HackerRank Beautiful Triplets problem solution

In this HackerRank Beautiful Triplets problem solution you have Given an increasing sequence of integers and the value of d, count the number of beautiful triplets in the sequence.

Function Description

Complete the beautifulTriplets function in the editor below.

beautifulTriplets has the following parameters:

• int d: the value to match
• int arr[n]: the sequence, sorted ascending

Returns

• int: the number of beautiful triplets

Input Format

The first line contains 2 space-separated integers, n and d, the length of the sequence and the beautiful difference.
The second line contains  space-separated integers arr[i].

Hackerrank Beautiful Triplets problem solution in Python.

n, d = [int(r) for r in input().split()]
a = [int(r) for r in input().split()]

triplets = 0
for i in range(n-2):
    for j in range(i + 1, n-1):
        if a[j] - a[i] == d:
            foundTrip = False
            for k in range(j + 1, n):
                if a[k] - a[j] == d:
                    triplets += 1
                    foundTrip = True
                    break
            if foundTrip == True:
                break
            
print(triplets)
            

Beautiful Triplets problem solution in Java.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int d = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        int ans = solve(n, d, a);
        System.out.println(ans);
    }

    private static int solve(int n, int d, int[] a) {
        Map<Integer, Integer> m1 = new HashMap<>();
        Map<Integer, Integer> m2 = new HashMap<>();

        int r = 0;
        for (int i = 0; i < n; i++) {
            if (m2.containsKey(a[i])) {
                int c = m2.remove(a[i]);
                r += c;
            }
            if (m1.containsKey(a[i])) {
                int c = m1.remove(a[i]);
                m2.put(a[i] + d, c);
            }
            add(m1, a[i]+d);
        }
        return r;
    }

    private static void add(Map<Integer, Integer> map, int key) {
        Integer old = map.get(key);
        if (old == null) {
            old = 0;
        }
        old++;
        map.put(key, old);
    }
}

Problem solution in C++ programming.

#include <bits/stdc++.h>
#define pb push_back
#define sqr(x) (x)*(x)
#define sz(a) int(a.size())
#define reset(a,b) memset(a,b,sizeof(a))
#define oo 1000000007

using namespace std;

typedef pair<int,int> pii;
typedef long long ll;

int a[111111],n,d;
set<int> mys;

int main(){
//    freopen("input.txt","r",stdin);
    cin>>n>>d;
    for(int i=1; i<=n; ++i){
        cin>>a[i];
        mys.insert(a[i]);
    }
    int res=0;
    for(int i=2; i<n; ++i) if(mys.count(a[i]-d) && mys.count(a[i]+d)) ++res;
    cout<<res<<endl;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int n, mid, d, i, j, k, *a, count = 0;
    
    scanf("%d %d", &n, &d);
    mid = n / 2;
    a = (int *) malloc(n * sizeof(int));
    
    //scanf("%d %d", &a[0], &a[1]);
    for (i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    
    for (i = 0; i < n - 2; i++) {
        j = 1;
        while (a[j] - a[i] <= d) {
            if (a[j] - a[i] == d) {
                k = j + 1;
                while (a[k] - a[j] <= d) {
                    if (a[k] - a[j] == d) {
                        count++;
                        break;
                    }
                    k++;
                }
            }
            j++;
        }
    }
    
    printf("%d", count);
    
    return 0;
}

Problem solution in JavaScript programming.

function processData(input) {
    var data = input.trim().split(/n/);
    var arr = (data[1].split(" ")).sort(function(a,b){return Number(a)-Number(b);});
    var cut = data[0].split(" ");
    var n = Number(cut[0]);
    var d = Number(cut[1]);
    var a;
    var b;
    var c;
    var trips = 0;

    if(n < 3){
        console.log(0);
    }
    else{
        for(var h = 0; h < n - 2; h++){
            a = arr[h];
            for(var i = h+1; i < n-1;i++){
                b = arr[i];
                if(b-a !== d){
                    continue;
                }
                for(var k = i+1; k < n; k++){

                    c = arr[k];
                    //console.log("round: "+h +"b-a=" + (b-a)+ "c-b=" + (c-b) + "c=" + c + "b = " + b);
                    if(c-b ===d){
                        trips++;
                    }
                }
            }
        }
        console.log(trips);
    }
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});