Count elements in an Array that can be represented as difference of two perfect squares

Given an array arr[], the task is to count the number of elements in the array that can be represented as in the form of the difference of two perfect square numbers. a^2 - b^2      
Examples: 
 

Input: arr[] = {1, 2, 3} 
Output: 2 
Explanation: 
There are two such elements that can be represented as 
difference of square of two numbers - 
Element 1 - 1^2 - 0^2 = 1      
Element 3 - 2^2 - 1^2 = 3      
Therefore, Count of such elements is 2.
Input: arr[] = {2, 5, 6} 
Output: 1 
Explanation: 
There is only one such element. That is - 
Element 5 - 3^2 - 2^2 = 9 - 4 = 5      
Therefore, Count of such elements is 1. 
 

Approach: The key observation in the problem is numbers which can be represented as the difference of the squares of two numbers never yield 2 as the remainder when divided by 4. 
For Example: 
 

N = 4 => 4^2 - 0^2      
N = 6 => Can't be represented as 6 \% 4 = 2      
N = 8 => 3^2 - 1^2      
N = 10 => Can't be represented as 10 \% 4 = 2      
 

Therefore, iterate over the array and count the number of such elements in the array.
Below is the implementation of the above approach:

// C++ implementation to count the
// number of elements which can be
// represented as the difference
// of the two square

#include <bits/stdc++.h>

using namespace std;

// Function to count of such elements
// in the array which can be represented
// as the difference of the two squares
int count_num(int arr[], int n)
{
    // Initialize count
    int count = 0;

    // Loop to iterate
    // over the array
    for (int i = 0; i < n; i++)

        // Condition to check if the
        // number can be represented
        // as the difference of squares
        if ((arr[i] % 4) != 2)
            count++;

    cout << count;
    return 0;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    count_num(arr, n);
    return 0;
}

// Java implementation to count the 
// number of elements which can be 
// represented as the difference 
// of the two square 
class GFG{

// Function to count of such elements 
// in the array which can be represented 
// as the difference of the two squares 
static void count_num(int []arr, int n) 
{ 
    
    // Initialize count 
    int count = 0; 
    
    // Loop to iterate 
    // over the array 
    for(int i = 0; i < n; i++) 
    {
       
       // Condition to check if the 
       // number can be represented 
       // as the difference of squares 
       if ((arr[i] % 4) != 2) 
           count++; 
    }
    System.out.println(count); 
} 
    
// Driver code 
public static void main (String[] args)
{ 
    int arr[] = { 1, 2, 3 }; 
    int n = arr.length; 
    
    count_num(arr, n); 
} 
}

// This code is contributed by AnkitRai01

# Python3 implementation to count the
# number of elements in the array
# which can be represented as difference 
# of the two elements 

# Function to return the
# Count of required count 
# of such elements
def count_num(arr, n):
    # Initialize count 
    count = 0
    
    # Loop to iterate over the 
    # array of elements
    for i in arr:
        
        # Condition to check if the 
        # number can be represented
        # as the difference 
        # of two squares
        if ((i % 4) != 2):
            count = count + 1
    
    return count

# Driver Code
if __name__ == "__main__":
    arr = [1, 2, 3]
    n = len(arr)
    
    # Function Call
    print(count_num(arr, n))

// C# implementation to count the 
// number of elements which can be 
// represented as the difference 
// of the two square 
using System;
class GFG{

// Function to count of such elements 
// in the array which can be represented 
// as the difference of the two squares 
static void count_num(int []arr, int n) 
{ 
    
    // Initialize count 
    int count = 0; 
    
    // Loop to iterate 
    // over the array 
    for(int i = 0; i < n; i++) 
    {
        
        // Condition to check if the 
        // number can be represented 
        // as the difference of squares 
        if ((arr[i] % 4) != 2) 
            count++; 
    }
    Console.WriteLine(count); 
} 
    
// Driver code 
public static void Main(string[] args)
{ 
    int []arr = { 1, 2, 3 }; 
    int n = arr.Length; 
    
    count_num(arr, n); 
} 
}

// This code is contributed by shivanisinghss2110

<script>

    // Javascript implementation to count the 
    // number of elements which can be 
    // represented as the difference 
    // of the two square 
    
    // Function to count of such elements 
    // in the array which can be represented 
    // as the difference of the two squares 
    function count_num(arr, n) 
    { 
        // Initialize count 
        let count = 0; 

        // Loop to iterate 
        // over the array 
        for (let i = 0; i < n; i++) 

            // Condition to check if the 
            // number can be represented 
            // as the difference of squares 
            if ((arr[i] % 4) != 2) 
                count++; 

        document.write(count); 
        return 0; 
    } 

    let arr = [ 1, 2, 3 ]; 
    let n = arr.length; 
    count_num(arr, n); 

</script>

2

Time complexity: O(n) where n is the number of elements in the given array
Auxiliary space: O(1)