The only way to make <code>x</code> larger is to increase it by <code>1</code> so if <code>y >= x</code> the answer is <code>y - x</code>.

For <code>y < x</code>, <code>x - y</code> is always a candidate answer since we can repeatedly decrease <code>x</code> by one to reach <code>y</code>.

We can also increase <code>x</code> and then use the division operations. For example, if <code>x = 10</code> and <code>y = 1</code>, we can increment <code>x</code> by <code>1</code> then divide it by <code>11</code>.

Find an upper bound <code>U</code> on the maximum value of <code>x</code> we will reach an optimal solution. Since all values of <code>x</code> will be in the range <code>[1, U]</code>, we can use BFS to find the answer.

One possible upper bound on <code>x</code> is <code>U = x + (x - y) </code>. To reach any number strictly greater than <code>U</code> from <code>x</code>, we will need more than <code>x - y</code> operations which is not optimal since we can always reach <code>y</code> in <code>x - y</code> operations.