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

 3 years ago
source link: https://www.geeksforgeeks.org/check-if-all-the-elements-can-be-made-equal-on-dividing-with-x-and-y/
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Check if all the elements can be made equal on dividing with X and Y
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Check if all the elements can be made equal on dividing with X and Y
  • Last Updated : 20 Nov, 2019

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:

  • Python3

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// 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;
}
Output: 
Yes

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