public static int beautifulTriplets(int d, List<Integer> arr) {
    int tripletcount = 0;
    for(int i = 0; i < arr.size(); i++) {
        for(int j=i+1; j< arr.size(); j++) {
            if(arr.get(j) - arr.get(i) == d && findThirdNumber(arr, j, d, arr.get(j))) {
                tripletcount++;
                break;
            }
        }
    }
    return tripletcount;
}

private static boolean findThirdNumber(List<Integer> arr, int j, int d, int jval) {
    for(int k = j+1; k < arr.size(); k++) {
        if(arr.get(k) - jval == d) return true;
    }
    return false;
}

Here is problem solution in Python, Java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-beautiful-triplets-problem-solution.html

Here is my O(n) c++ solution, you can watch the implementation here : https://youtu.be/vLD3N79nLSE

int beautifulTriplets(int d, vector<int> arr) {
    map<int, int> mp;
    int result = 0;
    for (int a : arr) {
        mp[a] += 1;
        result += mp[a-d]*mp[a-2*d];
    }
    return result;
}

i learned this elegant solution from a similar problem "count triplets", where it is a geometric series ak = aj *d, aj = ai *d:

func beautifulTriplets(d int32, arr []int32) int32 {

// Write your code here
pair, triple := map[int32]int32{}, map[int32]int32{}
var res int32
for _, a := range arr {
    if c, ok := triple[a]; ok {
        res += c
    }
    if c, ok := pair[a]; ok {
        triple[a+d] += c
    }
    pair[a+d]++
}
return res

}