Count Triplets


Problem Statement :


You are given an array and you need to find number of tripets of indices (i, j, k) such that the elements at those indices are in geometric progression for a given common ratio r and i < j < k . 

Function Description

Complete the countTriplets function in the editor below. It should return the number of triplets forming a geometric progression for a given

as an integer.

countTriplets has the following parameter(s):

    arr: an array of integers
    r: an integer, the common ratio

Input Format

The first line contains two space-separated integers n and r, the size of arr and the common ratio.
The next line contains n space-seperated integers arr[i] . 

Constraints

  1 <= n <= 10^5
  1 <= r <= 10^9
  1 <= arr[i] <= 10^9

Output Format

Return the count of triplets that form a geometric progression.


Solution :



title-img 




                        Solution in C++ :

In C++ :

#include<bits/stdc++.h>
using namespace std;

map<long long,long long> l,r;

int main()
{
	long long n, k,ans=0 ;
	scanf("%lld%lld",&n,&k);
	long long a[n];
	for(int i=0;i<n;i++)
		scanf("%lld",&a[i]);
	for(int i=0;i<n;i++)
		r[a[i]]++;
	for(int i=0;i<n;i++)
	{
		r[a[i]]--;
		if(a[i]%k==0)
		{
			ans+=l[a[i]/k]*r[a[i]*k];
		}
		l[a[i]]++;
	}
	printf("%lld\n",ans);
	return 0;
}






View More Similar Problems

Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

View Solution →

Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →