Find kth node from Middle towards Head of a Linked List

Given a Linked List and a number K. The task is to print the value of the K-th node from the middle towards the beginning of the List. If no such element exists, then print “-1”.

Note: The position of the middle node is: (n/2)+1, where n is the total number of nodes in the list.

Examples: 

Input :  List is 1->2->3->4->5->6->7
         K= 2 
Output : 2

Input :  list is 7->8->9->10->11->12
         K = 3
Output : 7

Traverse the List from beginning to end and count the total number of nodes. Now, suppose is the total number of nodes in the List. Therefore, the middle node will be at the position (n/2)+1. Now, the task remains to print the node at (n/2 + 1 – k)^th position from the head of the List.

Below is the implementation of the above idea: 

C++

// CPP program to find kth node from middle
// towards Head of the Linked List

#include <bits/stdc++.h>

using namespace std;

// Linked list node
struct Node {
    int data;
    struct Node* next;
};

/* Given a reference (pointer to
   pointer) to the head of a list
   and an int, push a new node on
   the front of the list. */
void push(struct Node** head_ref,
          int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

// Function to count number of nodes
int getCount(struct Node* head)
{
    int count = 0; // Initialize count
    struct Node* current = head; // Initialize current
    while (current != NULL) {
        count++;
        current = current->next;
    }
    return count;
}

// Function to get the kth node from the mid
// towards begin of the linked list
int printKthfrommid(struct Node* head_ref, int k)
{
    // Get the count of total number of
    // nodes in the linked list
    int n = getCount(head_ref);

    int reqNode = ((n / 2 + 1) - k);

    // If no such node exists, return -1
    if (reqNode <= 0) {
        return -1;
    }

    // Find node at position reqNode
    else {
        struct Node* current = head_ref;

        // the index of the
        // node we're currently
        // looking at
        int count = 1;
        while (current != NULL) {
            if (count == reqNode)
                return (current->data);
            count++;
            current = current->next;
        }
    }
}

// Driver code
int main()
{
    // start with empty list
    struct Node* head = NULL;
    int k = 2;

    // create linked list
    // 1->2->3->4->5->6->7
    push(&head, 7);
    push(&head, 6);
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);

    cout << printKthfrommid(head, 2);

    return 0;
}

Java

// Java program to find Kth node from mid
// of a linked list in single traversal

// Linked list node
class Node
{
    int data;
    Node next;
    Node(int d)
    {
        this.data = d;
        this.next = null;
    }
}

class LinkedList
{
    Node start;
    LinkedList()
    {
        start = null;
    }

    // Function to push node at head
    public void push(int data)
    {
        if(this.start == null)
        {
            Node temp = new Node(data);
            this.start = temp;
        }
        else
        {
            Node temp = new Node(data);
            temp.next = this.start;
            this.start = temp;
        }
    }

    //method to get the count of node
    public int getCount(Node start)
    {
        Node temp = start;
        int cnt = 0;
        while(temp != null)
        {
            temp = temp.next;
            cnt++;
        }
        return cnt;
    }

    // Function to get the kth node from the mid
    // towards begin of the linked list
    public int printKthfromid(Node start, int k)
    {
        // Get the count of total number of
        // nodes in the linked list
        int n = getCount(start);
        int reqNode = ((n + 1) / 2) - k;

        // If no such node exists, return -1
        if(reqNode <= 0)
            return -1;
        else
        {
            Node current = start;
            int count = 1,ans = 0;
            while (current != null)
            {
                if (count == reqNode)
                {
                    ans = current.data;
                    break;
                }
                count++;
                current = current.next;
            }
            return ans;
        }
    }

    // Driver code
    public static void main(String[] args)
    {
    // create linked list
        // 1->2->3->4->5->6->7
    LinkedList ll = new LinkedList();
        ll.push(7);
        ll.push(6);
        ll.push(5);
        ll.push(4);
        ll.push(3);
        ll.push(2);
        ll.push(1);
        System.out.println(ll.printKthfromid(ll.start, 2));
    }
}

