Make all numbers of an array equal

Given an array arr[], the task is to make all the array elements equal with the given operation. In a single operation, any element of the array can be either multiplied by 2 or by 3. If its possible to make all the array elements equal with the given operation then print Yes else print No.

Examples:

Input: arr[] = {50, 75, 100}
Output: Yes
{50 * 2 * 3, 75 * 2 * 2, 100 * 3} = {300, 300, 300}

Input: arr[] = {10, 14}
Output: No

Approach: Any positive integer number can be factorized and written as 2a * 3b * 5c * 7d * …..
We can multiply given numbers by 2 and 3 so we can increase a and b for them. So we can make all a and b equal by increasing them to the same big value (e.g. 100). But we can’t change powers of other prime numbers so they must be equal from the beginning. We can check it by diving all numbers from input by two and by three as many times as possible. Then all of them must be equal.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if all 
// the array elements can be made equal
// with the given operation
bool EqualNumbers(int a[], int n)
{
    for (int i = 0; i < n; i++) {
  
        // Divide number by 2
        while (a[i] % 2 == 0)
            a[i] /= 2;
  
        // Divide number by 3
        while (a[i] % 3 == 0)
            a[i] /= 3;
    }
  
    // Remaining numbers
    for (int i = 1; i < n; i++)
        if (a[i] != a[0]) {
            return false;
        }
  
    return true;
}
  
// Driver code
int main()
{
    int a[] = { 50, 75, 150 };
  
    int n = sizeof(a) / sizeof(a[0]);
  
    if (EqualNumbers(a, n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

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Java

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// Java implementation of above approach
class GFG 
{
  
    // Function that returns true if all 
    // the array elements can be made equal 
    // with the given operation 
    static boolean EqualNumbers(int a[], int n)
    {
        for (int i = 0; i < n; i++) 
        {
  
            // Divide number by 2 
            while (a[i] % 2 == 0
            {
                a[i] /= 2;
            }
  
            // Divide number by 3 
            while (a[i] % 3 == 0)
            {
                a[i] /= 3;
            }
        }
  
        // Remaining numbers 
        for (int i = 1; i < n; i++) 
        {
            if (a[i] != a[0])
            {
                return false;
            }
        }
  
        return true;
    }
  
    // Driver code 
    public static void main(String[] args) 
    {
        int a[] = {50, 75, 150};
  
        int n = a.length;
  
        if (EqualNumbers(a, n)) 
        {
            System.out.println("Yes");
        
        else 
        {
            System.out.println("No");
        }
    }
  
// This code is contributed by Rajput-JI

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Python3

# Python3 implementation of the approach

# Function that returns true if all
# the array elements can be made equal
# with the given operation
def EqualNumbers(a, n):

for i in range(0, n):

# Divide number by 2
while a[i] % 2 == 0:
a[i] //= 2

# Divide number by 3
while a[i] % 3 == 0:
a[i] //= 3

# Remaining numbers
for i in range(1, n):
if a[i] != a[0]:
return False

return True

# Driver code
if __name__ == “__main__”:

a = [50, 75, 150]
n = len(a)

if EqualNumbers(a, n):
print(“Yes”)
else:
print(“No”)

# This code is contributed by Rituraj Jain

C#

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// C# implementation of above approach 
using System;
  
class GFG 
  
    // Function that returns true if all 
    // the array elements can be made equal 
    // with the given operation 
    static bool EqualNumbers(int []a, int n) 
    
        for (int i = 0; i < n; i++) 
        
  
            // Divide number by 2 
            while (a[i] % 2 == 0) 
            
                a[i] /= 2; 
            
  
            // Divide number by 3 
            while (a[i] % 3 == 0) 
            
                a[i] /= 3; 
            
        
  
        // Remaining numbers 
        for (int i = 1; i < n; i++) 
        
            if (a[i] != a[0]) 
            
                return false
            
        
        return true
    
  
    // Driver code 
    public static void Main() 
    
        int []a = {50, 75, 150}; 
  
        int n = a.Length; 
  
        if (EqualNumbers(a, n)) 
        
            Console.WriteLine("Yes"); 
        
        else
        
            Console.WriteLine("No"); 
        
    
  
// This code is contributed by Ryuga 

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PHP

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<?php
// PHP implementation of the approach
// Function that returns true if all 
// the array elements can be made equal
// with the given operation
  
function EqualNumbers($a, $n)
{
    for ($i = 0; $i < $n; $i++) 
    {
  
        // Divide number by 2
        while ($a[$i] % 2 == 0)
            $a[$i] /= 2;
  
        // Divide number by 3
        while ($a[$i] % 3 == 0)
            $a[$i] /= 3;
    }
  
    // Remaining numbers
    for ($i = 1; $i < $n; $i++)
        if ($a[$i] != $a[0])
        {
            return false;
        }
  
    return true;
}
  
    // Driver code
    $a = array(50, 75, 150 );
    $n = sizeof($a) / sizeof($a[0]);
    if (EqualNumbers($a, $n))
        echo "Yes";
    else
        echo "No";
  
#This code is contributed by ajit..
?>

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Output:

Yes


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