# 1553. Minimum Number of Days to Eat N Oranges

## 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

• 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;
}
}

• 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;
}
};

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

return dfs(n)


• 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)
}

• 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);
}