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 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.
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 the beginning of the 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("") # This code is contributed by Pulkit |
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 refer complete article on Swap Kth node from beginning with Kth node from end in a Linked List for more details!
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