Swap Kth node from beginning with Kth node from end in a Linked List

Given a singly linked list, swap kth node from beginning with kth node from end. Swapping of data is not allowed, only pointers should be changed. This requirement may be logical in many situations where the linked list data part is huge (For example student details like Name, RollNo, Address, ..etc). The pointers are always fixed (4 bytes for most of the compilers)

Example:

Input: 5 -> 10 -> 8 -> 5 -> 9 -> 3, K = 2
Output: 5 -> 9 -> 8 -> 5 -> 10 -> 3 
Explanation: The 2nd node from 1st is 10 and 2nd node from last is 9, so swap them.

Input: 1 -> 2 -> 3 -> 4 -> 5, K = 5
Output: 5 -> 2 -> 3 -> 4 -> 1 
Explanation: The 5th node from 1st is 5 and 5th node from last is 1, so swap them.

Approach: To solve the problem follow the below idea:

The idea is very simple to find the kth node from the start and the kth node from the last is n-k+1th node from start. Swap both nodes
However, there are some corner cases, which must be handled

• Y is next to X
• X is next to Y
• X and Y are the same
• X and Y don’t exist (k is more than the number of nodes in the linked list)

Below is the implementation of the above approach:

C++

// A C++ program to swap Kth node from beginning with kth
// node from end
#include <bits/stdc++.h>
using namespace std;

// A Linked List node
typedef struct Node {
    int data;
    struct Node* next;
} Node;

// Utility function to insert a node at the beginning
void push(Node** head_ref, int new_data)
{
    Node* new_node = (Node*)malloc(sizeof(Node));
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

/* Utility function for displaying linked list */
void printList(Node* node)
{
    while (node != NULL) {
        cout << node->data << " ";
        node = node->next;
    }
    cout << endl;
}

/* Utility function for calculating
   length of linked list */
int countNodes(struct Node* s)
{
    int count = 0;
    while (s != NULL) {
        count++;
        s = s->next;
    }
    return count;
}

// Utility function for calculating length of linked list
void swapKth(struct Node** head_ref, int k)
{
    // Count nodes in linked list
    int n = countNodes(*head_ref);
    // Check if k is valid
    if (n < k)
        return;
    // If x (kth node from start) and y(kth node from end)
    // are same
    if (2 * k - 1 == n)
        return;
    // Find the kth node from the beginning of the linked
    // list. We also find previous of kth node because we
    // need to update next pointer of the previous.
    Node* x = *head_ref;
    Node* x_prev = NULL;
    for (int i = 1; i < k; i++) {
        x_prev = x;
        x = x->next;
    }
    // Similarly, find the kth node from end and its
    // previous. kth node from end is (n-k+1)th node from
    // beginning
    Node* y = *head_ref;
    Node* y_prev = NULL;
    for (int i = 1; i < n - k + 1; i++) {
        y_prev = y;
        y = y->next;
    }
    // If x_prev exists, then new next of it will be y.
    // Consider the case when y->next is x, in this case,
    // x_prev and y are same. So the statement "x_prev->next
    // = y" creates a self loop. This self loop will be
    // broken when we change y->next.
    if (x_prev)
        x_prev->next = y;
    // Same thing applies to y_prev
    if (y_prev)
        y_prev->next = x;
    // Swap next pointers of x and y. These statements also
    // break self loop if x->next is y or y->next is x
    Node* temp = x->next;
    x->next = y->next;
    y->next = temp;
    // Change head pointers when k is 1 or n
    if (k == 1)
        *head_ref = y;
    if (k == n)
        *head_ref = x;
}

// Driver code
int main()
{
    // Let us create the following linked list for testing
    // 1->2->3->4->5->6->7->8
    struct Node* head = NULL;
    for (int i = 8; i >= 1; i--)
        push(&head, i);
    cout << "Original Linked List: ";
    printList(head);
    for (int k = 1; k < 9; k++) {

        // Function call
        swapKth(&head, k);
        cout << "\nModified List for k = " << k << endl;
        printList(head);
    }
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// A C program to swap Kth node from beginning with kth
// node from end
#include <stdio.h>
#include <stdlib.h>

// A Linked List node
typedef struct Node {
    int data;
    struct Node* next;
} Node;

// Utility function to insert a node at the beginning
void push(Node** head_ref, int new_data)
{
    Node* new_node = (Node*)malloc(sizeof(Node));
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

// Utility function for displaying linked list
void printList(Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
    printf("\n");
}

// Utility function for calculating length of linked list
int countNodes(Node* s)
{
    int count = 0;
    while (s != NULL) {
        count++;
        s = s->next;
    }
    return count;
}

