Check if there exists some positive integers X and Y such that P^X is equal to Q^Y

Given two integers P and Q, the task is to check whether a pair (P, Q) is equal or not, and a pair is said to be equal if there exist some positive integers X and Y such that P^X = Q^Y.

Examples:

Input: P = 16 , Q = 4
Output: Yes
?Explanation: Let X = 2 and Y = 4. Thus, P^X = 16² = 256 and Q^Y = 4⁴ = 256 . Thus, the pair (16,4) is equal.

Input: P = 12 , Q = 24
Output: No

Approach: The problem can be solved based on the following observation:

For P^X = Q^Y to be true for some integer pair (X, Y),  any one of the below cases must be true:

1. There must exist some integer K, such that 
   • P = K^A
   • Q = K^B
2. X = Y = 0

Now to implement this, below algorithm can be used:

• Find maximum(max) and minimum(min) number for two integer.
• Iterate a loop and check if max and min is equal or max is divisible by min, then pair of integer is equal and break from the loop.
• Otherwise, pair of integer is not equal.

 Below is the implementation of the above approach.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to check whether a pair of
// integer is equal or not
void check(int p, int q)
{
  int maxi = max(p, q);
  int mini = min(p, q);
  while (true) {
    if (maxi == mini) {
      cout << "Yes";
      break;
    }
    if (maxi % mini != 0) {
      cout << "No";
      break;
    }
    int temp = maxi / mini;
    maxi = max(temp, mini);
    mini = min(temp, mini);
  }
}

// Driver Code
int main()
{
  int P = 16;
  int Q = 4;

  // Function call
  check(P, Q);

  return 0;
}

// This code is contributed by aarohirai2616.

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

public class GFG {

    // Function to check whether a pair of
    // integer is equal or not
    public static void check(int p, int q)
    {
        int max = Math.max(p, q);
        int min = Math.min(p, q);
        while (true) {
            if (max == min) {
                System.out.println("Yes");
                break;
            }
            if (max % min != 0) {
                System.out.println("No");
                break;
            }
            int temp = max / min;
            max = Math.max(temp, min);
            min = Math.min(temp, min);
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        int P = 16;
        int Q = 4;

        // Function call
        check(P, Q);
    }
}

Python3

# Python3 code to implement the above approach

# Function to check whether a pair of
# integer is equal or not
def check( p, q) :

  maxi = max(p, q);
  mini = min(p, q);
  while (True) :
    if (maxi == mini) :
      print("Yes");
      break;

    if (maxi % mini != 0) :
      print("No");
      break;

    temp = maxi // mini;
    maxi = max(temp, mini);
    mini = min(temp, mini);


# Driver Code
if __name__ == "__main__" :

  P = 16;
  Q = 4;

  # Function call
  check(P, Q);

  # This code is contributed by AnkThon

C#

// C# implementation
using System;
public class GFG{

  // Function to check whether a pair of
  // integer is equal or not
  public static void check(int p, int q)
  {
    int maxi = Math.Max(p, q);
    int mini = Math.Min(p, q);
    while (true) {
      if (maxi == mini) {
        Console.WriteLine("Yes");
        break;
      }
      if (maxi % mini != 0) {
        Console.WriteLine("No");
        break;
      }
      int temp = maxi / mini;
      maxi =Math.Max(temp, mini);
      mini =Math.Min(temp, mini);
    }
  }

  static public void Main (){
    int P = 16;
    int Q = 4;

    // Function call
    check(P, Q);
  }
}

// This code is contributed by ksam24000

Javascript

// js code to implement the approach

// Function to check whether a pair of
// integer is equal or not
function check (p, q)
{
  let maxi = Math.max(p, q);
  let mini = Math.min(p, q);
  while (true) {
    if (maxi == mini) {
      console.log("Yes");
      break;
    }
    if (maxi % mini != 0) {
      console.log("No");
      break;
    }
    let temp = Math.floor(maxi / mini);
    maxi = Math.max(temp, mini);
    mini = Math.min(temp, mini);
  }
}

// Driver Code

  let P = 16;
  let Q = 4;

  // Function call
  check(P, Q);

// This code is contributed by ksam24000.

Output

Yes

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