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

• Last Updated : 06 Jun, 2021

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:
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:
Explanation:
In the above-given example, there are 3 such numbers {2, 2, 5}

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 implementation 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 arrayvoid findAns(int arr[], int n, int x){    int ans;    unordered_set 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 increase the count variable        if (s.find(arr[i] + x) != s.end() || s.find(arr[i] - x) != s.end())            ans++;    }    cout << ans;    return;} // Driver Codeint 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 Arrayimport 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 arraystatic 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 increase the count variable        if (s.contains(arr[i] + x) ||            s.contains(arr[i] - x))            ans++;    }    System.out.print(ans);    return;} // Driver Codepublic 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 arraydef 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 Codearr = [ 2, 2, 4, 5, 6 ]n = len(arr)x = 3 # Function callfindAns(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 Arrayusing 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 arraystatic 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 increase the count variable       if (s.Contains(arr[i] + x) ||           s.Contains(arr[i] - x))           ans++;    }    Console.Write(ans);    return;}     // Driver Codepublic 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

Javascript


Output:
3

Performance Analysis:

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

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