
function insert( f,  s)
{
    var ptr = head;
    var temp = new Node();
    temp.first = f;
    temp.second = s;
    temp.next = null;
 
    if (head == null)
        head = temp;
    else {
        while (ptr.next != null)
            ptr = ptr.next;
 
        ptr.next = temp;
    }
}
 
// Function that returns true
// if the sum of element in a node = k
function checkK( head,  k)
{
 
    // Check every node of the linked list
    while (head != null) {
 
        // If sum of the data of the current node = k
        if ((head.first + head.second) == k)
            return true;
 
        // Get to next node in the list
        head = head.next;
    }
 
    // No matching node found
    return false;
}
 
// Driver Code
 
var head = null;
insert(1, 2);
insert(2, 3);
insert(3, 4);
insert(4, 5);
let k = 5;
 
if (checkK(head, k) == true)
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by jana_sayantn.
</script>

Output:

Yes

Time Complexity: O(N), where N is the number of nodes in the linked list.
Auxiliary Space: O(1), as constant space is being used.

Find if sum of elements of given Array is less than or equal to K

Premdeep Toppo

Improve

Article Tags :

Practice Tags :

Check whether the sum of element of a node is equal to the given key value

C++

Java

Python3

C#

Javascript

Similar Reads

Thank You!

What kind of Experience do you want to share?