Count triplets in an array such that i<j<k and a[j] – a[i] = a[k] – a[j] = D

Given an array arr and an integer D, the task is to count the number of triplets(arr[i], arr[j], arr[k]) in the array such that:

  1. i < j < k
  2. arr[j] – arr[i] = arr[k] – arr[j] = D

Examples:

Input: arr = {1, 6, 7, 7, 8, 10, 12, 13, 14}, D = 3
Output: 2
Explanation:
There are two triplets in the array that satisfies the given criteria.
-> 1st triplet(7, 10, 13) such that i = 3, j = 6 and k = 8, such that (i < j < k) and (10 – 7 = 13 – 10 = D (=3))
-> 2nd triplet(7, 10, 13) such that i = 4, j = 6 and k = 8, such that (i < j < k) and (10 – 7 = 13 – 10 = D (=3))

Input: arr = {6, 3, 4, 5}, D = 1
Output: 0

Naive Approach: The simplest approach is using three nested for loops and find the triplets that satisfy the given criteria. Below is the implementation of this approach.



C++

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// C++ program to count the triplets
#include<bits/stdc++.h>
using namespace std;
  
// Function to count the triplets
int CountTriplets(int arr[], int d, int n)
{
    int count = 0;
  
    // Three nested for loops to count the
    // triplets that satisfy the given criteria
    for(int i = 0; i < n - 2; i++)
    {
       for(int j = i + 1; j < n - 1; j++)
       {
          for(int k = j + 1; k < n; k++)
          {
             if ((arr[j] - arr[i] == d) && 
                 (arr[k] - arr[j] == d)) 
             {
                 count++;
             }
          }
       }
    }
    return count;
}
  
// Driver Code
int main()
{
    int A[] = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
    int D = 3;
    int n = sizeof(A) / sizeof(A[0]);
      
    cout << (CountTriplets(A, D, n));
}
  
// This code is contributed by chitranayal

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Java

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// Java program to count the triplets
  
class GFG {
  
    // Function to count the triplets
    static int CountTriplets(int[] arr, int d)
    {
        int count = 0;
        int n = arr.length;
  
        // Three nested for loops to count the
        // triplets that satisfy the given criteria
        for (int i = 0; i < n - 2; i++) {
            for (int j = i + 1; j < n - 1; j++) {
                for (int k = j + 1; k < n; k++) {
  
                    if ((arr[j] - arr[i] == d)
                        && (arr[k] - arr[j] == d)) {
                        count++;
                    }
                }
            }
        }
        return count;
    }
  
    // Driver Code
    public static void main(String args[])
    {
        int A[] = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
        int D = 3;
        System.out.println(CountTriplets(A, D));
    }
}

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Python3

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# Python3 program to count the triplets
  
# Function to count the triplets
def CountTriplets(arr, d):
      
    count = 0;
    n = len(arr);
  
    # Three nested for loops to count the
    # triplets that satisfy the given criteria
    for i in range(n - 1):
        for j in range(i + 1, n - 1):
            for k in range(j + 1, n):
  
                if ((arr[j] - arr[i] == d) and 
                    (arr[k] - arr[j] == d)):
                    count += 1;
    return count;
  
# Driver Code
if __name__ == '__main__':
      
    A = [ 1, 6, 7, 7, 8, 10, 12, 13, 14 ];
    D = 3;
      
    print(CountTriplets(A, D));
  
# This code is contributed by Rajput-Ji

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C#

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// C# program to count the triplets 
using System;
  
class GFG {
   
    // Function to count the triplets
    static int CountTriplets(int[] arr, int d)
    {
        int count = 0;
        int n = arr.Length;
   
        // Three nested for loops to count the
        // triplets that satisfy the given criteria
        for (int i = 0; i < n - 2; i++) {
            for (int j = i + 1; j < n - 1; j++) {
                for (int k = j + 1; k < n; k++) {
   
                    if ((arr[j] - arr[i] == d)
                        && (arr[k] - arr[j] == d)) {
                        count++;
                    }
                }
            }
        }
        return count;
    }
   
    // Driver Code
    public static void Main(String []args)
    {
        int []A = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
        int D = 3;
        Console.WriteLine(CountTriplets(A, D));
    }
}
  
// This code is contributed by PrinciRaj1992

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Output:

2

Time Complexity: O(N^3).

Efficient Approach:

  • An efficient approach for this problem is to use a map to store (key, values) pair where value will be count of key.
  • The idea is to traverse the array from 0 to N and do following:

    • check that A[i] – D and A[i] – 2*D are present in the map or not.
    • If it’s in the map then we will simply multiply their respective values and increase answer by that.
    • Now, we just have to put A[i] into the map and update the map.

Below is the implementation of the above approach.

