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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Beautiful Triplets
  5. Discussions

Beautiful Triplets

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1065 Discussions

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  • alban_tyrex
     + 0 comments

    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;
}

  • patrickgao2020
     + 0 comments

    Here is my Python solution!

    def beautifulTriplets(d, arr):
    beautiful = 0
    for i in range(len(arr) - 2):
        for j in range(i + 1, len(arr) - 1):
            if arr[j] - arr[i] == d:
                for k in range(j + 1, len(arr)):
                    if arr[k] - arr[j] == d:
                        beautiful += 1
    return beautiful

  • N1510361
     + 0 comments
     
 public static int beautifulTriplets(int d, List<Integer> arr) {
    // Write your code here
     int triplets=0;
     for(int i=0;i<arr.size();i++){
         int sum = d+arr.get(i);
         if(arr.contains(sum)){
            sum=sum+d;
            if(arr.contains(sum)){
                triplets++;
            }
         }
    }
        return triplets;
     }
    

    }

    `

  • akshay_vijay123
     + 1 comment

    My python code (only runs last for loop if needed):

    def beautifulTriplets(d, arr):
    ans = 0
    for a in range(len(arr)):
        for b in range(a + 1, len(arr)):
            if arr[b] - arr[a] == d:
                for c in range(b + 1, len(arr)):
                    if arr[c] - arr[b] == d:
                        ans += 1
    return ans

  • illogicalsleepi1
     + 0 comments
    def beautifulTriplets(d, arr):
    c=0
    n=len(arr)
    for i in range(n):
        for j in range(i+1,n):
            if arr[j]-arr[i]==d:
                for k in range(j+1,n):
                    if arr[k]-arr[j]==d:
                        c+=1
    return c