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

03 Jul, 2022
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 line Name, RollNo, Address, ..etc). The pointers are always fixed (4 bytes for most of the compilers).
Example: 
Input: 1 -> 2 -> 3 -> 4 -> 5, K = 2
Output: 1 -> 4 -> 3 -> 2 -> 5 
Explanation: The 2nd node from 1st is 2 and 
2nd node from last is 4, 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.
Illustration:  
 
Approach: The idea is very simple find the k th node from the start and the kth node from last is n-k+1 th node from start. Swap both the nodes. However there are some corner cases, which must be handled 
Y is next to XX is next to YX and Y are sameX and Y don’t exist (k is more than number of nodes in linked list)
Y is next to X
X is next to Y
X and Y are same
X and Y don’t exist (k is more than number of nodes in linked list)
Below is the implementation of the above approach.
C++
C
Java
Python3
C#
Javascript
// A C++ program to swap Kth node from beginning with kth// node from end#include <bits/stdc++.h>using namespace std;  // A Linked List nodetypedef struct Node {    int data;    struct Node* next;} Node;  // Utility function to insert a node at the beginningvoid 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 listvoid 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 program to test above functionsint 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++) {        swapKth(&head, k);        cout << "\nModified List for k = " << k << endl;        printList(head);    }    return 0;}  // This code is contributed by Aditya Kumar (adityakumar129)
// A C program to swap Kth node from beginning with kth// node from end#include <stdio.h>#include <stdlib.h>  // A Linked List nodetypedef struct Node {    int data;    struct Node* next;} Node;  // Utility function to insert a node at the beginningvoid 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 listvoid printList(Node* node){    while (node != NULL) {        printf("%d ", node->data);        node = node->next;    }    printf("\n");}  // Utility function for calculating length of linked listint 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// listvoid 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 program to test above functionsint 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++) {        swapKth(&head, k);        printf("\nModified List for k = %d\n", k);        printList(head);    }    return 0;}  // This code is contributed by Aditya Kumar (adityakumar129)
// 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 to test above    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++) {            llist.swapKth(i);            System.out.println("Modified List for k = " + i);            llist.printList();            System.out.println("");        }    }}  // This code is contributed by Aditya Kumar (adityakumar129)
"""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 Codellist = LinkedList()for i in range(8, 0, -1):    llist.push(i)llist.printList()    for i in range(1, 9):    llist.swapKth(i)    print("Modified List for k = ", i)    llist.printList()    print("\n")  # This code is contributed by Pulkit
// C# program to swap kth node from the beginning with// kth node from the endusing 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++) {            llist.swapKth(i);            Console.WriteLine("Modified List for k = " + i);            llist.printList();            Console.WriteLine("");        }    }}  // This code has been contributed by 29AjayKumar
<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>
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 

Complexity Analysis: 
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. 
29AjayKumar
agrawalpulkit1994
andrew1234
GauravRajput1
adityakumar129
hardikkoriintern
Amazon
Morgan Stanley
Rockstand
Arrays
Linked List
Morgan Stanley
Amazon
Rockstand
Linked List
Arrays
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