Scale Engineer

Minimum Number of Days to Eat N Oranges

Hard
#1430Time: Exponential. The function branches into two calls for `n/2` and `n/3`, leading to a large number of redundant computations. This will result in a "Time Limit Exceeded" error for the given constraints.Space: `O(log n)` due to the maximum depth of the recursion stack.
Patterns
Dynamic ProgrammingMemoization

Prompt

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.

Given the integer n, 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.

 

Constraints:

  • 1 <= n <= 2 * 109

Approaches

3 approaches with complexity analysis and trade-offs.

This is a naive approach that directly translates the problem's recurrence relation into a recursive function. The function calculates the minimum days for n by recursively finding the minimum days for the states reachable from n. The key insight is that to use the division operations, we must first have a number of oranges divisible by 2 or 3. We can always reach such a number by eating one orange at a time. This leads to an optimized recurrence which is then implemented without any caching, leading to severe performance issues.

Algorithm

    1. Define a recursive function minDays(n).
    1. Handle base cases: if n <= 1, return n.
    1. Recursively calculate the days required by taking the path towards n/2: (n % 2) + 1 + minDays(n / 2).
    1. Recursively calculate the days required by taking the path towards n/3: (n % 3) + 1 + minDays(n / 3).
    1. Return the minimum of the two results.

Walkthrough

The core idea is to define a function, say minDays(n), which computes the minimum days to eat n oranges.

  • The base cases are minDays(0) = 0 and minDays(1) = 1.
  • For any n > 1, we consider the optimal path. To use the powerful division actions, the number of oranges must be divisible by 2 or 3. The fastest way to get to a multiple of 2 is to eat n % 2 oranges one by one. This takes n % 2 days. Then, one more day is spent on the division action itself. A similar logic applies for division by 3.
  • This leads to the recurrence: minDays(n) = min( (n % 2) + 1 + minDays(n/2), (n % 3) + 1 + minDays(n/3) ).
  • A direct implementation of this recurrence without any optimization will lead to re-computation of the same subproblems. For instance, minDays(30) will call minDays(15) and minDays(10). minDays(15) will call minDays(5), and minDays(10) will also call minDays(5). The subproblem for 5 oranges is solved twice. This redundancy grows exponentially.
public int minDays(int n) {    if (n <= 1) {        return n;    }    // Option via division by 2    int res1 = (n % 2) + 1 + minDays(n / 2);    // Option via division by 3    int res2 = (n % 3) + 1 + minDays(n / 3);    return Math.min(res1, res2);}

Complexity

Time

Exponential. The function branches into two calls for `n/2` and `n/3`, leading to a large number of redundant computations. This will result in a "Time Limit Exceeded" error for the given constraints.

Space

`O(log n)` due to the maximum depth of the recursion stack.

Trade-offs

Pros

  • Simple to write and understand the logic.

Cons

  • Extremely inefficient due to repeated calculations of the same subproblems.

  • Not feasible for large values of n and will result in a Time Limit Exceeded (TLE) error.

Solutions

class Solution {private  Map<Integer, Integer> f = new HashMap<>();public  int minDays(int n) { return dfs(n); }private  int dfs(int n) {    if (n < 2) {      return n;    }    if (f.containsKey(n)) {      return f.get(n);    }    int res = 1 + Math.min(n % 2 + dfs(n / 2), n % 3 + dfs(n / 3));    f.put(n, res);    return res;  }}

Video walkthrough

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