Minimum Number Of Days To Eat N Oranges - Leetcode Solution

Problem Statement:

There are n oranges in a basket. You are tasked with eating exactly one orange per day until there are none left. However, you may only eat one orange on any given day, and you cannot leave any oranges uneaten overnight (i.e. you must eat them all on the same day). Given the number of oranges, n, determine the minimum number of days required to eat all n oranges.

Example 1:

Input: n = 10 Output: 4 Explanation: You can eat 1 orange on day 1, 2 oranges on day 2, 3 oranges on day 3, and 4 oranges on day 4 to finish eating all 10 oranges.

Example 2:

Input: n = 6 Output: 3 Explanation: You can eat 2 oranges on day 1, 2 oranges on day 2, and 2 oranges on day 3 to finish eating all 6 oranges.

Solution:

The approach to solve this problem is to use dynamic programming. Let's define dp[i] as the minimum number of days required to eat all i oranges.

Now, let's consider the base case where we have 0 or 1 orange. In both cases, the minimum number of days required to eat all oranges will be 0 or 1 respectively.

For all other cases, we have two options: either eat one orange on the first day and complete the remaining task in dp[i-1] days or eat two oranges on the first day and complete the remaining task in dp[i-2] days. We need to take the minimum of these two options as our answer for dp[i].

Let's see the implementation of this approach:

def minDays(n: int) -> int: dp = [0, 1] + [float('inf')] * (n - 1) for i in range(2, n+1): dp[i] = 1 + min(dp[i-1], i%2 + dp[i//2]) return dp[n]

We first initialize dp[0] to 0 and dp[1] to 1. We then loop through all numbers from 2 to n and calculate dp[i] using the formula mentioned above.

The i%2 + dp[i//2] part takes care of eating two oranges on the first day when i is even. Here, i//2 is the remaining oranges after eating two on the first day, and i%2 is 1 if i is odd and 0 if i is even.

We return dp[n] as our answer.

This approach has a time complexity of O(n) as we are iterating through all numbers from 2 to n only once. The space complexity is also O(n) as we are storing the minimum days required for all n values of oranges in an array.

Overall, this problem can be solved using dynamic programming in O(n) time and space complexity.

1