// Function for swapping kth nodes from both ends of linked
// list
void swapKth(Node** head_ref, int k)
{
    // Count nodes in linked list
    int n = countNodes(*head_ref);
    // Check if k is valid
    if (n < k)
        return;
    // If x (kth node from start) and y(kth node from end)
    // are same
    if (2 * k - 1 == n)
        return;
    // Find the kth node from the beginning of the linked
    // list. We also find previous of kth node because we
    // need to update next pointer of the previous.
    Node* x = *head_ref;
    Node* x_prev = NULL;
    for (int i = 1; i < k; i++) {
        x_prev = x;
        x = x->next;
    }
    // Similarly, find the kth node from end and its
    // previous. kth node from end is (n-k+1)th node from
    // beginning
    Node* y = *head_ref;
    Node* y_prev = NULL;
    for (int i = 1; i < n - k + 1; i++) {
        y_prev = y;
        y = y->next;
    }
    // If x_prev exists, then new next of it will be y.
    // Consider the case when y->next is x, in this case,
    // x_prev and y are same. So the statement "x_prev->next
    // = y" creates a self loop. This self loop will be
    // broken when we change y->next.
    if (x_prev)
        x_prev->next = y;
    // Same thing applies to y_prev
    if (y_prev)
        y_prev->next = x;
    // Swap next pointers of x and y. These statements also
    // break self loop if x->next is y or y->next is x
    Node* temp = x->next;
    x->next = y->next;
    y->next = temp;
    // Change head pointers when k is 1 or n
    if (k == 1)
        *head_ref = y;
    if (k == n)
        *head_ref = x;
}

// Driver code
int main()
{
    // Let us create the following linked list for testing
    // 1->2->3->4->5->6->7->8
    Node* head = NULL;
    for (int i = 8; i >= 1; i--)
        push(&head, i);
    printf("Original Linked List: ");
    printList(head);
    for (int k = 1; k < 9; k++) {

        // Function call
        swapKth(&head, k);
        printf("\nModified List for k = %d\n", k);
        printList(head);
    }
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// A Java program to swap kth node from the beginning with
// kth node from the end

class Node {
    int data;
    Node next;
    Node(int d)
    {
        data = d;
        next = null;
    }
}

class LinkedList {
    Node head;

    // Utility function to insert a node at the beginning
    void push(int new_data)
    {
        Node new_node = new Node(new_data);
        new_node.next = head;
        head = new_node;
    }

    /* Utility function for displaying linked list */
    void printList()
    {
        Node node = head;
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
        System.out.println("");
    }

    // Utility function for calculating length of linked
    // list
    int countNodes()
    {
        int count = 0;
        Node s = head;
        while (s != null) {
            count++;
            s = s.next;
        }
        return count;
    }

    // Function for swapping kth nodes from both ends of
    // linked list
    void swapKth(int k)
    {
        // Count nodes in linked list
        int n = countNodes();

        // Check if k is valid
        if (n < k)
            return;

        // If x (kth node from start) and
        // y(kth node from end) are same
        if (2 * k - 1 == n)
            return;

        // Find the kth node from beginning of linked list.
        // We also find previous of kth node because we need
        // to update next pointer of the previous.
        Node x = head;
        Node x_prev = null;
        for (int i = 1; i < k; i++) {
            x_prev = x;
            x = x.next;
        }

        // Similarly, find the kth node from end and its
        // previous. kth node from end is (n-k+1)th node
        // from beginning
        Node y = head;
        Node y_prev = null;
        for (int i = 1; i < n - k + 1; i++) {
            y_prev = y;
            y = y.next;
        }

        // If x_prev exists, then new next of it will be y.
        // Consider the case when y->next is x, in this
        // case, x_prev and y are same. So the statement
        // "x_prev->next = y" creates a self loop. This self
        // loop will be broken when we change y->next.
        if (x_prev != null)
            x_prev.next = y;

        // Same thing applies to y_prev
        if (y_prev != null)
            y_prev.next = x;

        // Swap next pointers of x and y. These statements
        // also break self loop if x->next is y or y->next
        // is x
        Node temp = x.next;
        x.next = y.next;
        y.next = temp;

        // Change head pointers when k is 1 or n
        if (k == 1)
            head = y;

        if (k == n)
            head = x;
    }

    // Driver code
    public static void main(String[] args)
    {
        LinkedList llist = new LinkedList();
        for (int i = 8; i >= 1; i--)
            llist.push(i);

        System.out.print("Original linked list: ");
        llist.printList();
        System.out.println("");

        for (int i = 1; i < 9; i++) {

            // Function call
            llist.swapKth(i);
            System.out.println("Modified List for k = "
                               + i);
            llist.printList();
            System.out.println("");
        }
    }
}

