Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 26 May, 2021

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.

 

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 <bits/stdc++.h>
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[0]);
 
    // 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<Integer,Integer> mp = new HashMap<Integer,Integer>();
  
    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<Integer,Integer> 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

Javascript




<script>
// Javascript 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
function count_element(N, K, arr)
{
 
    // Key in map will store elements
    // and value will store the
    // frequency of the elements
    let mp = new Map();
 
    for (let i = 0; i < N; ++i)
        if (mp.has(arr[i])) {
            mp.set(arr[i], mp.get(arr[i]) + 1)
        } else {
            mp.set(arr[i], 1)
        }
 
    let answer = 0;
 
    for (let i of mp) {
 
        // Find if i.first + K is
        // present in this map or not
        if (mp.has(i[0] + 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[1];
    }
 
    return answer;
}
 
// Driver code
 
// array initialisation
let arr = [3, 6, 2, 8, 7, 6, 5, 9];
 
// size of array
let N = arr.length;
 
// initialise k
let K = 2;
 
document.write(count_element(N, K, arr));
 
// This code is contributed by gfgking
</script>
Output: 
5

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :