Beautiful Hackerrank Solution in c

Given a sequence of integers , a triplet  is beautiful if:

Given an increasing sequenc of integers and the value of , count the number of beautiful triplets in the sequence.

For example, the sequence  and . There are three beautiful triplets, by index: . To test the first triplet,  and .

Function Description

Complete the beautifulTriplets function in the editor below. It must return an integer that represents the number of beautiful triplets in the sequence.

beautifulTriplets has the following parameters:

• d: an integer
• arr: an array of integers, sorted ascending

Input Format

The first line contains  space-separated integers  and , the length of the sequence and the beautiful difference.
The second line contains  space-separated integers .

Constraints

Output Format

Print a single line denoting the number of beautiful triplets in the sequence.

Sample Input

7 3
1 2 4 5 7 8 10

Sample Output

3

Explanation

The input sequence is , and our beautiful difference . There are many possible triplets , but our only beautiful triplets are  ,  and  by value not index. Please see the equations below:

Recall that a beautiful triplet satisfies the following equivalence relation:  where .

#include<stdio.h>

#include<stdlib.h>

#include<assert.h>

#include<math.h>

int main()

{

int n,d;

scanf("%d %d",&n,&d);

int a[n];

int c,p=0;

int i,j,k=0;

for(i=0;i<n;i++)

{

scanf("%d",&a[i]);

}

for(i=0;i<n;i++)

{

c=0;

for(j=i,k=j;j<n,k<n;j++,k++)

{

if((a[j]-a[i])==d)

{

k=j+1;

c=c+2;

for(k=j+1;k<n;k++)

{

if((a[k]-a[j])==d)

c=c+1;

}

}

}

if(c==3)

p=p+1;

}

printf("%d",p);

return 0;

}