# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

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.

## 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)); ` `    ``} ` `} `

## 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 `

Output:

```1
0
2
```

Time Complexity: O(N)

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