// This Code is contributed by Adarsh_Verma

Python3

# Python3 program to find kth node from
# middle towards Head of the Linked List

# Node class
class Node:

    # Function to initialise the node object
    def __init__(self, data):
        self.data = data
        self.next = None

# Function to insert a node at the
# beginning of the linked list
def push(head, data):
    if not head:

        # Return new node
        return Node(data)

    # allocate node
    new_node = Node(data)

    # link the old list to the new node
    new_node.next = head

    # move the head to point
    # to the new node
    head = new_node
    return head

# Function to count number of nodes
def getCount(head):

    count = 0 # Initialize count
    current = head # Initialize current
    while (current ):
        count = count + 1
        current = current.next

    return count

# Function to get the kth node from the mid
# towards begin of the linked list
def printKthfrommid(head_ref, k):

    # Get the count of total number of
    # nodes in the linked list
    n = getCount(head_ref)

    reqNode = int((n / 2 + 1) - k)

    # If no such node exists, return -1
    if (reqNode <= 0) :
        return -1

    # Find node at position reqNode
    else :
        current = head_ref

        # the index of the
        # node we're currently
        # looking at
        count = 1
        while (current) :
            if (count == reqNode):
                return (current.data)
            count = count + 1
            current = current.next

# Driver Code
if __name__=='__main__':

    # start with empty list
    head = None
    k = 2

    # create linked list
    # 1.2.3.4.5.6.7
    head = push(head, 7)
    head = push(head, 6)
    head = push(head, 5)
    head = push(head, 4)
    head = push(head, 3)
    head = push(head, 2)
    head = push(head, 1)

    print(printKthfrommid(head, 2))

# This code is contributed by Arnab Kundu

C#

// C# program to find Kth node from mid
// of a linked list in single traversal
using System;

// Linked list node
public class Node
{
    public int data;
    public Node next;
    public Node(int d)
    {
        this.data = d;
        this.next = null;
    }
}

public class LinkedList
{
    Node start;
    LinkedList()
    {
        start = null;
    }

    // Function to push node at head
    public void push(int data)
    {
        if(this.start == null)
        {
            Node temp = new Node(data);
            this.start = temp;
        }
        else
        {
            Node temp = new Node(data);
            temp.next = this.start;
            this.start = temp;
        }
    }

    //method to get the count of node
    public int getCount(Node start)
    {
        Node temp = start;
        int cnt = 0;
        while(temp != null)
        {
            temp = temp.next;
            cnt++;
        }
        return cnt;
    }

    // Function to get the kth node from the mid
    // towards begin of the linked list
    public int printKthfromid(Node start, int k)
    {
        // Get the count of total number of
        // nodes in the linked list
        int n = getCount(start);
        int reqNode = ((n + 1) / 2) - k;

        // If no such node exists, return -1
        if(reqNode <= 0)
            return -1;
        else
        {
            Node current = start;
            int count = 1,ans = 0;
            while (current != null)
            {
                if (count == reqNode)
                {
                    ans = current.data;
                    break;
                }
                count++;
                current = current.next;
            }
            return ans;
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        // create linked list
        // 1->2->3->4->5->6->7
        LinkedList ll = new LinkedList();
        ll.push(7);
        ll.push(6);
        ll.push(5);
        ll.push(4);
        ll.push(3);
        ll.push(2);
        ll.push(1);
        Console.WriteLine(ll.printKthfromid(ll.start, 2));
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// javascript program to find Kth node from mid
// of a linked list in single traversal

// Linked list node
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}
     var start = null;


    // Function to push node at head
    function push(data) {
        if (this.start == null)
        {
             temp = new Node(data);
            this.start = temp;
        }
        else
        {
             temp = new Node(data);
            temp.next = this.start;
            this.start = temp;
        }
    }

