Check if it is possible to reach (X, Y) from (1, 1) by given steps

Given two integers X and Y, the task is to check if it is possible to reach (X, Y) from (1, 1) by the following possible moves:

• From a point (a, b) such that b > a, move to the to point (a, b - a).
• From a point (a, b) such that a > b, move to the to point (a - b, b).
• Move from any point (a, b) to point (2 * a, b) or (a, 2 * b).

If it is possible to reach (X, Y), print "Yes". Otherwise, print "No".

Examples: 

Input: X = 4, Y = 7
Output: Yes
Explanation: Point (4, 7) can be reached by the following steps: (1, 1) -> (1, 2) -> (1, 4) -> (1, 8) -> (1, 7) -> (2, 7) -> (4, 7)

Input: X = 3, Y = 6
Output: No

Naive Approach: The simplest approach to solve the problem is to recursively check if (1, 1) can be reached from (X, Y) by making all possible moves from a point. If the point (1, 1) is reached at any point, print "Yes". Otherwise, print "No".

Time Complexity: O(N⁴) where N is the number of steps to reach the point (1, 1).
Auxiliary Space: O(1) 

Efficient Approach: The idea is to find the greatest common divisor and observe the following facts:

• By moving the point from (a, b) to the points (a, b - a) or (a - b, b), there is no change in the GCD of the pair
• By moving the point from (a, b) to the points (2 * a, b) or (a, 2 * b), GCD may get doubled or it may remain the same.
• Therefore, from the above observations, point (1, 1) can move to point (X, Y) if and only if gcd of (X, Y) is a power of 2.

Follow the steps below to solve the above approach:

1. Find the gcd of (X, Y) and store it in a variable, say val.
2. Check if val is a power of 2 by checking if (val & amp;(val-1)) is equal to 0 where & is bitwise AND.
3. If it is a power of two print "Yes". Otherwise, print "No".

Below is the implementation of the above approach:

// C++ program for the above approach 
#include <iostream>
using namespace std;

// Function to find the gcd of 
// two numbers
int gcd(int a, int b)
{
    
    // Base case
    if (a < b) 
    {
        int t = a;
        a = b;
        b = t;
    }
    if (a % b == 0)
        return b;
        
    // Recurse
    return gcd(b, a % b);
}

// Function to print the answer
void printAnswer(int x, int y)
{
    
    // GCD of X and Y
    int val = gcd(x, y);

    // If GCD is power of 2
    if ((val & (val - 1)) == 0)
        cout << "Yes";
    else
        cout << "No";
}

// Driver code 
int main()
{
    
    // Given X and Y
    int x = 4;
    int y = 7;

    // Function call
    printAnswer(x, y);

    return 0;
}

// This code is contributed by RohitOberoi

// Java program for the above approach

import java.io.*;

class GFG {

    // Function to find the gcd of two numbers
    public static int gcd(int a, int b)
    {

        // Base case
        if (a < b) {
            int t = a;
            a = b;
            b = t;
        }
        if (a % b == 0)
            return b;

        // Recurse
        return gcd(b, a % b);
    }

    // Function to print the answer
    static void printAnswer(int x, int y)
    {

        // GCD of X and Y
        int val = gcd(x, y);

        // If GCD is power of 2
        if ((val & (val - 1)) == 0)
            System.out.println("Yes");
        else
            System.out.println("No");
    }

    // Driver code
    public static void main(String[] args)
    {

        // Given X and Y
        int x = 4;
        int y = 7;

        // Function call
        printAnswer(x, y);
    }
}

# Python3 program for the
# above approach

# Function to find the gcd 
# of two numbers
def gcd(a, b):

    # Base case
    if (a < b):

        t = a
        a = b
        b = t

    if (a % b == 0):
        return b

    # Recurse
    return gcd(b, a % b)

# Function to print the 
# answer
def printAnswer(x, y):

    # GCD of X and Y
    val = gcd(x, y)

    # If GCD is power of 2
    if ((val & 
        (val - 1)) == 0):
        print("Yes")
    else:
        print("No")

# Driver code
if __name__ == "__main__":

    # Given X and Y
    x = 4
    y = 7

    # Function call
    printAnswer(x, y)

# This code is contributed by Chitranayal

// C# program for the above approach
using System;

class GFG{

// Function to find the gcd of two numbers
public static int gcd(int a, int b)
{
    
    // Base case
    if (a < b) 
    {
        int t = a;
        a = b;
        b = t;
    }
    if (a % b == 0)
        return b;
        
    // Recurse
    return gcd(b, a % b);
}

// Function to print the answer
static void printAnswer(int x, int y)
{
    
    // GCD of X and Y
    int val = gcd(x, y);

    // If GCD is power of 2
    if ((val & (val - 1)) == 0)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}

// Driver code
public static void Main()
{
    
    // Given X and Y
    int x = 4;
    int y = 7;

    // Function call
    printAnswer(x, y);
}
}

// This code is contributed by bgangwar59

<script>
// JavaScript program for the above approach

// Function to find the gcd of
// two numbers
function gcd(a, b)
{
    
    // Base case
    if (a < b)
    {
        let t = a;
        a = b;
        b = t;
    }
    if (a % b == 0)
        return b;
        
    // Recurse
    return gcd(b, a % b);
}

// Function to print the answer
function printAnswer(x, y)
{
    
    // GCD of X and Y
    let val = gcd(x, y);

    // If GCD is power of 2
    if ((val & (val - 1)) == 0)
        document.write("Yes");
    else
        document.write("No");
}

// Driver code
    
    // Given X and Y
    let x = 4;
    let y = 7;

    // Function call
    printAnswer(x, y);



// This code is contributed by Surbhi Tyagi.
</script>

Yes

Time Complexity: O(Log(min(X, Y))) where (X, Y) is the given point. 
Auxiliary Space: O(1)