Check if product of array containing prime numbers is a perfect square

Given an array arr[] containing only prime numbers, the task is to check if the product of the array elements is a perfect square or not.
Examples:

Input: arr[] = {2, 2, 7, 7} 
Output: Yes 
2 * 2 * 7 * 7 = 196 = 14²
Input: arr[] = {3, 3, 3, 5, 5} 
Output: No 

Naive approach: Multiply all the elements of the array and check whether the product is a perfect square or not.
Efficient approach: Count the frequencies of all the elements of the array, if frequency of all the elements are even then the product will be a perfect square else print No.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true if
// the product of all the array elements
// is a perfect square
bool isPerfectSquare(int arr[], int n)
{
    unordered_map<int, int> umap;

    // Update the frequencies of
    // all the array elements
    for (int i = 0; i < n; i++)
        umap[arr[i]]++;

    unordered_map<int, int>::iterator itr;
    for (itr = umap.begin(); itr != umap.end(); itr++)

        // If frequency of some element
        // in the array is odd
        if ((itr->second) % 2 == 1)
            return false;

    return true;
}

// Driver code
int main()
{
    int arr[] = { 2, 2, 7, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);

    if (isPerfectSquare(arr, n))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{
    // Function that returns true if
    // the product of all the array elements
    // is a perfect square
    static boolean isPerfectSquare(int [] arr, int n)
    {

        HashMap<Integer, Integer> umap = new HashMap<>();

        // Update the frequencies of
        // all the array elements
        for (int i = 0; i < n; i++)
        {
            if(umap.containsKey(arr[i]))
                umap.put(arr[i], umap.get(arr[i]) + 1 );
            else
                umap.put(arr[i], 1);

        }

        Iterator<Map.Entry<Integer, Integer> >
                iterator = umap.entrySet().iterator();

        while(iterator.hasNext())
        {
            Map.Entry<Integer, Integer> entry = iterator.next();

            // If frequency of some element
            // in the array is odd
            if (entry.getValue() % 2 == 1)
                return false;
        }
        return true;
    }

    // Driver code
    public static void main (String[] args)
    {

        int arr [] = { 2, 2, 7, 7 };
        int n = arr.length;

        if (isPerfectSquare(arr, n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by ihritik

Python3

# Python3 implementation of the approach

# Function that returns true if
# the product of all the array elements
# is a perfect square
def isPerfectSquare(arr, n) :

    umap = dict.fromkeys(arr, n);

    # Update the frequencies of
    # all the array elements
    for key in arr :
        umap[key] += 1;

    for key in arr :

        # If frequency of some element
        # in the array is odd
        if (umap[key] % 2 == 1) :
            return False;

    return True;

# Driver code
if __name__ == "__main__" :

    arr = [ 2, 2, 7, 7 ];
    n = len(arr)

    if (isPerfectSquare(arr, n)) :
        print("Yes");
    else :
        print("No");

# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{
    // Function that returns true if
    // the product of all the array elements
    // is a perfect square
    static bool isPerfectSquare(int [] arr, int n)
    {

        Dictionary<int, int> umap = new Dictionary<int, int>();

        // Update the frequencies of
        // all the array elements
        for (int i = 0; i < n; i++)
        {
            if(umap.ContainsKey(arr[i]))
                umap[arr[i]]++;
            else
                umap[arr[i]] = 1;

        }

        Dictionary<int, int>.ValueCollection valueColl =
                                                umap.Values;
        foreach(int val in valueColl)
        {
            // If frequency of some element
            // in the array is odd
            if (val % 2 == 1)
                return false;
        }
        return true;
    }

    // Driver code
    public static void Main ()
    {

        int [] arr = { 2, 2, 7, 7 };
        int n = arr.Length;

        if (isPerfectSquare(arr, n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by ihritik

Javascript

<script>
// Javascript implementation of the approach

// Function that returns true if
// the product of all the array elements
// is a perfect square
function isPerfectSquare(arr, n)
{
    let umap = new Map();

    // Update the frequencies of
    // all the array elements
    for (let i = 0; i < n; i++){
        umap[arr[i]]++;
        if(umap.has(arr[i])){
            umap.set(arr[i], umap.get(arr[i]) + 1)
        }else{
            umap.set(arr[i], 1)
        }
    }

    for (let itr of umap)

        // If frequency of some element
        // in the array is odd
        if ((itr[1]) % 2 == 1)
            return false;

    return true;
}

// Driver code
    let arr = [ 2, 2, 7, 7 ];
    let n = arr.length;

    if (isPerfectSquare(arr, n))
        document.write("Yes");
    else
        document.write("No");

// This code is contributed by _saurabh_jaiswal
</script>

Output:

Yes

Time Complexity: O(n)

Auxiliary Space: O(n)