// This code is contributed by Aditya Kumar (adityakumar129)

Python3

"""
A Python3 program to swap kth node from
the beginning with kth node from the end
"""


class Node:
    def __init__(self, data, next=None):
        self.data = data
        self.next = next


class LinkedList:

    def __init__(self, *args, **kwargs):
        self.head = Node(None)
    """
    Utility function to insert a node at the beginning
    @args:
        data: value of node
    """

    def push(self, data):
        node = Node(data)
        node.next = self.head
        self.head = node

    # Print linked list
    def printList(self):
        node = self.head
        while node.next is not None:
            print(node.data, end=" ")
            node = node.next

    # count number of node in linked list
    def countNodes(self):
        count = 0
        node = self.head
        while node.next is not None:
            count += 1
            node = node.next
        return count

    """
    Function for swapping kth nodes from
    both ends of linked list
    """

    def swapKth(self, k):

        # Count nodes in linked list
        n = self.countNodes()

        # check if k is valid
        if n < k:
            return

        """
        If x (kth node from start) and
        y(kth node from end) are same
        """
        if (2 * k - 1) == n:
            return

        """
        Find the kth node from beginning of linked list.
        We also find previous of kth node because we need
        to update next pointer of the previous.
        """
        x = self.head
        x_prev = Node(None)
        for i in range(k - 1):
            x_prev = x
            x = x.next

        """
        Similarly, find the kth node from end and its
        previous. kth node from end is (n-k + 1)th node
        from beginning
        """
        y = self.head
        y_prev = Node(None)
        for i in range(n - k):
            y_prev = y
            y = y.next

        """
        If x_prev exists, then new next of it will be y.
        Consider the case when y->next is x, in this case,
        x_prev and y are same. So the statement
        "x_prev->next = y" creates a self loop. This self
        loop will be broken when we change y->next.
        """
        if x_prev is not None:
            x_prev.next = y

        # Same thing applies to y_prev
        if y_prev is not None:
            y_prev.next = x

        """
        Swap next pointers of x and y. These statements
        also break self loop if x->next is y or y->next
        is x
        """
        temp = x.next
        x.next = y.next
        y.next = temp

        # Change head pointers when k is 1 or n
        if k == 1:
            self.head = y

        if k == n:
            self.head = x


# Driver Code
if __name__ == "__main__":
    llist = LinkedList()
    for i in range(8, 0, -1):
        llist.push(i)
    llist.printList()
    print("\n")

    for i in range(1, 9):

            # Function call
        llist.swapKth(i)
        print("Modified List for k = ", i)
        llist.printList()
        print("\n")

# This code is contributed by Pulkit

C#

// C# program to swap kth node from the beginning with
// kth node from the end
using System;

public class Node {
    public int data;
    public Node next;
    public Node(int d)
    {
        data = d;
        next = null;
    }
}

public class LinkedList {
    Node head;

    /* Utility function to insert
    a node at the beginning */
    void push(int new_data)
    {
        Node new_node = new Node(new_data);
        new_node.next = head;
        head = new_node;
    }

    /* Utility function for displaying linked list */
    void printList()
    {
        Node node = head;
        while (node != null) {
            Console.Write(node.data + " ");
            node = node.next;
        }
        Console.WriteLine("");
    }

    /* Utility function for calculating
    length of linked list */
    int countNodes()
    {
        int count = 0;
        Node s = head;
        while (s != null) {
            count++;
            s = s.next;
        }
        return count;
    }

    /* Function for swapping kth nodes from
    both ends of linked list */
    void swapKth(int k)
    {
        // Count nodes in linked list
        int n = countNodes();

        // Check if k is valid
        if (n < k)
            return;

        // If x (kth node from start) and y(kth node from
        // end) are same
        if (2 * k - 1 == n)
            return;

        // Find the kth node from beginning of linked list.
        // We also find previous of kth node because we need
        // to update next pointer of the previous.
        Node x = head;
        Node x_prev = null;
        for (int i = 1; i < k; i++) {
            x_prev = x;
            x = x.next;
        }

        // Similarly, find the kth node from end and its
        // previous. kth node from end is (n-k+1)th node
        // from beginning
        Node y = head;
        Node y_prev = null;
        for (int i = 1; i < n - k + 1; i++) {
            y_prev = y;
            y = y.next;
        }

        // If x_prev exists, then new next of it will be y.
        // Consider the case when y->next is x, in this
        // case, x_prev and y are same. So the statement
        // "x_prev->next = y" creates a self loop. This self
        // loop will be broken when we change y->next.
        if (x_prev != null)
            x_prev.next = y;

