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}

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:

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// C++ implementation to count of
// elements such that its sum/difference
// with X also exists in the Array
  
#include <bits/stdc++.h>
  
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;
}
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// 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<Integer> s = new HashSet<Integer>();
  
    // 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
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# 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
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// 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
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Output:
3

Performance Analysis:

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