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 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

  1. Y is next to X
  2. X is next to Y
  3. X and Y are same
  4. 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++

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// 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
struct Node {
    int data;
    struct Node* next;
};
  
/* Utility function to insert
   a node at the beginning */
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node
        = (struct Node*)malloc(
            sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}
  
/* Utility function for displaying linked list */
void printList(struct 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;
}
  
/* Function for swapping kth nodes 
   from both ends 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 program to test above functions
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++) {
        swapKth(&head, k);
        cout << "\nModified List for k = " << k << endl;
        printList(head);
    }
  
    return 0;
}

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Java

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// 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("");
        }
    }
}

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Python3

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"""
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
llist = 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

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C#

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// 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++) {
            llist.swapKth(i);
            Console.WriteLine("Modified List for k = " + i);
            llist.printList();
            Console.WriteLine("");
        }
    }
}
  
// This code has been contributed by 29AjayKumar

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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

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.

Thanks to Chandra Prakash for initial solution. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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