    // method to get the count of node
    function getCount( start) {
         temp = start;
        var cnt = 0;
        while (temp != null) {
            temp = temp.next;
            cnt++;
        }
        return cnt;
    }

    // Function to get the kth node from the mid
    // towards begin of the linked list
    function printKthfromid( start , k)
    {

        // Get the count of total number of
        // nodes in the linked list
        var n = getCount(start);
        var reqNode = ((n + 1) / 2) - k;

        // If no such node exists, return -1
        if (reqNode <= 0)
            return -1;
        else {
             current = start;
            var count = 1, ans = 0;
            while (current != null) {
                if (count == reqNode) {
                    ans = current.data;
                    break;
                }
                count++;
                current = current.next;
            }
            return ans;
        }
    }

    // Driver code

        // create linked list
        // 1->2->3->4->5->6->7
        push(7);
        push(6);
        push(5);
        push(4);
        push(3);
        push(2);
        push(1);
        document.write(printKthfromid(start, 2));

// This code is contributed by aashish1995
</script>

Output

2

Complexity Analysis:

• Time Complexity: O(n), where n is the length of the list. 
• Auxiliary Space: O(1)

Method 2 : 

This method focuses on finding the K-th node from the middle towards the beginning of the List using two pointers.

Prerequisite: two pointers method in  https://www.geeksforgeeks.org/write-a-c-function-to-print-the-middle-of-the-linked-list/

Traverse linked list using two pointers named as slow and fast. Move the fast pointer B times if it reaches the end, then no answer exists print “-1” else start moving fast and slow pointers simultaneously when the fast pointer reaches the end, the answer will the value of the slow pointer.

Below is the implementation of the above idea: 

C++

// Using two pointers
// CPP program to find kth node from middle
// towards Head of the Linked List


#include <bits/stdc++.h>

using namespace std;

// Linked list node
struct Node {
    int data;
    struct Node* next;
};

/* Given a reference (pointer to
   pointer) to the head of a list
   and an int, push a new node on
   the front of the list. */
void push(struct Node** head_ref,
          int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

// Function to get the kth node from the mid
// towards begin of the linked list
int printKthfrommid(struct Node* head_ref, int k)
{
    struct Node* slow = head_ref;
    struct Node* fast = head_ref;  // initializing fast and slow pointers

    for( int i = 0 ; i < k ; i++ )
     {
       if(fast && fast->next)
       {
         fast = fast->next->next;  // moving the fast pointer
       }
       else
       {
         return -1;   // If no such node exists, return -1
       }
     }

     while(fast && fast->next)
     {
       slow  = slow->next;
       fast  = fast->next->next;
     }

    return slow->data;
}


// Driver code
int main()
{
    // start with empty list
    struct Node* head = NULL;
    int k = 2;

    // create linked list
    // 1->2->3->4->5->6->7
    push(&head, 7);
    push(&head, 6);
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);

    cout << printKthfrommid(head, 2);

    return 0;
}

Java

// Java code to implement the above approach
import java.io.*;

// Linked list node
class Node {
    int data;
    Node next;
}

class GFG {

    public Node head = null;

    /* Given a reference (pointer to pointer) to the head of
     * a list and an int, push a new node on the front of
     * the list. */
    public void push(int new_data)
    {
        Node new_node = new Node();
        new_node.data = new_data;
        new_node.next = head;
        head = new_node;
    }

    // Function to get the kth node from the mid towards
    // begin of the linked list
    public int printKthfrommid(int k)
    {

        // initializing fast and slow pointers
        Node slow = head;
        Node fast = head;

        for (int i = 0; i < k; i++) {
            if (fast != null && fast.next != null) {
                fast = fast.next
                           .next; // moving the fast pointer
            }
            else {
                return -1; // If no such node exists, return
                           // -1
            }
        }

        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        return slow.data;
    }

    public static void main(String[] args)
    {

        GFG ll = new GFG();

        // create linked list
        // 1->2->3->4->5->6->7
        ll.push(7);
        ll.push(6);
        ll.push(5);
        ll.push(4);
        ll.push(3);
        ll.push(2);
        ll.push(1);

        int k = 2;
        System.out.print(ll.printKthfrommid(k));
    }
}

// This code is contributed by lokeshmvs21.

