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

• Difficulty Level : Medium
• Last Updated : 10 Jan, 2023

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.

Recommended Practice

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:

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



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.

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