706071. Triplets in Sorted Linked List
Count Triplets
Summary
This problem involves counting triplets in a sorted linked list. The triplets are formed by considering every three consecutive nodes and checking if they form an arithmetic progression (AP). The input is a linked list where each node contains an integer value, and the output is the count of all possible triplets that can be formed from these values.
Detailed Explanation
The algorithm to solve this problem involves iterating over the linked list and considering every three consecutive nodes as potential triplets. For each triplet, check if it forms an AP by verifying if the difference between the first two elements and the last element is the same.
Here's a step-by-step breakdown of the solution:
- Initialize a variable
countto store the count of triplets. - Iterate over the linked list using a pointer
p. - For each node, consider the next two nodes as a potential triplet (
p,p->next, andp->next->next). - Check if the difference between the first two elements (
p->dataandp->next->data) is equal to the difference between the last element (p->next->next->data) and the middle element (p->next->data). If it is, increment thecount. - Repeat steps 3-4 until the end of the linked list.
- Return the
countas the final output.
Time complexity: O(n), where n is the number of nodes in the linked list. Space complexity: O(1), as we only use a constant amount of space to store the count and pointers.
Optimized Solutions
Java
public int countTriplets(Node head) {
int count = 0;
Node p = head;
while (p != null && p.next != null && p.next.next != null) {
if ((p.data - p.next.data) == (p.next.data - p.next.next.data)) {
count++;
}
p = p.next.next;
}
return count;
}