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Minimum Number of Days to Eat N Oranges

Number: 1676

Difficulty: Hard

Paid? No

Companies: Google, Amazon

Problem Description

Given n oranges, each day you can choose one of three actions: eat one orange, if n is divisible by 2 then eat n/2 oranges, or if n is divisible by 3 then eat 2*(n/3) oranges. The goal is to determine the minimum number of days required to eat all the oranges.

Key Insights

  • The problem asks for a minimum path (number of days) to reduce n to 0 with different operations available.
  • Greedy approaches do not always yield the optimal solution, so we use dynamic programming with recursion and memoization.
  • The state is defined by the number of remaining oranges.
  • At each state, reduce the problem by either performing a division (if applicable) or performing subtraction operations to set up for division.
  • Use memoization to cache results for states that have already been computed.

Space and Time Complexity

Time Complexity: Approximately O(log n), as the operations drastically reduce n using division, and memoization prevents recomputation. Space Complexity: O(log n) for the memoization recursion tree depth and cache storage.

Solution

We solve this problem using recursion with memoization. The recursion function dp(n) represents the minimum number of days needed to eat n oranges. For any n, if it is not divisible by 2 or 3, we subtract oranges one by one until it becomes divisible by 2 or 3, and then apply the division operations. The recurrence is:
dp(n) = 1 + min( (n mod 2) + dp(n // 2), (n mod 3) + dp(n // 3) )
The additional (n mod divisor) accounts for the days we waste subtracting oranges so that n becomes divisible by dividend, ensuring we can use the operation that reduces n faster.

Code Solutions

# Python solution with detailed comments

class Solution:
    def minDays(self, n: int) -> int:
        # Memoization dictionary to store computed results for n
        memo = {}
        
        def dp(remaining):
            # Base case: if no oranges left, return 0 days
            if remaining == 0:
                return 0
            # If only one orange is left, take one day to eat it.
            if remaining == 1:
                return 1
            # If result is already computed, return it.
            if remaining in memo:
                return memo[remaining]
            
            # Compute cost if making n divisible by 2: subtract extra oranges until it is divisible
            # The cost to subtract extra functionality plus recursive call after division.
            days_by_2 = remaining % 2 + dp(remaining // 2)
            # Compute cost if making n divisible by 3 similarly.
            days_by_3 = remaining % 3 + dp(remaining // 3)
            
            # 1 more day for the current operation.
            memo[remaining] = 1 + min(days_by_2, days_by_3)
            return memo[remaining]
        
        return dp(n)

# Example usage:
# sol = Solution()
# print(sol.minDays(10))
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