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Hard
2048
Weekly Contest 202 Q4
Memoization
Dynamic Programming

1553. Minimum Number of Days to Eat N Oranges

中文文档

Description

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

Solutions

Solution 1: Memoization Search

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

  1. Decrease $n$ by $1$;
  2. If $n$ can be divided by $2$, divide the value of $n$ by $2$;
  3. If $n$ can be divided by $3$, divide the value of $n$ by $3$.

Therefore, the problem is equivalent to finding the minimum number of days to reduce $n$ to $0$ through the above three ways.

We design a function $dfs(n)$, which represents the minimum number of days to reduce $n$ to $0$. The execution process of the function $dfs(n)$ is as follows:

  1. If $n &lt; 2$, return $n$;
  2. Otherwise, we can first reduce $n$ to a multiple of $2$ by $n \bmod 2$ operations of $1$, and then perform operation $2$ to reduce $n$ to $n/2$; we can also first reduce $n$ to a multiple of $3$ by $n \bmod 3$ operations of $1$, and then perform operation $3$ to reduce $n$ to $n/3$. We choose the minimum of the above two ways, that is, $1 + \min(n \bmod 2 + dfs(n/2), n \bmod 3 + dfs(n/3))$.

To avoid repeated calculations, we use the method of memoization search and store the calculated values of $dfs(n)$ in a hash table.

The time complexity is $O(\log^2 n)$, and the space complexity is $O(\log^2 n)$.

Python3

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)

Java

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;
    }
}

C++

class Solution {
public:
    unordered_map<int, int> f;

    int minDays(int n) {
        return dfs(n);
    }

    int dfs(int n) {
        if (n < 2) {
            return n;
        }
        if (f.count(n)) {
            return f[n];
        }
        int res = 1 + min(n % 2 + dfs(n / 2), n % 3 + dfs(n / 3));
        f[n] = res;
        return res;
    }
};

Go

func minDays(n int) int {
	f := map[int]int{0: 0, 1: 1}
	var dfs func(int) int
	dfs = func(n int) int {
		if v, ok := f[n]; ok {
			return v
		}
		res := 1 + min(n%2+dfs(n/2), n%3+dfs(n/3))
		f[n] = res
		return res
	}
	return dfs(n)
}

TypeScript

function minDays(n: number): number {
    const f: Record<number, number> = {};
    const dfs = (n: number): number => {
        if (n < 2) {
            return n;
        }
        if (f[n] !== undefined) {
            return f[n];
        }
        f[n] = 1 + Math.min((n % 2) + dfs((n / 2) | 0), (n % 3) + dfs((n / 3) | 0));
        return f[n];
    };
    return dfs(n);
}