Given an array arr[], the task is to find the count of array elements whose squares are already present in the array.
Examples:
Input: arr[] = {2, 4, 5, 20, 16}
Output: 2
Explanation:
{2, 4} has their squares {4, 16} present in the array.Input: arr[] = {1, 30, 3, 8, 64}
Output: 2
Explanation:
{1, 8} has their squares {1, 64} present in the array.
Naive Approach: Follow the steps below to solve the problem:
- Initialize a variable, say, count, to store the required count.
-
Traverse the array and for each and every array element, perform the following operations:
- Traverse the array and search if the square of the current element is present in the array.
- If the square found increment the count.
- Print count as the answer.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the count of elements whose // squares are already present int the array void countSquares( int arr[], int N)
{ // Stores the required count
int count = 0;
// Traverse the array
for ( int i = 0; i < N; i++) {
// Square of the element
int square = arr[i] * arr[i];
// Traverse the array
for ( int j = 0; j < N; j++) {
// Check whether the value
// is equal to square
if (arr[j] == square) {
// Increment count
count = count + 1;
}
}
}
// Print the count
cout << count;
} // Driver Code int main()
{ // Given array
int arr[] = { 2, 4, 5, 20, 16 };
// Size of the array
int N = sizeof (arr) / sizeof (arr[0]);
countSquares(arr, N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to find the count of elements whose // squares are already present int the array static void countSquares( int arr[], int N)
{ // Stores the required count
int count = 0 ;
// Traverse the array
for ( int i = 0 ; i < N; i++)
{
// Square of the element
int square = arr[i] * arr[i];
// Traverse the array
for ( int j = 0 ; j < N; j++)
{
// Check whether the value
// is equal to square
if (arr[j] == square)
{
// Increment count
count = count + 1 ;
}
}
}
// Print the count
System.out.print(count);
} // Driver Code public static void main(String[] args)
{ // Given array
int arr[] = { 2 , 4 , 5 , 20 , 16 };
// Size of the array
int N = arr.length;
countSquares(arr, N);
} } // This code is contributed by shikhasingrajput |
# Python program for the above approach # Function to find the count of elements whose # squares are already present the array def countSquares(arr, N):
# Stores the required count
count = 0 ;
# Traverse the array
for i in range (N):
# Square of the element
square = arr[i] * arr[i];
# Traverse the array
for j in range (N):
# Check whether the value
# is equal to square
if (arr[j] = = square):
# Increment count
count = count + 1 ;
# Print count
print (count);
# Driver Code if __name__ = = '__main__' :
# Given array
arr = [ 2 , 4 , 5 , 20 , 16 ];
# Size of the array
N = len (arr);
countSquares(arr, N);
# This code is contributed by shikhasingrajput |
// C# program of the above approach using System;
class GFG{
// Function to find the count of elements whose // squares are already present int the array static void countSquares( int [] arr, int N)
{ // Stores the required count
int count = 0;
// Traverse the array
for ( int i = 0; i < N; i++)
{
// Square of the element
int square = arr[i] * arr[i];
// Traverse the array
for ( int j = 0; j < N; j++)
{
// Check whether the value
// is equal to square
if (arr[j] == square)
{
// Increment count
count = count + 1;
}
}
}
// Print the count
Console.WriteLine(count);
} // Driver code static void Main()
{ // Given array
int [] arr = { 2, 4, 5, 20, 16 };
// Size of the array
int N = arr.Length;
countSquares(arr, N);
} } // This code is contributed by divyeshrabadiya07 |
<script> // Javascript program for the above approach // Function to find the count of elements whose // squares are already present int the array function countSquares(arr, N)
{ // Stores the required count
var count = 0;
// Traverse the array
for ( var i = 0; i < N; i++) {
// Square of the element
var square = arr[i] * arr[i];
// Traverse the array
for ( var j = 0; j < N; j++) {
// Check whether the value
// is equal to square
if (arr[j] == square) {
// Increment count
count = count + 1;
}
}
}
// Print the count
document.write( count);
} // Driver Code // Given array var arr = [2, 4, 5, 20, 16];
// Size of the array var N = arr.length;
countSquares(arr, N); </script> |
2
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The optimal idea is to use unordered_map to keep the count of visited elements and update the variable count accordingly. Below are the steps:
- Initialize a Map to store the frequency of array elements and initialize a variable, say, count.
