There are n oranges in the kitchen and you decided to eat some of these oranges every day as follows:
Eat one orange.
If the number of remaining oranges (n) is divisible by 2 then you can eat  n/2 oranges.
If the number of remaining oranges (n) is divisible by 3 then you can eat  2*(n/3) oranges.
You can only choose one of the actions per day.
Return the minimum number of days to eat n oranges.
Example 1:
Input: n = 10
Output: 4
Explanation: 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.
Example 2:
Input: n = 6
Output: 3
Explanation: 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.
Example 3:
Input: n = 1
Output: 1
Example 4:
Input: n = 56
Output: 6
Constraints:
1 <= n <= 2*10^9
class Solution:
    def minDays(self, n: int) -> int:
        ONE_LIMIT = 50
        TWO_LIMIT = 23
        f = {1:1}
        def search(n, one, two):
            # if n == 1 return 1
            if one > ONE_LIMIT:
                return 1000
            if two > TWO_LIMIT:
                return 1000
            if n in f:
                return f[n]
            ans = 1000
            if n % 3 == 0:
                ans = min(ans, search(n//3, one, two))
            if n % 2 == 0:
                ans = min(ans, search(n//2, one, two + 1))
            ans = min(ans, search(n-1, one+1, two))
            f[n] = 1 + ans
            return f[n]
        
        search(n,0,0)
        
        return f[n]
class Solution:
    def minDays(self, n: int) -> int:
        ONE_LIMIT = 50
        TWO_LIMIT = 23
        f = {1:1}
        def search(n, one, two):
            # if n == 1 return 1
            if one > ONE_LIMIT:
                return 1000
            if two > TWO_LIMIT:
                return 1000
            if n in f:
                return f[n]
            ans = 1000
            if n % 3 == 0:
                ans = min(ans, search(n//3, one, two))
            if n % 2 == 0:
                ans = min(ans, search(n//2, one, two + 1))
            ans = min(ans, search(n-1, one+1, two))
            f[n] = 1 + ans
            return f[n]
        
        search(n,0,0)
        
        return f[n]