Related Articles

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

• Last Updated : 12 Aug, 2021

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:
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:

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++

 `// C++ program to count the triplets``#include``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);``    ` `    ``cout << (CountTriplets(A, D, n));``}` `// This code is contributed by chitranayal`

## Java

 `// 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));``    ``}``}`

## Python3

 `# 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`

## C#

 `// 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`

## Javascript

 ``
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

 `// C++14 program to count the number``// of triplets from an array.``#include``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`

## Java

 `// 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 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));``    ``}``}`

## Python3

 `# Python3 program to count the number``# of triplets from an array.` `# Function to count the triplets``def` `countTriplets (arr, d, n):``    ` `    ``count ``=` `-``1``    ` `    ``# Create a map to store (key, values) pair.``    ``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` `i ``in` `range` `(``0``, n):``        ``if` `((arr[i] ``-` `d) ``in` `set``.keys() ``and``            ``(arr[i] ``-` `2` `*` `d) ``in` `set``.keys()):``            ``count ``+``=` `(``set``[arr[i] ``-` `d] ``*``                      ``set``[arr[i] ``-` `2` `*` `d])``                      ` `        ``# Update the map``        ``if` `(arr[i]) ``in` `set``.keys():``            ``set``[arr[i]] ``+``=` `1``        ``else``:``            ``set``[arr[i]] ``=` `1``            ` `    ``print``(count)``    ` `# Driver Code ` `# Test Case 1:``a1 ``=` `[ ``1``, ``6``, ``7``, ``7``, ``8``, ``10``, ``12``, ``13``, ``14` `]``d1 ``=` `3``countTriplets(a1, d1, ``9``)` `# Test Case 2:``a2 ``=` `[ ``6``, ``3``, ``4``, ``5` `]``d2 ``=` `1``countTriplets(a2, d2, ``4``)` `# Test Case 3:``a3 ``=` `[ ``1``, ``2``, ``4``, ``5``, ``7``, ``8``, ``10` `]``d3 ``=` `3``countTriplets(a3, d3, ``7``)` `# This code is contributed by Stream_Cipher`

## C#

 `// 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`

## Javascript

 ``
Output:
```1
0
2```

Time Complexity: O(NlogN)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up