- Traverse the array and for each element, increment its count in the Map.
- Again traverse the array and for each element check for the frequency of the square of the element in the map and add it to the variable count.
- Print count as the number of elements whose square is already present in the array.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the count of elements whose // squares is already present int the array int countSquares( int arr[], int N)
{ // Stores the count of array elements
int count = 0;
// Stores frequency of visited elements
unordered_map< int , int > m;
// Traverse the array
for ( int i = 0; i < N; i++) {
m[arr[i]] = m[arr[i]] + 1;
}
for ( int i = 0; i < N; i++) {
// Square of the element
int square = arr[i] * arr[i];
// Update the count
count += m[square];
}
// Print the count
cout << count;
} // Driver Code int main()
{ // Given array
int arr[] = { 2, 4, 5, 20, 16 };
// Size of the array
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
countSquares(arr, N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG
{ // Function to find the count of elements whose // squares is already present int the array static void countSquares( int arr[], int N)
{ // Stores the count of array elements
int count = 0 ;
// Stores frequency of visited elements
HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
// Traverse the array
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 );
}
}
for ( int i = 0 ; i < N; i++)
{
// Square of the element
int square = arr[i] * arr[i];
// Update the count
count += mp.containsKey(square)?mp.get(square): 0 ;
}
// Print the count
System.out.print(count);
} // Driver Code public static void main(String[] args)
{ // Given array
int arr[] = { 2 , 4 , 5 , 20 , 16 };
// Size of the array
int N = arr.length;
// Function Call
countSquares(arr, N);
} } // This code is contributed by 29AjayKumar |
# Python 3 program for the above approach from collections import defaultdict
# Function to find the count of elements whose # squares is already present int the array def countSquares( arr, N):
# Stores the count of array elements
count = 0 ;
# Stores frequency of visited elements
m = defaultdict( int );
# Traverse the array
for i in range (N):
m[arr[i]] = m[arr[i]] + 1
for i in range ( N ):
# Square of the element
square = arr[i] * arr[i]
# Update the count
count + = m[square]
# Print the count
print (count)
# Driver Code if __name__ = = "__main__" :
# Given array
arr = [ 2 , 4 , 5 , 20 , 16 ]
# Size of the array
N = len (arr)
# Function Call
countSquares(arr, N);
# This code is contributed by chitranayal. |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG
{ // Function to find the count of elements whose // squares is already present int the array static void countSquares( int []arr, int N)
{ // Stores the count of array elements
int count = 0;
// Stores frequency of visited elements
Dictionary< int , int > mp =
new Dictionary< int , int >();
// Traverse the array
for ( int i = 0; i < N; i++)
{
if (mp.ContainsKey(arr[i]))
{
mp.Add(arr[i], mp[arr[i]] + 1);
}
else
{
mp.Add(arr[i], 1);
}
}
for ( int i = 0; i < N; i++)
{
// Square of the element
int square = arr[i] * arr[i];
// Update the count
count += mp.ContainsKey(square)?mp[square]:0;
}
// Print the count
Console.Write(count);
} // Driver Code public static void Main()
{ // Given array
int []arr = { 2, 4, 5, 20, 16 };
// Size of the array
int N = arr.Length;
// Function Call
countSquares(arr, N);
} } // This code is contributed by Samim Hossain Mondal |
<script> // Javascript program for the above approach // Function to find the count of elements whose // squares is already present int the array function countSquares(arr, N)
{ // Stores the count of array elements
var count = 0;
// Stores frequency of visited elements
var m = new Map();
// Traverse the array
for ( var i = 0; i < N; i++) {
if (m.has(arr[i]))
m.set(arr[i], m.get(arr[i])+1)
else m.set(arr[i], 1);
}
for ( var i = 0; i < N; i++) {
// Square of the element
var square = arr[i] * arr[i];
// Update the count
if (m.has(square))
count += m.get(square);
}
// Print the count
document.write( count);
} // Driver Code // Given array var arr = [2, 4, 5, 20, 16];
// Size of the array var N = arr.length;
// Function Call countSquares(arr, N); </script> |
2
Time Complexity: O(N)
Auxiliary Space: O(N)