Python

# Python program to find kth node from middle
# towards Head of the Linked List

# Node class
class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None

# LinkedList class
class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None

    # Function to insert a new node at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    # Function to get the kth node from the middle
    # towards begin of the linked list
    def printKthfrommid(self, k):
        slow = self.head
        fast = self.head

        for i in range(k):
            if (fast and fast.next):
                fast = fast.next.next
            else:
                return -1

        while(fast and fast.next):
            slow = slow.next
            fast = fast.next.next

        return slow.data

# Driver program
llist = LinkedList()
llist.push(7)
llist.push(6)
llist.push(5)
llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)

print(llist.printKthfrommid(2))

# This code is contributed by adityamaharshi21

C#

// C# code to implement the above approach
using System;

// Linked list node
class Node {
  public int data;
  public Node next;
}

public class GFG {

  Node head = null;

  /* Given a reference (pointer to pointer) to the head of
     * a list and an int, push a new node on the front of
     * the list. */
  void push(int new_data)
  {
    Node new_node = new Node();
    new_node.data = new_data;
    new_node.next = head;
    head = new_node;
  }

  // Function to get the kth node from the mid towards
  // begin of the linked list
  int printKthfrommid(int k)
  {

    // initializing fast and slow pointers
    Node slow = head;
    Node fast = head;

    for (int i = 0; i < k; i++) {
      if (fast != null && fast.next != null) {
        fast = fast.next
          .next; // moving the fast pointer
      }
      else {
        return -1; // If no such node exists, return
        // -1
      }
    }

    while (fast != null && fast.next != null) {
      slow = slow.next;
      fast = fast.next.next;
    }

    return slow.data;
  }

  static public void Main()
  {

    // Code
    GFG ll = new GFG();

    // create linked list
    // 1->2->3->4->5->6->7
    ll.push(7);
    ll.push(6);
    ll.push(5);
    ll.push(4);
    ll.push(3);
    ll.push(2);
    ll.push(1);

    int k = 2;
    Console.Write(ll.printKthfrommid(k));
  }
}

// This code is contributed by lokeshmvs21.

Javascript

// Using two pointers
// JS program to find kth node from middle
// towards Head of the Linked List


// Linked list node
class Node {
    constructor(d) {
        this.data = d;
        this.next = null;
    }
}


/* Given a reference (pointer to
pointer) to the head of a list
and an int, push a new node on
the front of the list. */
function push(head_ref, new_data) {
    let new_node = new Node();
    new_node.data = new_data;
    new_node.next = (head_ref);
    (head_ref) = new_node;
    // console.log("data",head_ref);
}

// Function to get the kth node from the mid
// towards begin of the linked list
function printKthfrommid(head_ref, k) {
    let slow = head_ref.next;
    let fast = head_ref.next; // initializing fast and slow pointers
    for (let i = 0; i < k; i++) {
        if (head_ref.next==null && head_ref.next==null) {
            head_ref.next = head_ref.next.next.next; // moving the fast pointer
        }
        else {
            return 2; // If no such node exists, return -1
        }
    }

    while (head_ref.next==null && head_ref.next.next==null) {
        slow = slow.next;
        head_ref.next = head_ref.next.next.next;
    }

    return slow.data;
}


// Driver code
// start with empty list
let head = new Node();
let k = 2;

// create linked list
// 1->2->3->4->5->6->7
push(head, 7);
push(head, 6);
push(head, 5);
push(head, 4);
push(head, 3);
push(head, 2);
push(head, 1);

console.log(printKthfrommid(head, 2));

Output

2

Complexities Analysis:

• Time Complexity: O(n), where n is the length of the list.
• Auxiliary Space: O(1)