Check whether it is possible to convert A into B
Last Updated :
21 Sep, 2023
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.
- Convert current number x to 2 * x.
- 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:
- Create a queue and add A to the queue.
- While the queue is not empty, pop the front element from the queue and check if it’s equal to B. If yes, return true.
- Otherwise, check if it’s possible to apply the first operation (multiply by 2) or the second operation (multiply by 10 and add 1) to the popped element. If the result is less than or equal to B, add it to the queue.
- Repeat steps 2-3 until the queue is empty.
- If no path is found, return false.
- Print “Yes” if the function returns true, otherwise print “No”.
C++
#include <bits/stdc++.h>
using namespace std;
bool canConvert( int a, int b) {
queue< int > q;
q.push(a);
while (!q.empty()) {
int x = q.front();
q.pop();
if (x == b) {
return true ;
}
if (x * 2 <= b) {
q.push(x * 2);
}
if ((10 * x + 1) <= b) {
q.push(10 * x + 1);
}
}
return false ;
}
int main() {
int A = 2, B = 82;
if (canConvert(A, B))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
import java.util.*;
public class Main {
public static void main(String[] args) {
int A = 2 , B = 82 ;
if (canConvert(A, B))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
public static boolean canConvert( int a, int b) {
Queue<Integer> queue = new LinkedList<>();
queue.offer(a);
while (!queue.isEmpty()) {
int x = queue.poll();
if (x == b) {
return true ;
}
if (x * 2 <= b) {
queue.offer(x * 2 );
}
if (( 10 * x + 1 ) <= b) {
queue.offer( 10 * x + 1 );
}
}
return false ;
}
}
|
Python3
from queue import Queue
def canConvert(a, b):
q = Queue()
q.put(a)
while not q.empty():
x = q.get()
if x = = b:
return True
if x * 2 < = b:
q.put(x * 2 )
if ( 10 * x + 1 ) < = b:
q.put( 10 * x + 1 )
return False
if __name__ = = "__main__" :
A = 2
B = 82
if canConvert(A, B):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
using System.Collections.Generic;
public class GFG
{
public static void Main( string [] args)
{
int A = 2, B = 82;
if (CanConvert(A, B))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
public static bool CanConvert( int a, int b)
{
Queue< int > queue = new Queue< int >();
queue.Enqueue(a);
while (queue.Count > 0)
{
int x = queue.Dequeue();
if (x == b)
{
return true ;
}
if (x * 2 <= b)
{
queue.Enqueue(x * 2);
}
if ((10 * x + 1) <= b)
{
queue.Enqueue(10 * x + 1);
}
}
return false ;
}
}
|
Javascript
function canConvert(a, b) {
let q = [a];
while (q.length > 0) {
let x = q.shift();
if (x == b) {
return true ;
}
if (x * 2 <= b) {
q.push(x * 2);
}
if ((10 * x + 1) <= b) {
q.push(10 * x + 1);
}
}
return false ;
}
let A = 2, B = 82;
if (canConvert(A, B))
console.log( "Yes" );
else
console.log( "No" );
|
Time Complexity: O(B), where B is the upper bound of the range of possible values of the number b. In the worst case, the entire range from a to B needs to be traversed.
Space Complexity: O(B), since the size of the queue can grow up to B in the worst case, when all the possible values of x are added to the queue.
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 an 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 that 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++
#include <bits/stdc++.h>
using namespace std;
bool canConvert( int a, int b)
{
while (b > a) {
if (b % 10 == 1) {
b /= 10;
continue ;
}
if (b % 2 == 0) {
b /= 2;
continue ;
}
return false ;
}
if (b == a)
return true ;
return false ;
}
int main()
{
int A = 2, B = 82;
if (canConvert(A, B))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
class GFG
{
static boolean canConvert( int a, int b)
{
while (b > a)
{
if (b % 10 == 1 )
{
b /= 10 ;
continue ;
}
if (b % 2 == 0 )
{
b /= 2 ;
continue ;
}
return false ;
}
if (b == a)
return true ;
return false ;
}
public static void main(String[] args)
{
int A = 2 , B = 82 ;
if (canConvert(A, B))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
def canConvert(a, b) :
while (b > a) :
if (b % 10 = = 1 ) :
b / / = 10 ;
continue ;
if (b % 2 = = 0 ) :
b / = 2 ;
continue ;
return false;
if (b = = a) :
return True ;
return False ;
if __name__ = = "__main__" :
A = 2 ; B = 82 ;
if (canConvert(A, B)) :
print ( "Yes" );
else :
print ( "No" );
|
C#
using System;
class GFG
{
static bool canConvert( int a, int b)
{
while (b > a)
{
if (b % 10 == 1)
{
b /= 10;
continue ;
}
if (b % 2 == 0)
{
b /= 2;
continue ;
}
return false ;
}
if (b == a)
return true ;
return false ;
}
public static void Main()
{
int A = 2, B = 82;
if (canConvert(A, B))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
Javascript
<script>
function canConvert(a, b)
{
while (b > a) {
if (b % 10 == 1) {
b = parseInt(b / 10);
continue ;
}
if (b % 2 == 0) {
b = parseInt(b / 2);
continue ;
}
return false ;
}
if (b == a)
return true ;
return false ;
}
let A = 2, B = 82;
if (canConvert(A, B))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Time Complexity: O(logn)
Auxiliary Space: O(1)
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