LeetCode 1553. Minimum Number of Days to Eat N Oranges Solution in Java, C++, Python & More | Explanation + Code

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 < 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).

PythonJavaC++GoTypeScript

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)(code-box)

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; } }(code-box)

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; } };(code-box)

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) }(code-box)

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); }(code-box)