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.
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 –
Element 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 –
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 =>
N = 6 => Can’t be represented as
N = 8 =>
N = 10 => Can’t be represented as
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> |
Output:
2
Time complexity: O(n) where n is the number of elements in the given array
Auxiliary space: O(1)