Given an array arr[] and an integer X, the task is to count the elements of the array such that their exist a element
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 = 4Input: 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
Below is the implemenation of the above approach:
// 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;
} |
// 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 |
# 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# 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 |
3
Performance Analysis:
- Time Complexity: O(N)
- Auxiliary Space Complexity: O(N)
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