chore: add daily leetcode post 2335. 装满杯子需要的最短总时长_translated

Author: longsizhuo

app/docs/CommunityShare/Leetcode/2335. 装满杯子需要的最短总时长_translated.md

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+---
+title: 2335. The shortest total time to be filled with a cup One question daily
+date: '2024.01.01 0:00'
+tags:
+  - - Python
+  - - answer
+  - - dp
+  - - One question daily
+  - - golang
+abbrlink: 4400daa1
+---
+# topic：
+
+
+[2335. The shortest total time to be filled with a cup.md](https://leetcode.cn/problems/minimum-amount-of-time-to-fill-cups/description/)
+
+# Thought：
+
+1. This question is very simple，但是我好像没用贪心的Thought。Look at the second test case and find it found，In fact, the minimum number of seconds is to avoid a certain number as much as possible0。
+   So keep sorting，Always operate the two largest numbers。This sending can also be used for DoriamountCase，But because of sorting，I don't know if it can be used for a large number。
+	Look atylbBig，和我一样但是Big细节处理得很好，Two fewer judgments than me，Then it seems that the classification discussion method below may be a solution to the difficult situation。
+2. mathematical method？Sort the number of drinks from small to large，Set the quantity x，y，z。Our goal is to match the different drinks as much as possible。
+   like$x+y<=z$，The answer isz。like反之，Then set$t=(x+y-z)$，t是偶数The answer is
+   $ \frac{t−1}{2} +z$Plus one
+
+# Code：
+
+```python
+class Solution:
+    def fillCups(self, amount: List[int]) -> int:
+        amount.sort()
+        count = 0
+        # Try to avoid returning0
+        while amount[-1] > 0:
+            if amount[-1] > 0 and amount[1] > 0:
+                amount[-1] -= 1
+                amount[1] -= 1
+                count += 1
+            if amount[-1] > 0 and amount[1] == 0:
+                return count + amount[-1]
+            amount.sort()
+        return count
+```
+```go
+import "sort"
+
+func fillCups(amount []int) int {
+	ans := 0
+	for amount[0] + amount[1] + amount[2] > 0 {
+		sort.Ints(amount)
+		ans ++
+		amount[2] --
+		if amount[1] > 0{
+			amount[1] --
+		}
+	}
+	return ans
+}
+```