C++14

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// C++14 program to count the number 
// of triplets from an array.
#include<bits/stdc++.h>
using namespace std; 
  
// Function to count the triplets 
int countTriplets (int arr[], int d, int n)
{
    int count = -1;
      
    // Create a map to store (key, values) pair. 
    map<int, int> set;
  
    // Traverse the array and check that we 
    // have already put (a[i]-d and a[i]-2*d) 
    // into map or not. If yes we have to get 
    // the values of both the keys and 
    // multiply them, else put a[i] into the map. 
    for(int i = 0; i < n; i++)
    {
        if ((set.find(arr[i] - d) != set.end()) &&
            (set.find(arr[i] - 2 * d) != set.end()))
        {
            count += (set[arr[i] - d] * 
                      set[arr[i] - 2 * d]);
        }
  
        // Update the map
        if (set.find(arr[i]) == set.end())
            set[arr[i]] = 1;
        else
            set[arr[i]] += 1;
    }
    return count;
}
  
// Driver Code
int main() 
      
    // Test Case 1: 
    int a1[] = { 1, 6, 7, 7, 8, 10, 12, 13, 14 }; 
    int d1 = 3; 
    cout << countTriplets(a1, d1, 9) << endl; 
  
    // Test Case 2: 
    int a2[] = { 6, 3, 4, 5 }; 
    int d2 = 1; 
    cout << countTriplets(a2, d2, 4) << endl; 
  
    // Test Case 3: 
    int a3[] = { 1, 2, 4, 5, 7, 8, 10 }; 
    int d3 = 3; 
    cout << countTriplets(a3, d3, 7) << endl; 
  
    return 0; 
  
// This code is contributed by himanshu77

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Java

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// Java program to count the number
// of triplets from an array.
  
import java.util.*;
  
class GFG {
  
    // Function to count the triplets
    static int countTriplets(int[] arr, int d)
    {
        int count = -1;
  
        // Create a map to store (key, values) pair.
        Map<Integer, Integer> set = new HashMap<>();
  
        // Traverse the array and check that we
        // have already put (a[i]-d and a[i]-2*d)
        // into map or not. If yes we have to get
        // the values of both the keys and
        // multiply them, else put a[i] into the map.
        for (int i = 0; i < arr.length; i++) {
  
            if (set.get(arr[i] - d) != null
                && set.get(arr[i] - 2 * d) != null) {
                count += (set.get(arr[i] - d)
                          * set.get(arr[i] - 2 * d));
            }
  
            // Update the map.
            if (set.get(arr[i]) == null) {
                set.put(arr[i], 1);
            }
            else {
                set.put(arr[i], set.get(arr[i]) + 1);
            }
        }
  
        return count;
    }
  
    // Driver Code
    public static void main(String args[])
    {
  
        // Test Case 1:
        int a1[] = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
        int d1 = 3;
        System.out.println(countTriplets(a1, d1));
  
        // Test Case 2:
        int a2[] = { 6, 3, 4, 5 };
        int d2 = 1;
        System.out.println(countTriplets(a2, d2));
  
        // Test Case 3:
        int a3[] = { 1, 2, 4, 5, 7, 8, 10 };
        int d3 = 3;
        System.out.println(countTriplets(a3, d3));
    }
}

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C#

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// C# program to count the number
// of triplets from an array.
using System;
using System.Collections.Generic;
  
class GFG {
   
    // Function to count the triplets
    static int countTriplets(int[] arr, int d)
    {
        int count = -1;
   
        // Create a map to store (key, values) pair.
        Dictionary<int, int> set = new Dictionary<int, int>();
   
        // Traverse the array and check that we
        // have already put (a[i]-d and a[i]-2*d)
        // into map or not. If yes we have to get
        // the values of both the keys and
        // multiply them, else put a[i] into the map.
        for (int i = 0; i < arr.Length; i++) {
   
            if (set.ContainsKey(arr[i] - d)
                && set.ContainsKey(arr[i] - 2 * d)) {
                count += (set[arr[i] - d]
                          * set[arr[i] - 2 * d]);
            }
   
            // Update the map.
            if (!set.ContainsKey(arr[i])) {
                set.Add(arr[i], 1);
            }
            else {
                set[arr[i]] = set[arr[i]] + 1;
            }
        }
   
        return count;
    }
   
    // Driver Code
    public static void Main(String []args)
    {
   
        // Test Case 1:
        int []a1 = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
        int d1 = 3;
        Console.WriteLine(countTriplets(a1, d1));
   
        // Test Case 2:
        int []a2 = { 6, 3, 4, 5 };
        int d2 = 1;
        Console.WriteLine(countTriplets(a2, d2));
   
        // Test Case 3:
        int []a3 = { 1, 2, 4, 5, 7, 8, 10 };
        int d3 = 3;
        Console.WriteLine(countTriplets(a3, d3));
    }
}
  
// This code is contributed by Princi Singh

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Output:

1
0
2

Time Complexity: O(N)

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