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Check if all the elements can be made equal on dividing with X and Y
  • Last Updated : 20 Nov, 2019
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Given an array arr[] and two integers X and Y. The task is to check whether it is possible to make all the elements equal by dividing them with X and Y any number of times including 0.

Examples:

Input: arr[] = {2, 4, 6, 8}, X = 2, Y = 3
Output: Yes
2 -> 2
4 -> (4 / X) = (4 / 2) = 2
6 -> (6 / Y) = (6 / 3) = 2
8 -> (8 / X) = (8 / 2) = 4 and 4 -> (4 / X) = (4 / 2) = 2

Input: arr[] = {2, 4, 10}, X = 11, Y = 12
Output: No

Approach: Find the gcd of all the elements from the given array because this gcd is the value which can we get by dividing all the elements with some arbitrary constants say gcd = arr[0] / k1 or arr[1] / k2 or … or arr[n-1] / kn. Now the task is to find whether these constants k1, k2, k3, …, kn are of the form X * X * X * … * Y Y Y * ….. If yes then it is possible to make all the elements equal with the given operation else it isn’t.



Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if num
// is of the form x*x*x*...*y*y*...
bool isDivisible(int num, int x, int y)
{
  
    // While num divisible is divible
    // by either x or y, keep dividing
    while (num % x == 0 || num % y == 0) {
        if (num % x == 0)
            num /= x;
        if (num % y == 0)
            num /= y;
    }
  
    // If num > 1, it means it cannot be
    // further divided by either x or y
    if (num > 1)
        return false;
  
    return true;
}
  
// Function that returns true if all
// the array elements can be made
// equal with the given operation
bool isPossible(int arr[], int n, int x, int y)
{
  
    // To store the gcd of the array elements
    int gcd = arr[0];
    for (int i = 1; i < n; i++)
        gcd = __gcd(gcd, arr[i]);
  
    // For every element of the array
    for (int i = 0; i < n; i++) {
  
        // Check if k is of the form x*x*..*y*y*...
        // where (gcd * k = arr[i])
        if (!isDivisible(arr[i] / gcd, x, y))
            return false;
    }
    return true;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 4, 6, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 2, y = 3;
  
    if (isPossible(arr, n, x, y))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java




// Java implementation of the approach 
  
class GFG 
{
  
    // Function that returns true if num
    // is of the form x*x*x*...*y*y*...
    public static boolean isDivisible(int num, int x, int y) 
    {
  
        // While num divisible is divible
        // by either x or y, keep dividing
        while (num % x == 0 || num % y == 0)
        {
            if (num % x == 0)
                num /= x;
            if (num % y == 0)
                num /= y;
        }
  
        // If num > 1, it means it cannot be
        // further divided by either x or y
        if (num > 1)
            return false;
  
        return true;
    }
  
    // Funcion to calculate gcd of two numbers
    // using Euclid's algorithm
    public static int _gcd(int a, int b) 
    {
        while (a != b) 
        {
            if (a > b)
                a = a - b;
            else
                b = b - a;
        }
  
        return a;
    }
  
    // Function that returns true if all
    // the array elements can be made
    // equal with the given operation
    public static boolean isPossible(int[] arr, int n, 
                                        int x, int y)
    {
          
        // To store the gcd of the array elements
        int gcd = arr[0];
        for (int i = 1; i < n; i++)
            gcd = _gcd(gcd, arr[i]);
  
        // For every element of the array
        for (int i = 0; i < n; i++)
        {
  
            // Check if k is of the form x*x*..*y*y*...
            // where (gcd * k = arr[i])
            if (!isDivisible(arr[i] / gcd, x, y))
                return false;
        }
        return true;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int[] arr = { 2, 4, 6, 8 };
        int n = arr.length;
        int x = 2, y = 3;
        if (isPossible(arr, n, x, y))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
  
// This code is contributed by
// sanjeev2552

Python3




# Python3 implementation of the approach
from math import gcd as __gcd
  
# Function that returns True if num
# is of the form x*x*x*...*y*y*...
def isDivisible(num, x, y):
  
    # While num divisible is divible
    # by either x or y, keep dividing
    while (num % x == 0 or num % y == 0):
        if (num % x == 0):
            num //= x
        if (num % y == 0):
            num //= y
  
    # If num > 1, it means it cannot be
    # further divided by either x or y
    if (num > 1):
        return False
  
    return True
  
# Function that returns True if all
# the array elements can be made
# equal with the given operation
def isPossible(arr, n, x, y):
  
    # To store the gcd of the array elements
    gcd = arr[0]
    for i in range(1,n):
        gcd = __gcd(gcd, arr[i])
  
    # For every element of the array
    for i in range(n):
  
        # Check if k is of the form x*x*..*y*y*...
        # where (gcd * k = arr[i])
        if (isDivisible(arr[i] // gcd, x, y) == False):
            return False
    return True
  
  
# Driver code
  
arr = [2, 4, 6, 8]
n = len(arr)
x = 2
y = 3
  
if (isPossible(arr, n, x, y) == True):
    print("Yes")
else:
    print("No")
      
# This code is contributed by mohit kumar 29

C#




// C# implementation of the approach 
using System;
  
class GFG 
{
  
    // Function that returns true if num
    // is of the form x*x*x*...*y*y*...
    public static bool isDivisible(int num, int x, int y) 
    {
  
        // While num divisible is divible
        // by either x or y, keep dividing
        while (num % x == 0 || num % y == 0)
        {
            if (num % x == 0)
                num /= x;
            if (num % y == 0)
                num /= y;
        }
  
        // If num > 1, it means it cannot be
        // further divided by either x or y
        if (num > 1)
            return false;
  
        return true;
    }
  
    // Funcion to calculate gcd of two numbers
    // using Euclid's algorithm
    public static int _gcd(int a, int b) 
    {
        while (a != b) 
        {
            if (a > b)
                a = a - b;
            else
                b = b - a;
        }
  
        return a;
    }
  
    // Function that returns true if all
    // the array elements can be made
    // equal with the given operation
    public static bool isPossible(int[] arr, int n, 
                                        int x, int y)
    {
          
        // To store the gcd of the array elements
        int gcd = arr[0];
        for (int i = 1; i < n; i++)
            gcd = _gcd(gcd, arr[i]);
  
        // For every element of the array
        for (int i = 0; i < n; i++)
        {
  
            // Check if k is of the form x*x*..*y*y*...
            // where (gcd * k = arr[i])
            if (!isDivisible(arr[i] / gcd, x, y))
                return false;
        }
        return true;
    }
  
    // Driver code
    public static void Main() 
    {
        int[] arr = { 2, 4, 6, 8 };
        int n = arr.Length;
        int x = 2, y = 3;
        if (isPossible(arr, n, x, y))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
  
// This code is contributed by
// anuj_67..
Output:
Yes

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