Check if product of every pair exists in an array

Given a array of n integers, we need to check whether for every pair of numbers a[i] & a[j] there exists a a[k] such that a[k] = a[i]*a[j] where k can be equal to i or j too.

Examples :

Input : arr[] = {0. 1} 
Output : Yes
Here a[0]*a[1] is equal to a[0]

Input : arr[] = {5, 6}
Output : No

An array will satisfy the problem conditions if the array follows all the below mentioned conditions : 

• Condition 1 : The array must have number of elements other than 1, 0, -1 less than or equal to 1 because if it has more than 1 such elements there will be no element present in the array whose product will be equal to the largest of those two (or more) elements. Suppose that number of such elements be 2 and their values are 5, 6 so there is no element equal to 5*6 = 30 in the array.
• Condition 2 : If the array has an other number say x (other than 0, 1 and -1) and -1 is also present, then also answer is false. Because presence of -1 makes it required that both x and -x should be present, but this violates condition 1.
• Condition 3 : if there are more than one “-1” and no one in the array then also answer will be no because the product of two “-1” is equal to 1.

Below is the implementation of above conditions.

C++

// C++ program to find if
// product of every pair
// is present in array.
#include<bits/stdc++.h>
using namespace std;

// Returns true if product
// of every pair in arr[]
// is present in arr[]
bool checkArray(int arr[] , int n)
{
    // variable to store number
    //  of zeroes, ones, minus
    // one and other numbers.
    int zero = 0, one = 0,
        minusone = 0, other=0;
    for (int i = 0; i < n; i++)
    {
        // incrementing the
        // variable values
        if (arr[i] == 0)
            zero++;
        else if (arr[i] == 1)
            one++;
        else if (arr[i] == -1)
            minusone++;
        else
            other++;
    }

    // checking the conditions
    if (other > 1)
        return false;
    else if (other != 0 &&
             minusone != 0)
        return false;
    else if (minusone > 1 &&
             one == 0)
        return false;

    return true;
}

// Driver Code
int main()
{
    int arr[] = {0, 1, 1, 10};
    int n = sizeof(arr) / sizeof(arr[0]);
    if (checkArray(arr, n))
    cout << "Yes";
    else
    cout << "No";
    return 0;
}

Java

// Java program to find
// if product of every pair
// is present in array.

class GFG
{
    // Returns true if product
    // of every pair in arr[]
    // is present in arr[]
    static boolean checkArray(int arr[] ,
                              int n)
    {
        // variable to store number
        //  of zeroes, ones, minus
        // one and other numbers.
        int zero = 0, one = 0,
            minusone = 0, other=0;
        for (int i = 0; i < n; i++)
        {
            // incrementing the
            // variable values
            if (arr[i] == 0)
                zero++;
            else if (arr[i] == 1)
                one++;
            else if (arr[i] == -1)
                minusone++;
            else
                other++;
        }

        // checking the conditions
        if (other > 1)
            return false;
        else if (other != 0 &&
                 minusone != 0)
            return false;
        else if (minusone > 1 &&
                 one == 0)
            return false;

        return true;
    }

    // Driver Code
    public static void main (String[] args)
    {
        int arr[] = {0, 1, 1, 10};
        int n = arr.length;
        if (checkArray(arr, n))
        System.out.println("Yes");
        else
        System.out.println("No");
    }
}

// This code is contributed by Harsh Agarwal

Python3

# Python3 program to find
# if product of every pair
# is present in array.

# Returns True if product
# of every pair in arr[] is
# present in arr[]
def checkArray(arr, n):

    # variable to store number
    # of zeroes, ones, minus
    # one and other numbers.
    zero = 0; one = 0;
    minusone = 0; other = 0
    for i in range(0, n):

        # incrementing the
        # variable values
        if (arr[i] == 0):
            zero += 1
        elif (arr[i] == 1):
            one += 1
        elif (arr[i] == -1):
            minusone += 1
        else:
            other += 1

