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

Beautiful Triplets

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1070 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;
}

  • dahuaxie
     + 0 comments

    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

    }

  • herlianaNoviar
     + 0 comments

    My PHP solution:

    function beautifulTriplets($d, $arr) {
    // Write your code here
    $hasil = 0;
    
    for ($i=0; $i < count($arr); $i++) {
        
        for ($j=$i+1; $j < count($arr); $j++) {
            
            if (($arr[$j] - $arr[$i]) == $d) {
                
                for ($k=$j+1; $k < count ($arr); $k++) {
                    
                    if (($arr[$k] - $arr[$j]) == $d) {
                        $hasil++;
                        break 2;
                    }
                }
            }
        }
    }
    return $hasil;
}

  • akash_prajapati6
     + 0 comments
    def beautifulTriplets(d, arr):
    count=0
    for i in range(0,len(arr)-2):
        for j in range(i+1,len(arr)-1):
            if(abs(arr[i]-arr[j])==d):
                for k in range(j+1,len(arr)):
                    if(abs(arr[j]-arr[k])==d):
                        print(i,j,k)
                        count=count+1
    return count

  • aksaini2282003
     + 0 comments

    This is the simple approach in the python language 

    		l=len(arr)
    count=0 
    for i in range(l-2):
        if arr[i]+d in arr and arr[i]+(2*d) in arr:
            count+=1 
    return count