Minimum Number of Operations to Make X and Y Equal
<p>You are given two positive integers <code>x</code> and <code>y</code>.</p>

<p>In one operation, you can do one of the four following operations:</p>

<ol>
	<li>Divide <code>x</code> by <code>11</code> if <code>x</code> is a multiple of <code>11</code>.</li>
	<li>Divide <code>x</code> by <code>5</code> if <code>x</code> is a multiple of <code>5</code>.</li>
	<li>Decrement <code>x</code> by <code>1</code>.</li>
	<li>Increment <code>x</code> by <code>1</code>.</li>
</ol>

<p>Return <em>the <strong>minimum</strong> number of operations required to make </em> <code>x</code> <i>and</i> <code>y</code> equal.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> x = 26, y = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can make 26 equal to 1 by applying the following operations: 
1. Decrement x by 1
2. Divide x by 5
3. Divide x by 5
It can be shown that 3 is the minimum number of operations required to make 26 equal to 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> x = 54, y = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> We can make 54 equal to 2 by applying the following operations: 
1. Increment x by 1
2. Divide x by 11 
3. Divide x by 5
4. Increment x by 1
It can be shown that 4 is the minimum number of operations required to make 54 equal to 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> x = 25, y = 30
<strong>Output:</strong> 5
<strong>Explanation:</strong> We can make 25 equal to 30 by applying the following operations: 
1. Increment x by 1
2. Increment x by 1
3. Increment x by 1
4. Increment x by 1
5. Increment x by 1
It can be shown that 5 is the minimum number of operations required to make 25 equal to 30.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= x, y &lt;= 10<sup>4</sup></code></li>
</ul>
Breadth-First Search; Memoization; Dynamic Programming
class Solution:
    def minimumOperationsToMakeEqual(self, x: int, y: int) -> int:
        @cache
        def dfs(x: int) -> int:
            if y >= x:
                return y - x
            ans = x - y
            ans = min(ans, x % 5 + 1 + dfs(x // 5))
            ans = min(ans, 5 - x % 5 + 1 + dfs(x // 5 + 1))
            ans = min(ans, x % 11 + 1 + dfs(x // 11))
            ans = min(ans, 11 - x % 11 + 1 + dfs(x // 11 + 1))
            return ans

        return dfs(x)
