Count of elements such that its sum/difference with X also exists in the Array

Given an array arr[] and an integer X, the task is to count the elements of the array such that their exist a element or in the array.

Examples:

Input: arr[] = {3, 4, 2, 5}, X = 2
Output: 4
Explanation:
In the above-given example, there are 4 such numbers –
For Element 3: Possible numbers are 1, 5, whereas 5 is present in array
For Element 4: Possible numbers are 2, 6, whereas 2 is present in array
For Element 2: Possible numbers are 0, 4, whereas 4 is present in array
For Element 5: Possible numbers are 3, 7, whereas 3 is present in array
Therefore, Total count = 4

Input: arr[] = {2, 2, 4, 5, 6}, X = 3
Output: 3
Explanation:
In the above-given example, there are 3 such numbers {2, 2, 5}

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

Approach: The idea is to use hash-map to check that an element is present in the hash-map or not in O(1) time. Then, iterate over the elements of the array and for each element check that or is present in the array. If yes, then increment the count of such elements by 1.

Below is the implemenation of the above approach:

C++

 `// C++ implementation to count of ` `// elements such that its sum/difference ` `// with X also exists in the Array ` ` `  `#include ` ` `  `using` `namespace` `std; ` ` `  `// Function to find the count of ` `// elements in the array such that ` `// element at the difference at X ` `// is present in the array ` `void` `findAns(``int` `arr[], ``int` `n, ``int` `x) ` `{ ` `    ``int` `ans; ` `    ``unordered_set<``int``> s; ` ` `  `    ``// Loop to insert the elements ` `    ``// of the array into the set ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``s.insert(arr[i]); ` ` `  `    ``ans = 0; ` ` `  `    ``// Loop to iterate over the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// if any of the elements are there ` `        ``// then increse the count variable ` `        ``if` `(s.find(arr[i] + x) != s.end() || s.find(arr[i] - x) != s.end()) ` `            ``ans++; ` `    ``} ` `    ``cout << ans; ` `    ``return``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 2, 4, 5, 6 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` `    ``int` `x = 3; ` ` `  `    ``findAns(arr, n, x); ` ` `  `    ``return` `0; ` `} `

Java

 `// Java implementation to count of ` `// elements such that its sum/difference ` `// with X also exists in the Array ` `import` `java.util.*; ` `class` `GFG{ ` ` `  `// Function to find the count of ` `// elements in the array such that ` `// element at the difference at X ` `// is present in the array ` `static` `void` `findAns(``int` `arr[], ` `                    ``int` `n, ``int` `x) ` `{ ` `    ``int` `ans; ` `    ``HashSet s = ``new` `HashSet(); ` ` `  `    ``// Loop to insert the elements ` `    ``// of the array into the set ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``s.add(arr[i]); ` ` `  `    ``ans = ``0``; ` ` `  `    ``// Loop to iterate over the array ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// if any of the elements are there ` `        ``// then increse the count variable ` `        ``if` `(s.contains(arr[i] + x) || ` `            ``s.contains(arr[i] - x)) ` `            ``ans++; ` `    ``} ` `    ``System.out.print(ans); ` `    ``return``; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``2``, ``2``, ``4``, ``5``, ``6` `}; ` `    ``int` `n = arr.length; ` `    ``int` `x = ``3``; ` ` `  `    ``findAns(arr, n, x); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Python3

 `# Python3 implementation to count of  ` `# elements such that its sum/difference  ` `# with X also exists in the array  ` ` `  `# Function to find the count of  ` `# elements in the array such that  ` `# element at the difference at X  ` `# is present in the array  ` `def` `findAns(arr, n, x): ` `     `  `    ``s ``=` `set``() ` `     `  `    ``# Loop to insert the elements  ` `    ``# of the array into the set ` `    ``for` `i ``in` `range``(n): ` `        ``s.add(arr[i]) ` `         `  `    ``ans ``=` `0` ` `  `    ``# Loop to iterate over the array  ` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# If any of the elements are there  ` `        ``# then increase the count variable ` `        ``if` `arr[i] ``+` `x ``in` `s ``or` `arr[i] ``-` `x ``in` `s: ` `            ``ans ``=` `ans ``+` `1` ` `  `    ``print``(ans) ` ` `  `# Driver Code  ` `arr ``=` `[ ``2``, ``2``, ``4``, ``5``, ``6` `] ` `n ``=` `len``(arr) ` `x ``=` `3` ` `  `# Function call ` `findAns(arr, n, x)  ` ` `  `# This code is contributed by ishayadav181 `

C#

 `// C# implementation to count of ` `// elements such that its sum/difference ` `// with X also exists in the Array ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG{ ` ` `  `// Function to find the count of ` `// elements in the array such that ` `// element at the difference at X ` `// is present in the array ` `static` `void` `findAns(``int``[] arr, ` `                    ``int` `n, ``int` `x) ` `{ ` `    ``int` `ans; ` `    ``HashSet<``int``> s = ``new` `HashSet<``int``>(); ` `     `  `    ``// Loop to insert the elements ` `    ``// of the array into the set ` `    ``for``(``int` `i = 0; i < n; i++) ` `       ``s.Add(arr[i]); ` `     `  `    ``ans = 0; ` `     `  `    ``// Loop to iterate over the array ` `    ``for``(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        `  `       ``// if any of the elements are there ` `       ``// then increse the count variable ` `       ``if` `(s.Contains(arr[i] + x) || ` `           ``s.Contains(arr[i] - x)) ` `           ``ans++; ` `    ``} ` `    ``Console.Write(ans); ` `    ``return``; ` `} ` `     `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int``[] arr = { 2, 2, 4, 5, 6 }; ` `    ``int` `n = arr.Length; ` `    ``int` `x = 3; ` `     `  `    ``findAns(arr, n, x); ` `} ` `} ` ` `  `// This code is contributed by ShubhamCoder `

Output:

```3
```

Performance Analysis:

• Time Complexity: O(N)
• Auxiliary Space Complexity: O(N)

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