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Check if the Product of all Array elements is a Perfect Square or not

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  • Last Updated : 17 Dec, 2021

Given an array arr[] consisting of N positive integers, the task is to check if the product of all the elements 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: The product of all the numbers is 1 x 4 x 100 = 400 which is a perfect square. Therefore, print “Yes”.

Input: arr[] = {1, 3}
Output: No

 

Naive Approach: Find the product of all the elements of the array and try to find if this is a perfect square or not. But the problem with this approach is that the product can be so large that we cannot store it and hence this approach will fail.
Time Complexity: O(N) 
Auxiliary Space: O(1) 

Efficient Approach: This approach is based on mathematical observation. A number is a perfect square if all the prime factors of the number are raised to powers that are even. We will be using this concept to find if the product is a perfect square or not. Follow the steps mentioned below:

  • Create a frequency array to store the powers of prime factors.
  • Start traversing the array.
  • For each element arr[i] use Sieve of Eratosthenes to find the powers of the prime factors of arr[i] and add that in the frequency array.
  • After array is traversed, start traversing the frequency array.
  • If any element (except 1) has odd frequency then return false, otherwise, return true.

Below is the implementation of the above approach:

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the product
// of all elements is perfect square or not
bool isPerfectSquare(vector<int>& arr)
{
    // Map to store the power of prime factors
    map<int, int> freq;
 
    // Loop to implement the concept
    // of Sieve of Eratosthenes
    for (int x : arr) {
        for (int i = 2; i <= sqrt(x); i++) {
            while (x > 1 and x % i == 0) {
                freq[i]++;
                x /= i;
            }
        }
        if (x >= 2)
            freq[x]++;
    }
 
    // Loop to check if all the prime factors
    // have even power
    for (auto it = freq.begin();
         it != freq.end(); it++)
        if (it->second % 2)
            return false;
 
    return true;
}
 
// Driver code
int main()
{
    vector<int> arr = { 1, 4, 100 };
 
    isPerfectSquare(arr)
        ? cout << "YES\n"
        : cout << "NO\n";
    return 0;
}

Java




import java.util.HashMap;
 
class GFG {
 
    // Function to check if the product
    // of all elements is perfect square or not
    public static boolean isPerfectSquare(int[] arr)
    {
       
        // Map to store the power of prime factors
        HashMap<Integer, Integer> freq = new HashMap<Integer, Integer>();
 
        // Loop to implement the concept
        // of Sieve of Eratosthenes
        for (int x : arr) {
            for (int i = 2; i <= Math.sqrt(x); i++) {
                while (x > 1 && x % i == 0) {
                    if (freq.containsKey(i)) {
                        freq.put(i, freq.get(i) + 1);
                    } else {
                        freq.put(i, 1);
                    }
                    x /= i;
                }
            }
            if (x >= 2) {
                if (freq.containsKey(x)) {
                    freq.put(x, freq.get(x) + 1);
                } else {
                    freq.put(x, 1);
                }
            }
        }
 
        // Loop to check if all the prime factors
        // have even power
        for (int it : freq.values())
            if (it % 2 > 0)
                return false;
 
        return true;
    }
 
    // Driver code
    public static void main(String args[]) {
        int[] arr = { 1, 4, 100 };
 
        if (isPerfectSquare(arr) == true)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}
 
// This code is contributed by gfgking.

Python3




# Python Program to implement
# the above approach
import math
 
# Function to check if the product
# of all elements is perfect square or not
def isPerfectSquare(arr):
 
    # Map to store the power of prime factors
    freq = dict()
 
    # Loop to implement the concept
    # of Sieve of Eratosthenes
    for x in arr:
        for i in range(2, math.floor(math.sqrt(x)) + 1):
            while (x > 1 and x % i == 0):
                if (i in freq):
                    freq[i] += + 1
                else:
                    freq[i] = 1
 
                x = x // i
        if (x >= 2):
            freq[x] += 1
     
    # Loop to check if all the prime factors
    # have even power
    for value in freq.values():
        if (value % 2 == 1):
            return False
    return True
 
# Driver code
arr = [1, 4, 100]
print("YES") if isPerfectSquare(arr) else print("NO")
 
# This code is contributed by gfgking

C#




using System;
using System.Collections.Generic;
 
public class GFG {
 
    // Function to check if the product
    // of all elements is perfect square or not
    public static bool isPerfectSquare(int[] arr)
    {
       
        // Map to store the power of prime factors
        Dictionary<int, int> freq = new Dictionary<int, int>();
 
        // Loop to implement the concept
        // of Sieve of Eratosthenes
        int new_x = 0;
        foreach (int x in arr) {
            new_x = x;
            for (int i = 2; i <= Math.Sqrt(new_x); i++) {
                while (new_x > 1 && x % i == 0) {
                    if (freq.ContainsKey(i)) {
                        freq[i] = freq[i] + 1;
                    } else {
                        freq.Add(i, 1);
                    }
                    new_x = new_x/i;
                }
            }
            if (new_x >= 2) {
                if (freq.ContainsKey(new_x)) {
                    freq[new_x] = freq[new_x] + 1;
                } else {
                    freq.Add(new_x, 1);
                }
            }
        }
 
        // Loop to check if all the prime factors
        // have even power
        foreach (int it in freq.Values)
            if (it % 2 > 0)
                return false;
 
        return true;
    }
 
    // Driver code
    public static void Main(String []args) {
        int[] arr = { 1, 4, 100 };
 
        if (isPerfectSquare(arr) == true)
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
        // JavaScript Program to implement
        // the above approach
 
        // Function to check if the product
        // of all elements is perfect square or not
        function isPerfectSquare(arr)
        {
         
            // Map to store the power of prime factors
            let freq = new Map();
 
            // Loop to implement the concept
            // of Sieve of Eratosthenes
            for (let x of arr) {
                for (let i = 2; i <= Math.floor(Math.sqrt(x)); i++) {
                    while (x > 1 && x % i == 0) {
                        if (freq.has(i))
                            freq.set(i, freq.get(i) + 1);
                        else
                            freq.set(i, 1);
 
                        x = Math.floor(x / i);
                    }
                }
                if (x >= 2)
                    freq.set(x, freq.get(x) + 1)
            }
 
            // Loop to check if all the prime factors
            // have even power
            for (let value of freq.values()) {
                if (value % 2 == 1)
                    return false;
            }
 
            return true;
        }
 
        // Driver code
        let arr = [1, 4, 100];
 
        isPerfectSquare(arr)
            ? document.write("YES" + '<br>')
            : document.write("NO" + '<br>');
 
    // This code is contributed by Potta Lokesh
    </script>
Output
YES

Time Complexity: O(N * log X) where X is the largest element of the array
Auxiliary Space: O(N)


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