        // Same thing applies to y_prev
        if (y_prev != null)
            y_prev.next = x;

        // Swap next pointers of x and y. These statements
        // also break self loop if x->next is y or y->next
        // is x
        Node temp = x.next;
        x.next = y.next;
        y.next = temp;

        // Change head pointers when k is 1 or n
        if (k == 1)
            head = y;

        if (k == n)
            head = x;
    }

    // Driver code
    public static void Main(String[] args)
    {
        LinkedList llist = new LinkedList();
        for (int i = 8; i >= 1; i--)
            llist.push(i);

        Console.Write("Original linked list: ");
        llist.printList();
        Console.WriteLine("");

        for (int i = 1; i < 9; i++) {

            // Function call
            llist.swapKth(i);
            Console.WriteLine("Modified List for k = " + i);
            llist.printList();
            Console.WriteLine("");
        }
    }
}

// This code has been contributed by 29AjayKumar

Javascript

<script>
// A javascript program to swap kth
// node from the beginning with
// kth node from the end
 class Node {
        constructor(val) {
            this.data = val;
            this.next = null;
        }
    }

    var head;

    /*
     * Utility function to insert a node at the beginning
     */
    function push(new_data) {
         new_node = new Node(new_data);
        new_node.next = head;
        head = new_node;
    }

    /* Utility function for displaying linked list */
    function printList()
    {
         node = head;
        while (node != null)
        {
            document.write(node.data + " ");
            node = node.next;
        }
        document.write("");
    }

    /*
     * Utility function for calculating length of linked list
     */
    function countNodes()
    {
        var count = 0;
         s = head;
        while (s != null)
        {
            count++;
            s = s.next;
        }
        return count;
    }

    /*
     * Function for swapping kth nodes from both ends of linked list
     */
    function swapKth(k)
    {

        // Count nodes in linked list
        var n = countNodes();

        // Check if k is valid
        if (n < k)
            return;

        // If x (kth node from start) and
        // y(kth node from end) are same
        if (2 * k - 1 == n)
            return;

        // Find the kth node from beginning of linked list.
        // We also find previous of kth node because we need
        // to update next pointer of the previous.
         x = head;
         x_prev = null;
        for (i = 1; i < k; i++)
        {
            x_prev = x;
            x = x.next;
        }

        // Similarly, find the kth node from end and its
        // previous. kth node from end is (n-k+1)th node
        // from beginning
         y = head;
         y_prev = null;
        for (i = 1; i < n - k + 1; i++)
        {
            y_prev = y;
            y = y.next;
        }

        // If x_prev exists, then new next of it will be y.
        // Consider the case when y->next is x, in this case,
        // x_prev and y are same. So the statement
        // "x_prev->next = y" creates a self loop. This self
        // loop will be broken when we change y->next.
        if (x_prev != null)
            x_prev.next = y;

        // Same thing applies to y_prev
        if (y_prev != null)
            y_prev.next = x;

        // Swap next pointers of x and y. These statements
        // also break self loop if x->next is y or y->next
        // is x
         temp = x.next;
        x.next = y.next;
        y.next = temp;

        // Change head pointers when k is 1 or n
        if (k == 1)
            head = y;

        if (k == n)
            head = x;
    }

    // Driver code to test above
        for (let i = 8; i >= 1; i--)
            push(i);

        document.write("Original linked list: <br/>");
        printList();
        document.write("<br/>");

        for (let i = 1; i < 9; i++)
        {
            swapKth(i);
            document.write("Modified List for k = " + i + "<br/>");
            printList();
            document.write("<br/>");
        }

// This code is contributed by gauravrajput1
</script>

Output

Original Linked List: 1 2 3 4 5 6 7 8 

Modified List for k = 1
8 2 3 4 5 6 7 1 

Modified List for k = 2
8 7 3 4 5 6 2 1 

Modified List for k = 3
8 7 6 4 5 3 2 1 

Modified List for k = 4
8 7 6 5 4 3 2 1 

Modified List for k = 5
8 7 6 4 5 3 2 1 

Modified List for k = 6
8 7 3 4 5 6 2 1 

Modified List for k = 7
8 2 3 4 5 6 7 1 

Modified List for k = 8
1 2 3 4 5 6 7 8 

Time Complexity: O(N), where N is the length of the list. One traversal of the list is needed.
Auxiliary Space: O(1). No extra space is required.

Please note that the above code runs three separate loops to count nodes, find x and x prev, and to find y and y_prev. These three things can be done in a single loop. The code uses three loops to keep things simple and readable.