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Check if the array has an element which is equal to product of remaining elements

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  • Difficulty Level : Easy
  • Last Updated : 11 Jul, 2022

Given an array of N elements, the task is to check if the array has an element which is equal to the product of all the remaining elements.

Examples: 

Input: arr[] = {1, 2, 12, 3, 2}
Output: YES
12 is the product of all the remaining elements 
i.e. 1 * 2 * 3 * 2 = 12

Input: arr[] = {1, 2, 3}
Output: NO

Method-1:  

  1. First, take the product of all the element of the array.
  2. Now traverse the whole array again.
  3. For any element a[i] check if it is equal to the product of all elements divided by that element.
  4. Print Yes if at least one such element is found.

Below is the implementation of the above approach:  

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to Check if the array
// has an element which is equal to
// product of all the remaining elements
bool CheckArray(int arr[], int n)
{
    int prod = 1;
 
    // Calculate the product of all the elements
    for (int i = 0; i < n; ++i)
        prod *= arr[i];
 
    // Return true if any such element is found
    for (int i = 0; i < n; ++i)
        if (arr[i] == prod / arr[i])
            return true;
 
    // If no element is found
    return false;
}
 
int main()
{
    int arr[] = { 1, 2, 12, 3, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    if (CheckArray(arr, n))
        cout << "YES";
 
    else
        cout << "NO";
 
    return 0;
}

Java



// Java implementation of the above approach
 
import java.io.*;
 
class GFG {
 
 
// Function to Check if the array
// has an element which is equal to
// product of all the remaining elements
static boolean CheckArray(int arr[], int n)
{
    int prod = 1;
 
    // Calculate the product of all the elements
    for (int i = 0; i < n; ++i)
        prod *= arr[i];
 
    // Return true if any such element is found
    for (int i = 0; i < n; ++i)
        if (arr[i] == prod / arr[i])
            return true;
 
    // If no element is found
    return false;
}
 
 
 
    public static void main (String[] args) {
            int arr[] = { 1, 2, 12, 3, 2 };
    int n =arr.length;
 
    if (CheckArray(arr, n))
        System.out.println("YES");
 
    else
        System.out.println("NO");
 
    }
}
// This code is contributed by shs..
Python3



# Python 3 implementation of the above approach
 
# Function to Check if the array
# has an element which is equal to
# product of all the remaining elements
def CheckArray(arr, n):
    prod = 1
 
    # Calculate the product of all
    # the elements
    for i in range(0, n, 1):
        prod *= arr[i]
 
    # Return true if any such element
    # is found
    for i in range(0, n, 1):
        if (arr[i] == prod / arr[i]):
            return True
 
    # If no element is found
    return False
 
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 12, 3, 2]
    n = len(arr)
 
    if (CheckArray(arr, n)):
        print("YES")
 
    else:
        print("NO")
 
# This code is contributed by
# Surendra_Gangwar
C#



// C# implementation of the above approach
class GFG
{
// Function to Check if the array
// has an element which is equal to
// product of all the remaining elements
static bool CheckArray(int[] arr, int n)
{
    int prod = 1;
 
    // Calculate the product of
    // all the elements
    for (int i = 0; i < n; ++i)
        prod *= arr[i];
 
    // Return true if any such
    // element is found
    for (int i = 0; i < n; ++i)
        if (arr[i] == prod / arr[i])
            return true;
 
    // If no element is found
    return false;
}
 
// Driver Code
public static void Main ()
{
    int[] arr = new int[] { 1, 2, 12, 3, 2 };
    int n = arr.Length;
     
    if (CheckArray(arr, n))
        System.Console.WriteLine("YES");
     
    else
        System.Console.WriteLine("NO");
}
}
 
// This code is contributed by mits
PHP



<?php
// PHP implementation of the above approach
 
// Function to Check if the array
// has an element which is equal to
// product of all the remaining elements
function CheckArray($arr, $n)
{
    $prod = 1;
 
