# Find the number of elements X such that X + K also exists in the array

Given an array a[] and an integer k, find the number of elements x in this array such that the sum of x and k is also present in the array.

Examples:

```Input:  { 3, 6, 2, 8, 7, 6, 5, 9 } and k = 2
Output: 5
Explanation:
Elements {3, 6, 7, 6, 5} in this array have x + 2 value that is
{5, 8, 9, 8, 7} present in this array.

Input:  { 1, 2, 3, 4, 5} and k = 1
Output: 4
Explanation:
Elements {1, 2, 3, 4} in this array have x + 2 value that is
{2, 3, 4, 5} present in this array.
```

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

The problem is similar to finding the count of pairs with a given sum in an Array.

Naive Approach:

The solve the problem mentioned above we can use the brute force approach and find the result. Iterate two loops, check whether x+2 exists in an array or not for each x in this array. If it exists then increment the count and finally return the count.

Efficient Approach:

The efficient approach to solve the problem is to use a HashMap, the key in map will act as the unique element in this array and the corresponding value will tell us the frequency of this element.

Iterate over this map and find for each key x in this map whether the key x + k exists in the map or not, if exist then add this frequency to the answer i.e for this element x we have x+k in this array. Finally, return the answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of find number ` `// of elements x in this array ` `// such x+k also present in this array. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the ` `// count of element x such that ` `// x+k also lies in this array ` `int` `count_element(``int` `N, ``int` `K, ``int``* arr) ` `{ ` `    ``// Key in map will store elements ` `    ``// and value will store the ` `    ``// frequency of the elements ` `    ``map<``int``, ``int``> mp; ` ` `  `    ``for` `(``int` `i = 0; i < N; ++i) ` `        ``mp[arr[i]]++; ` ` `  `    ``int` `answer = 0; ` ` `  `    ``for` `(``auto` `i : mp) { ` ` `  `        ``// Find if i.first + K is ` `        ``// present in this map or not ` `        ``if` `(mp.find(i.first + K) != mp.end()) ` ` `  `            ``// If we find i.first or key + K in this map ` `            ``// then we have to increase in answer ` `            ``// the frequency of this element ` `            ``answer += i.second; ` `    ``} ` ` `  `    ``return` `answer; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// array initialisation ` `    ``int` `arr[] = { 3, 6, 2, 8, 7, 6, 5, 9 }; ` ` `  `    ``// size of array ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// initialise k ` `    ``int` `K = 2; ` ` `  `    ``cout << count_element(N, K, arr); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of find number ` `// of elements x in this array ` `// such x+k also present in this array. ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to return the ` `// count of element x such that ` `// x+k also lies in this array ` `static` `int` `count_element(``int` `N, ``int` `K, ``int``[] arr) ` `{ ` `    ``// Key in map will store elements ` `    ``// and value will store the ` `    ``// frequency of the elements ` `    ``HashMap mp = ``new` `HashMap(); ` `  `  `    ``for` `(``int` `i = ``0``; i < N; ++i) ` `        ``if``(mp.containsKey(arr[i])){ ` `            ``mp.put(arr[i], mp.get(arr[i])+``1``); ` `        ``}``else``{ ` `            ``mp.put(arr[i], ``1``); ` `    ``} ` `  `  `    ``int` `answer = ``0``; ` `  `  `    ``for` `(Map.Entry i : mp.entrySet()) { ` `  `  `        ``// Find if i.first + K is ` `        ``// present in this map or not ` `        ``if` `(mp.containsKey(i.getKey() + K) ) ` `  `  `            ``// If we find i.first or key + K in this map ` `            ``// then we have to increase in answer ` `            ``// the frequency of this element ` `            ``answer += i.getValue(); ` `    ``} ` `  `  `    ``return` `answer; ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``// array initialisation ` `    ``int` `arr[] = { ``3``, ``6``, ``2``, ``8``, ``7``, ``6``, ``5``, ``9` `}; ` `  `  `    ``// size of array ` `    ``int` `N = arr.length; ` `  `  `    ``// initialise k ` `    ``int` `K = ``2``; ` `  `  `    ``System.out.print(count_element(N, K, arr)); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## Python3

 `# Python3 implementation of find number ` `# of elements x in this array ` `# such x+k also present in this array. ` ` `  `# Function to return the ` `# count of element x such that ` `# x+k also lies in this array ` `def` `count_element(N, K, arr): ` `     `  `    ``# Key in map will store elements ` `    ``# and value will store the ` `    ``# frequency of the elements ` `    ``mp ``=` `dict``() ` ` `  `    ``for` `i ``in` `range``(N): ` `        ``mp[arr[i]] ``=` `mp.get(arr[i], ``0``) ``+` `1` ` `  `    ``answer ``=` `0` ` `  `    ``for` `i ``in` `mp: ` ` `  `        ``# Find if i.first + K is ` `        ``# present in this map or not ` `        ``if` `i ``+` `K ``in` `mp: ` ` `  `            ``# If we find i.first or key + K in this map ` `            ``# then we have to increase in answer ` `            ``# the frequency of this element ` `            ``answer ``+``=` `mp[i] ` ` `  `    ``return` `answer ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``# array initialisation ` `    ``arr``=``[``3``, ``6``, ``2``, ``8``, ``7``, ``6``, ``5``, ``9``] ` ` `  `    ``# size of array ` `    ``N ``=` `len``(arr) ` ` `  `    ``# initialise k ` `    ``K ``=` `2` ` `  `    ``print``(count_element(N, K, arr)) ` `     `  `# This code is contributed by mohit kumar 29     `

## C#

 `// C# implementation of find number ` `// of elements x in this array ` `// such x+k also present in this array. ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `public` `class` `GFG{ ` `   `  `// Function to return the ` `// count of element x such that ` `// x+k also lies in this array ` `static` `int` `count_element(``int` `N, ``int` `K, ``int``[] arr) ` `{ ` `    ``// Key in map will store elements ` `    ``// and value will store the ` `    ``// frequency of the elements ` `    ``Dictionary<``int``,``int``> mp = ``new` `Dictionary<``int``,``int``>(); ` `   `  `    ``for` `(``int` `i = 0; i < N; ++i) ` `        ``if``(mp.ContainsKey(arr[i])){ ` `            ``mp[arr[i]] = mp[arr[i]]+1; ` `        ``}``else``{ ` `            ``mp.Add(arr[i], 1); ` `    ``} ` `   `  `    ``int` `answer = 0; ` `   `  `    ``foreach` `(KeyValuePair<``int``,``int``> i ``in` `mp) { ` `   `  `        ``// Find if i.first + K is ` `        ``// present in this map or not ` `        ``if` `(mp.ContainsKey(i.Key + K) ) ` `   `  `            ``// If we find i.first or key + K in this map ` `            ``// then we have to increase in answer ` `            ``// the frequency of this element ` `            ``answer += i.Value; ` `    ``} ` `   `  `    ``return` `answer; ` `} ` `   `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``// array initialisation ` `    ``int` `[]arr = { 3, 6, 2, 8, 7, 6, 5, 9 }; ` `   `  `    ``// size of array ` `    ``int` `N = arr.Length; ` `   `  `    ``// initialise k ` `    ``int` `K = 2; ` `   `  `    ``Console.Write(count_element(N, K, arr)); ` `} ` `} ` `// This code contributed by Princi Singh `

Output:

```5
```

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