1676. Minimum Number Of Days To Eat N Oranges¶

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1676. Minimum Number of Days to Eat N Oranges

Hard

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

Solution¶

import java.util.HashMap;

class Solution {
    private HashMap<Integer, Integer> memo = new HashMap<>();
    public int minDays(int n) {
        if (n == 0)
            return 0;
        if (n == 1)
            return 1;
        if (memo.containsKey(n))
            return memo.get(n);

        int op1 = (n % 2) + 1 + minDays(n / 2);
        int op2 = (n % 3) + 1 + minDays(n / 3);
        int result = Math.min(op1, op2);
        memo.put(n, result);

        return result;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here