    // Calculate the product of
    // all the elements
    for ($i = 0; $i < $n; ++$i)
        $prod *= $arr[$i];
 
    // Return true if any such element
    // is found
    for ($i = 0; $i < $n; ++$i)
        if ($arr[$i] == $prod / $arr[$i])
            return true;
 
    // If no element is found
    return false;
}
 
// Driver Code
$arr = array(1, 2, 12, 3, 2);
$n = sizeof($arr);
 
if (CheckArray($arr, $n))
    echo "YES";
else
    echo "NO";
 
// This code is contributed
// by Akanksha Rai
Javascript



<script>
// Java script implementation of the above approach
 
// Function to Check if the array
// has an element which is equal to
// product of all the remaining elements
function CheckArray(arr,n)
{
    let prod = 1;
 
    // Calculate the product of all the elements
    for (let i = 0; i < n; ++i)
        prod *= arr[i];
 
    // Return true if any such element is found
    for (let i = 0; i < n; ++i)
        if (arr[i] == prod / arr[i])
            return true;
 
    // If no element is found
    return false;
}
 
 
 
    let arr = [ 1, 2, 12, 3, 2 ];
    let n =arr.length;
 
    if (CheckArray(arr, n))
        document.write("YES");
 
    else
        document.write("NO");
 
// This code is contributed by sravan kumar Gottumukkala
</script>
Output: 
YES
 
Time Complexity: O(n)
Auxiliary Space: O(1)
Method-2: The approach is to find the product of all the elements of the array and check if it is a perfect square or not. If it is a perfect square then check if the square root of the product exists in the array or not. If exists then print Yes else print No.
According to the problem statement, a * b = N 
where b is the product of all the remaining elements of the array except a i.e arr[i] 
And it is also mentioned that find the index such that a = b. 
So, it simply means that a*a = N i.e. N is a perfect square. and a is its square root.
Below is the implementation of the above approach:  
C++



// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to Check if the array
// has an element which is equal to
// product of all the remaining elements
bool CheckArray(int arr[], int n)
{
    int prod = 1;
 
    // Storing frequency in map
    unordered_set<int> freq;
 
    // Calculate the product of all the elements
    for (int i = 0; i < n; ++i) {
        freq.insert(arr[i]);
        prod *= arr[i];
    }
 
    int root = sqrt(prod);
 
    // If the prod is a perfect square
    if (root * root == prod)
 
        // then check if its square root
        // exist in the array or not
        if (freq.find(root) != freq.end())
            return true;
 
    return false;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 12, 3, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    if (CheckArray(arr, n))
        cout << "YES";
 
    else
        cout << "NO";
 
    return 0;
}
Java



import java.util.ArrayList;
 
// Java implementation of the above approach
class GFG {
 
// Function to Check if the array
// has an element which is equal to
// product of all the remaining elements
    static boolean CheckArray(int arr[], int n) {
        int prod = 1;
 
        // Storing frequency in map
        ArrayList<Integer> freq = new ArrayList<>();
 
        // Calculate the product of all the elements
        for (int i = 0; i < n; ++i) {
            freq.add(arr[i]);
            prod *= arr[i];
        }
 
        int root = (int) Math.sqrt(prod);
 
        // If the prod is a perfect square
        if (root * root == prod) // then check if its square root
        // exist in the array or not
        {
            if (freq.contains(root) & freq.lastIndexOf(root) != (freq.size())) {
                return true;
            }
        }
 
        return false;
    }
// Driver code
 
    public static void main(String[] args) {
 
        int arr[] = {1, 2, 12, 3, 2};
        int n = arr.length;
 
        if (CheckArray(arr, n)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
 
    }
}
//This code is contributed by 29AjayKumar
Python3



# Python3 implementation of the above approach
  
import math
 
# Function to Check if the array
# has an element which is equal to
# product of all the remaining elements
def CheckArray( arr, n):
 
    prod = 1
  
    # Storing frequency in map
    freq = []
  
    # Calculate the product of all the elements
    for i in range(n) :
        freq.append(arr[i])
        prod *= arr[i]
  
    root = math.sqrt(prod)
  
