Check whether the product of every subsequence is a perfect square or not

Given an array arr[] consisting of N positive integers, the task is to check if the product of elements of every subsequence of the given array arr[] is a perfect square or not. If found to be true, then print Yes. Otherwise, print No.

Examples:

Input: arr[] = {1, 4, 100}
Output: Yes
Explanation:
Following are the subsequences of the given array arr[]:

1. {1}, the product is equal to 1, and is a perfect square.
2. {1, 4}, the product is equal to 4, and is a perfect square.
3. {1, 100}, the product is equal to 100 and is a perfect square.
4. {1, 4, 100}, the product is equal to 400 and is a perfect square.
5. {4}, the product is equal to 4 and is a perfect square.
6. {4, 100}, the product is equal to 400 and is a perfect square.
7. {100}, the product is equal to 100 and is a perfect square.

Therefore, print “Yes”.

Input: arr[] = {1, 3}
Output: No  

Naive Approach: The simplest approach to solve the given problem is to generate all possible subsequences of the given array and if the product elements of every subsequence is a perfect square, then print Yes. Otherwise, print No.

Time Complexity: O(N*2^N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized using the fact that the product of two perfect square numbers will also be a perfect square. Therefore, to check if the individual product of elements of all subsequences is a perfect square or not, the idea is to check if all the array elements are perfect squares or not. If found to be true, then print Yes. Otherwise, print No.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if the product of
// every subsequence of the array is a
// perfect square or not
string perfectSquare(int arr[], int N)
{
    // Traverse the given array
    for (int i = 0; i < N; i++) {

        // If arr[i] is a perfect
        // square or not
        int p = sqrt(arr[i]);

        if (p * p != arr[i]) {
            return "No";
        }
    }

    // Return "Yes"
    return "Yes";
}

// Driver Code
int main()
{
    int arr[] = { 1, 4, 100 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << perfectSquare(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;

class GFG{

// Function to check if the product of
// every subsequence of the array is a
// perfect square or not
static String perfectSquare(int arr[], int N)
{

    // Traverse the given array
    for(int i = 0; i < N; i++)
    {

        // If arr[i] is a perfect
        // square or not
        int p = (int)Math.sqrt(arr[i]);

        if (p * p != arr[i])
        {
            return "No";
        }
    }

    // Return "Yes"
    return "Yes";
}

// Driver Code
public static void main (String[] args)
{
    int arr[] = { 1, 4, 100 };
    int N = arr.length;

    System.out.println(perfectSquare(arr, N));
}
}

// This code is contributed by Potta Lokesh

Python3

# Python 3 program for the above approach
from math import sqrt

# Function to check if the product of
# every subsequence of the array is a
# perfect square or not
def perfectSquare(arr, N):

    # Traverse the given array
    for i in range(N):

        # If arr[i] is a perfect
        # square or not
        p = sqrt(arr[i])

        if (p * p != arr[i]):
            return "No"

    # Return "Yes"
    return "Yes"

# Driver Code
if __name__ == '__main__':
    arr = [1, 4, 100]
    N = len(arr)
    print(perfectSquare(arr, N))

    # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if the product of
// every subsequence of the array is a
// perfect square or not
static String perfectSquare(int[] arr, int N)
{

    // Traverse the given array
    for(int i = 0; i < N; i++)
    {

        // If arr[i] is a perfect
        // square or not
        int p = (int)Math.Sqrt(arr[i]);

        if (p * p != arr[i])
        {
            return "No";
        }
    }

    // Return "Yes"
    return "Yes";
}

// Driver Code
public static void Main()
{
    int[] arr = { 1, 4, 100 };
    int N = arr.Length;

    Console.WriteLine(perfectSquare(arr, N));
}
}

// This code is contributed by subham348

Javascript

<script>

// Javascript program for the above approach

// Function to check if the product of
// every subsequence of the array is a
// perfect square or not
function perfectSquare(arr, N)
{

    // Traverse the given array
    for(let i = 0; i < N; i++)
    {

        // If arr[i] is a perfect
        // square or not
        let p = Math.sqrt(arr[i]);

        if (p * p != arr[i])
        {
            return "No";
        }
    }

    // Return "Yes"
    return "Yes";
}

// Driver Code
let arr = [ 1, 4, 100 ];
let N = arr.length;

document.write(perfectSquare(arr, N));

// This code is contributed by target_2

</script>

Output:

Yes

Time Complexity: O(N*log(A_i)) where N is the size of the array.
Auxiliary Space: O(1)