    # checking the conditions
    if (other > 1):
        return false
    elif (other != 0 and
          minusone != 0):
        return false
    elif (minusone > 1 and
          one == 0):
        return false

    return True


# Driver Code
arr = [0, 1, 1, 10]
n = len(arr)
if (checkArray(arr, n)):
    print("Yes")
else:
    print("No")

# This code is contributed
# by Smitha Dinesh Semwal.

C#

// C# program to find if
// product of every pair
// is present in array.
using System;

class GFG
{
    // Returns true if product
    // of every pair in arr[]
    // is present in arr[]
    static Boolean checkArray(int []arr ,
                              int n)
    {
        // variable to store number
        // of zeroes, ones, minus
        // one and other numbers.
        int zero = 0, one = 0,
            minusone = 0, other=0;
        for (int i = 0; i < n; i++)
        {
            // incrementing the
            // variable values
            if (arr[i] == 0)
                zero++;
            else if (arr[i] == 1)
                one++;
            else if (arr[i] == -1)
                minusone++;
            else
                other++;
        }

        // checking the conditions
        if (other > 1)
            return false;
        else if (other != 0 &&
                 minusone != 0)
            return false;
        else if (minusone > 1 &&
                 one == 0)
            return false;

        return true;
    }

    // Driver Code
    public static void Main (String[] args)
    {
        int []arr = {0, 1, 1, 10};
        int n = arr.Length;
        if (checkArray(arr, n))
        Console.Write("Yes");
        else
        Console.Write("No");
    }
}

// This code is contributed by parashar....

PHP

<?php
// PHP program to find if
// product of every pair
// is present in array.

// Returns true if product
// of every pair in arr[]
// is present in arr[]
function checkArray($arr , $n)
{
    // variable to store number
    // of zeroes, ones, minus
    // one and other numbers.
    $zero = 0; $one = 0;
    $minusone = 0; $other=0;
    for ($i = 0; $i < $n; $i++)
    {
        // incrementing the
        // variable values
        if ($arr[$i] == 0)
            $zero++;
        else if ($arr[$i] == 1)
            $one++;
        else if ($arr[$i] == -1)
            $minusone++;
        else
            $other++;
    }

    // checking the conditions
    if ($other > 1)
        return false;
    else if ($other != 0 &&
             $minusone != 0)
        return false;
    else if ($minusone > 1 &&
             $one == 0)
        return false;

    return true;
}

// Driver Code
{
    $arr = array(0, 1, 1, 10);
    $n = sizeof($arr) / sizeof($arr[0]);
    if (checkArray($arr, $n))
    echo "Yes";
    else
    echo "No";
    return 0;
}

//This code is contributed
// by nitin mittal.
?>

Javascript

<script>
//javascript program to find if
// product of every pair
// is present in array.
// Returns true if product
// of every pair in arr[]
// is present in arr[]
function checkArray(arr,n)
{
    // variable to store number
    // of zeroes, ones, minus
    // one and other numbers.
    let zero = 0, one = 0,
        minusone = 0, other=0;
    for (let i = 0; i < n; i++)
    {
        // incrementing the
        // variable values
        if (arr[i] == 0)
            zero++;
        else if (arr[i] == 1)
            one++;
        else if (arr[i] == -1)
            minusone++;
        else
            other++;
    }

    // checking the conditions
    if (other > 1)
        return false;
    else if (other != 0 &&
            minusone != 0)
        return false;
    else if (minusone > 1 &&
            one == 0)
        return false;

    return true;
}

    let arr = [0, 1, 1, 10];
    let n = arr.length;
    if (checkArray(arr, n))
    document.write("Yes");
    else
    document.write("No");

// This code is contributed by vaibhavrabadiya117.
</script>

Output

Yes

Time Complexity: O(n)
Auxiliary Space: O(1)