<p>There are <code>n</code> oranges in the kitchen and you decided to eat some of these oranges every day as follows:</p>

<ul>
	<li>Eat one orange.</li>
	<li>If the number of remaining oranges <code>n</code> is divisible by <code>2</code> then you can eat <code>n / 2</code> oranges.</li>
	<li>If the number of remaining oranges <code>n</code> is divisible by <code>3</code> then you can eat <code>2 * (n / 3)</code> oranges.</li>
</ul>

<p>You can only choose one of the actions per day.</p>

<p>Given the integer <code>n</code>, return <em>the minimum number of days to eat</em> <code>n</code> <em>oranges</em>.</p>

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

<pre>
<strong>Input:</strong> n = 10
<strong>Output:</strong> 4
<strong>Explanation:</strong> You have 10 oranges.
Day 1: Eat 1 orange,  10 - 1 = 9.  
Day 2: Eat 6 oranges, 9 - 2*(9/3) = 9 - 6 = 3. (Since 9 is divisible by 3)
Day 3: Eat 2 oranges, 3 - 2*(3/3) = 3 - 2 = 1. 
Day 4: Eat the last orange  1 - 1  = 0.
You need at least 4 days to eat the 10 oranges.
</pre>

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

<pre>
<strong>Input:</strong> n = 6
<strong>Output:</strong> 3
<strong>Explanation:</strong> You have 6 oranges.
Day 1: Eat 3 oranges, 6 - 6/2 = 6 - 3 = 3. (Since 6 is divisible by 2).
Day 2: Eat 2 oranges, 3 - 2*(3/3) = 3 - 2 = 1. (Since 3 is divisible by 3)
Day 3: Eat the last orange  1 - 1  = 0.
You need at least 3 days to eat the 6 oranges.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minDays(self, n: int) -> int:
        q = [n]
        steps = {n: 0}
        
        while q:
            cur = q.pop(0)
            step = steps[cur]
            
            if cur == 0:
                return step
            
            if cur % 2 == 0 and cur // 2 not in steps:
                steps[cur // 2] = step + 1
                q.append(cur // 2)
            if cur % 3 == 0 and cur // 3 not in steps:
                steps[cur // 3] = step + 1
                q.append(cur // 3)
            if cur - 1 not in steps:
                steps[cur - 1] = step + 1
                q.append(cur - 1)

        return -1
92