    # If the prod is a perfect square
    if (root * root == prod):
  
        # then check if its square root
        # exist in the array or not
        if root in freq:
            return True
  
    return False
  
# Driver code
if __name__ == "__main__":
 
    arr = [1, 2, 12, 3, 2 ]
    n = len(arr)
  
    if (CheckArray(arr, n)):
        print ("YES")
  
    else:
        print ("NO")
C#



// C# implementation of above approach
using System;
using System.Collections;
 
class GFG
{
 
    // Function to Check if the array
    // has an element which is equal to
    // product of all the remaining elements
    static bool CheckArray(int []arr, int n)
    {
        int prod = 1;
 
        // Storing frequency in map
        ArrayList freq = new ArrayList();
 
        // Calculate the product of all the elements
        for (int i = 0; i < n; ++i)
        {
            freq.Add(arr[i]);
            prod *= arr[i];
        }
 
        int root = (int) Math.Sqrt(prod);
 
        // If the prod is a perfect square
        if (root * root == prod) // then check if its square root
        // exist in the array or not
        {
            if (freq.Contains(root) & freq.LastIndexOf(root) != (freq.Count))
            {
                return true;
            }
        }
 
        return false;
    }
     
    // Driver code
    public static void Main()
    {
 
        int []arr = {1, 2, 12, 3, 2};
        int n = arr.Length;
 
        if (CheckArray(arr, n))
        {
            Console.WriteLine("YES");
        }
        else
        {
            Console.WriteLine("NO");
        }
    }
}
 
/* This code contributed by PrinciRaj1992 */
PHP



<?php
// PHP implementation of the above approach
 
// Function to Check if the array
// has an element which is equal to
// product of all the remaining elements
function CheckArray($arr, $n)
{
    $prod = 1;
 
    // Storing frequency in map
    $freq = array();
 
    // Calculate the product of all the elements
    for ($i = 0; $i < $n; ++$i)
    {
        array_push($freq, $arr[$i]);
        $prod *= $arr[$i];
    }
    $freq = array_unique($freq);
    $root = (int)(sqrt($prod));
 
    // If the prod is a perfect square
    if ($root * $root == $prod)
 
        // then check if its square root
        // exist in the array or not
        if (in_array($root, $freq))
            return true;
 
    return false;
}
 
// Driver code
$arr = array( 1, 2, 12, 3, 2 );
$n = count($arr);
 
if (CheckArray($arr, $n))
    echo "YES";
 
else
    echo "NO";
 
// This code is contributed by mits
?>
Javascript



<script>
 
// JavaScript implementation of
// the above approach
 
    // Function to Check if the array
// has an element which is equal to
// product of all the remaining elements
    function CheckArray(arr,n)
    {
        let prod = 1;
  
        // Storing frequency in map
        let freq = [];
  
        // Calculate the product of all the elements
        for (let i = 0; i < n; ++i) {
            freq.push(arr[i]);
            prod *= arr[i];
        }
  
        let root =  Math.floor(Math.sqrt(prod));
  
        // If the prod is a perfect square
        // then check if its square root
        if (root * root == prod)
        // exist in the array or not
        {
            if (freq.includes(root) &
            freq.lastIndexOf(root) != (freq.length))
            {
                return true;
            }
        }
  
        return false;
    }
     
    // Driver code
    let arr=[1, 2, 12, 3, 2];
    let n = arr.length;
     
    if (CheckArray(arr, n)) {
        document.write("YES");
    } else {
        document.write("NO");
    }
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>
Output: 
YES
 
Time Complexity: O(n)
Auxiliary Space: O(n)

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Article Contributed By :
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imdhruvgupta
@imdhruvgupta
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