Introduction
The linked list is one of the most important concepts and data structures to learn while preparing for interviews. Having a good grasp of Linked Lists can be a huge plus point in a coding interview.
Problem Statement
In this problem, we have been given a linked list and positive integer k. Our task is to write a function that accepts the head node of the linked list as a parameter and returns the value of the node present at (ceil(n/k))th position in the linked list.
Here, n is the count of the number of nodes in the linked list.
Problem Statement Understanding
Let’s try to understand the problem with help of examples.
Suppose we are given a linked list
Now according to the problem statement we have to find the node at (ceil(n/k))th position in the linked list and return its value.
For the above given linked list n = 6, as there are 6 nodes in the linked list, so
- (ceil(n/k)) = (ceil(6/4)) = 2, it means that we have to return the value of 2nd node of the linked list as output.
Output: B
Explanation: We will return B in output as B is the value of (ceil(n/k))th node of the above given linked list.
If the linked list is
For the above linked list n = 9 (as there are 9 nodes in the linked list), then the value of (ceil(n/k)) is:
- (ceil(n/k)) = (ceil(9/3)) = 3, so we will return the value of the 3rd node of the linked list as output.
Output: C
Note: Here in this problem we are considering 1 based indexing.
Some more Examples
Input:
Output: 5
Input:
Output: 60
Input:
Output: 18
Now I think from the above examples the problem is clear, and now we have to think how can we approach it.
Approach 1
A simple approach is:
- First, loop through the linked list to find the count of number of nodes n present in the linked list.
- Then calculate the value of (ceil(n/k)).
- Now starting from the first node of the list traverse to this position given by (ceil(n/k)) and return the data of the node at this position.
Time Complexity: O(n), as we are traversing the complete length of the list.
Space Complexity: O(1)
This approach traverses the linked list 2 times. Let’s try to think, can we do better than this? Can we find out the (ceil(n/k))th node of the linked list in one single traversal?
Let’s see the approach 2 to get the answers to the above questions.
Approach 2
In this approach, we can get the required node by traversing the list once only. The complete algorithm for this approach is explained below.
Algorithm
- If k<=0 or our head==NULL, then return NULL.
- We will take two pointers temp and req_node and initialize temp with head and req_node with NULL.
- For every k jumps of the temp pointer, we will make one jump of the req_node pointer.
- Now when temp will be at NULL, req_node will be at (ceil(n/k))th node of the linked list.
- Finally, we will return the req_node->data as output.
Dry Run
Code Implementation:
#include#include struct Node { int data; struct Node* next; }; struct Node* newNode(int x) { struct Node* node = malloc(sizeof(struct Node*)); node->data = x; node->next = NULL; return node; } struct Node* fractionalNodes(struct Node* head, int k) { if (k <= 0 || head == NULL) return NULL; struct Node* req_node = NULL; int i = 0; for (struct Node* temp = head; temp != NULL; temp = temp->next) { if (i % k == 0) { if (req_node == NULL) req_node = head; else req_node = req_node->next; } i++; } return req_node; } void printList(struct Node* node) { while (node != NULL) { printf("%d ", node->data); node = node->next; } printf("\n"); } int main(void) { struct Node* head = newNode(3); head->next = newNode(5); head->next->next = newNode(7); head->next->next->next = newNode(9); head->next->next->next->next = newNode(11); int k = 2; printf("List is "); printList(head); struct Node* answer = fractionalNodes(head, k); printf("\nFractional node is "); printf("%d\n", answer->data); return 0; }
#includeusing namespace std; struct Node { int data; Node* next; }; Node* newNode(int data) { Node* new_node = new Node; new_node->data = data; new_node->next = NULL; return new_node; } Node* fractionalNodes(Node* head, int k) { if (k <= 0 || head == NULL) return NULL; Node* req_node = NULL; int i = 0; for (Node* temp = head; temp != NULL; temp = temp->next) { if (i % k == 0) { if (req_node == NULL) req_node = head; else req_node = req_node->next; } i++; } return req_node; } void printList(Node* node) { while (node != NULL) { printf("%d ", node->data); node = node->next; } printf("\n"); } int main(void) { Node* head = newNode(3); head->next = newNode(5); head->next->next = newNode(7); head->next->next->next = newNode(9); head->next->next->next->next = newNode(11); int k = 2; printf("List is "); printList(head); Node* answer = fractionalNodes(head, k); printf("\nFractional node is "); printf("%d\n", answer->data); return 0; }
public class FractionalNode
{
static class Node{
int data;
Node next;
Node (int data){
this.data = data;
}
}
static Node fractionalNodes(Node head, int k)
{
// Corner cases
if (k <= 0 || head == null)
return null;
Node fractionalNode = null;
// Traverse the given list
int i = 0;
for (Node temp = head; temp != null;temp = temp.next)
{
if (i % k == 0){
// First time we see a multiple of k
if (fractionalNode == null)
fractionalNode = head;
else
fractionalNode = fractionalNode.next;
}
i++;
}
return fractionalNode;
}
static void printList(Node node)
{
while (node != null)
{
System.out.print(node.data+" ");
node = node.next;
}
System.out.println();
}
public static void main(String[] args) {
Node head = new Node(3);
head.next = new Node(5);
head.next.next = new Node(7);
head.next.next.next = new Node(9);
head.next.next.next.next = new Node(11);
int k =2;
System.out.print("List is ");
printList(head);
Node answer = fractionalNodes(head, k);
System.out.println("Fractional node is "+answer.data);
}
}
class Node:
def __init__(self, data):
self.data = data
self.next = None
def newNode(data):
new_node = Node(data)
new_node.data = data
new_node.next = None
return new_node
def fractionalNodes(head, k):
if (k <= 0 or head == None):
return None
fractionalNode = None
i = 0
temp = head
while (temp != None):
if (i % k == 0):
if (fractionalNode == None):
fractionalNode = head
else:
fractionalNode = fractionalNode.next
i = i + 1
temp = temp.next
return fractionalNode
def printList(node):
while (node != None):
print(node.data, end = ' ')
node = node.next
if __name__ == '__main__':
head = newNode(3)
head.next = newNode(5)
head.next.next = newNode(7)
head.next.next.next = newNode(9)
head.next.next.next.next = newNode(11)
k = 2
print("List is", end = ' ')
printList(head)
answer = fractionalNodes(head, k)
print("\nFractional node is", end = ' ')
print(answer.data)
Output
List is 3 5 7 9 11
Fractional node is 7
Time Complexity: O(n), as we are traversing the complete length of the list
[forminator_quiz id=”4337″]
So, in this article, you have learnt how to get the value of the node present at (ceil(n/k))th position in the Linked List. If you want to solve more questions on Linked List, which are curated by our expert mentors at PrepBytes, you can follow this link Linked List.