1553. Minimum Number of Days to Eat N Oranges.md

1553. Minimum Number of Days to Eat N Oranges

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the last one in Weekly Contest 202.

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Difficulty : Hard

Related Topics : Dynamic Programming

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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.

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.

Example 3:

Input: n = 1
Output: 1

Example 4:

Input: n = 56
Output: 6

Constraints:

• 1 <= n <= 2*10^9

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Solution

• mine
  • Java
    • Memory Limit Exceeded

      //O(N)time
      //O(N)space
      public int minDays(int n) {
          int[] dp = new int[n + 1];
          for(int i = 1; i <= n; i++){
              dp[i] = dp[i - 1] + 1;
              if(i % 2 == 0) dp[i] = Math.min(dp[i], dp[i/2] + 1);
              if(i % 3 == 0) dp[i] = Math.min(dp[i], dp[i/3] + 1);
          }
          return dp[n];
      }
      

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• the most votes

• Runtime: 3 ms, faster than 98.84%, Memory Usage: 38.7 MB, less than 82.31% of Java online submissions

  // O(N)time
  // O(N)space
  Map<Integer, Integer> dp = new HashMap<>();
  public int minDays(int n) {
      if (n <= 1)
          return n;
      if (!dp.containsKey(n))
          dp.put(n, 1 + Math.min(n % 2 + minDays(n / 2), n % 3 + minDays(n / 3)));
      return dp.get(n);
  }
  

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