Check if all the elements can be made equal on dividing with X and Y

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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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