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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);
    }
    
    

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