<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:
        hm = {}

        def dfs(i):
            if i == 0:
                return 0
            
            if i == 1:
                return 1
            
            if i in hm:
                return hm[i]
            
            a = float('inf')
            if i % 2 == 0:
                a = min(a,dfs(i//2)+1)
            
            if i % 3 == 0:
                a = min(a,dfs(i//3)+1)
            
            if i % 3 or i % 2:
                a = min(a,dfs(i-1)+1)
            
            hm[i] = a
            return a
        return dfs(n)
        
60