#1553

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Number of Days to Eat N Oranges

Find the minimum number of days to eat n oranges using state transition dynamic programming with memoization.

Problem Statement

You are given an integer n, representing the number of oranges in your kitchen. Each day, you must choose one of the following actions:

Eat 1 orange. 2. Divide the oranges by 2 and eat half (only if divisible by 2). 3. Divide the oranges by 3 and eat a third (only if divisible by 3). The goal is to determine the minimum number of days required to eat all the oranges.

Examples

Example 1

Input: n = 10

Output: 4

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

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.

Constraints

1 <= n <= 2 * 109

Solution Approach

State Transition Dynamic Programming

This problem follows a state transition dynamic programming approach. For each day, we choose the optimal action: either eating 1 orange, dividing by 2, or dividing by 3. The transition is captured with the recursive function minOranges = 1 + min( (n%2) + f(n/2), (n%3) + f(n/3) ).

Memoization for Efficiency

Memoization helps by storing results of already computed subproblems, preventing redundant calculations and improving performance. This allows solving the problem efficiently even for large values of n.

Base Case Handling

Handle the base case where n equals 1, which immediately returns 1 day as the solution. This case is essential for the recursion to terminate correctly.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the approach, but memoization reduces redundant calculations. The space complexity is determined by the number of states stored in the memoization table, which is proportional to n in the worst case.

What Interviewers Usually Probe

Understand state transition and recursive problem-solving techniques.
Test for efficiency and correct handling of large inputs.
Evaluate the candidate's understanding of memoization in dynamic programming.

Common Pitfalls or Variants

Common pitfalls

Incorrect base case handling could result in infinite recursion or incorrect results.
Not implementing memoization leads to redundant computations, significantly affecting performance.
For large n, failing to optimize recursion depth or memoization can lead to stack overflow or slow execution.

Follow-up variants

Modify the problem to allow for other divisibility conditions (e.g., divide by 4 or 5).
Alter the minimum action (e.g., allow eating multiple oranges each day).
Extend the problem to allow additional actions like skipping days.

FAQ

What is the main algorithmic approach for solving the Minimum Number of Days to Eat N Oranges?

The main approach is state transition dynamic programming combined with memoization, minimizing the number of days based on the optimal actions of eating 1, dividing by 2, or dividing by 3.

What are the common pitfalls when solving this problem?

Common pitfalls include incorrect base case handling, not implementing memoization, and failing to optimize recursion for large values of n.

How does memoization optimize the solution for Minimum Number of Days to Eat N Oranges?

Memoization optimizes the solution by storing previously computed results, preventing redundant calculations and improving performance for larger values of n.

What is the time complexity of the solution to the Minimum Number of Days to Eat N Oranges?

The time complexity depends on the memoization table, with each state being computed only once, making it O(n) in the worst case.

What is the base case in the Minimum Number of Days to Eat N Oranges problem?

The base case occurs when n equals 1, where only 1 day is needed to eat the last orange.

terminal

Solution

Solution 1: Memoization Search

According to the problem description, for each $n$, we can choose one of three ways:

1

2

3

4

5

6

7

8

9

class Solution:
    def minDays(self, n: int) -> int:
        @cache
        def dfs(n: int) -> int:
            if n < 2:
                return n
            return 1 + min(n % 2 + dfs(n // 2), n % 3 + dfs(n // 3))

        return dfs(n)

Continue Practicing

#1569 number of ways to reorder array to get same bst #1575 count all possible routes #1611 minimum one bit operations to make integers zero