Check whether it is possible to convert A into B

Given two integers A and B. The task is to check whether it is possible to convert A into B by performing below operations any number of times.

  1. Convert current number x to 2 * x.
  2. Convert current number x to (10 * x) + 1.

Examples:

Input: A = 2, B = 82
Output: Yes
2 -> 4 -> 41 -> 82

Input: A = 2, B = 5
Output: No

Approach: Let’s solve this problem in a reverse way – try to get the number A from B.

Note, that if B ends with 1 the last operation was to append the digit 1 to the right of the current number. Because of that let’s delete the last digit of B and move to the new number.

If the last digit is even then the last operation was to multiply the current number by 2. Because of that let’s divide B by 2 and move to the new number.

In the other cases (if B ends with odd digit except 1) the answer is No.

We need to repeat the described algorithm after every time we get a new number. If at some point, we get a number which is equal to A then the answer is Yes, and if the new number is less than A then the answer is No.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if A can be
// converted to B with the given operations
bool canConvert(int a, int b)
{
    while (b > a) {
  
        // If the current number ends with 1
        if (b % 10 == 1) {
            b /= 10;
            continue;
        }
  
        // If the current number is divisible by 2
        if (b % 2 == 0) {
            b /= 2;
            continue;
        }
  
        // If above two conditions fail
        return false;
    }
  
    // If is is possible to convert A to B
    if (b == a)
        return true;
    return false;
}
  
// Driver code
int main()
{
    int A = 2, B = 82;
  
    if (canConvert(A, B))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
  
    // Function that returns true if A can be
    // converted to B with the given operations
    static boolean canConvert(int a, int b)
    {
        while (b > a) 
        {
  
            // If the current number ends with 1
            if (b % 10 == 1
            {
                b /= 10;
                continue;
            }
  
            // If the current number is divisible by 2
            if (b % 2 == 0)
            {
                b /= 2;
                continue;
            }
  
            // If above two conditions fail
            return false;
        }
  
        // If is is possible to convert A to B
        if (b == a)
            return true;
        return false;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
  
        int A = 2, B = 82;
  
        if (canConvert(A, B))
            System.out.println("Yes");
        else
            System.out.println("No");
  
    }
}
  
// This code contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach 
  
# Function that returns true if A can be 
# converted to B with the given operations 
def canConvert(a, b) : 
  
    while (b > a) :
  
        # If the current number ends with 1 
        if (b % 10 == 1) :
            b //= 10
            continue
          
        # If the current number is divisible by 2 
        if (b % 2 == 0) :
            b /= 2
            continue
  
        # If above two conditions fail 
        return false; 
      
    # If is is possible to convert A to B 
    if (b == a) :
        return True;
          
    return False
  
# Driver code 
if __name__ == "__main__"
  
    A = 2; B = 82
  
    if (canConvert(A, B)) :
        print("Yes"); 
    else :
        print("No"); 
      
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
class GFG
{
  
    // Function that returns true if A can be
    // converted to B with the given operations
    static bool canConvert(int a, int b)
    {
        while (b > a) 
        {
  
            // If the current number ends with 1
            if (b % 10 == 1) 
            {
                b /= 10;
                continue;
            }
  
            // If the current number is divisible by 2
            if (b % 2 == 0)
            {
                b /= 2;
                continue;
            }
  
            // If above two conditions fail
            return false;
        }
  
        // If is is possible to convert A to B
        if (b == a)
            return true;
        return false;
    }
  
    // Driver code
    public static void Main() 
    {
  
        int A = 2, B = 82;
  
        if (canConvert(A, B))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
  
    }
}
  
// This code is contributed by anuj_67..

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Output:

Yes


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Improved By : Rajput-Ji, vt_m, AnkitRai01