chore: add daily leetcode post 2341. 数组能形成多少数对_translated

Author: longsizhuo

app/docs/CommunityShare/Leetcode/2341. 数组能形成多少数对_translated.md

@@ -0,0 +1,47 @@
+---
+title: 2341. How much can the array be formed  One question daily
+tags:
+  - - Python
+  - - answer
+abbrlink: f953c753
+date: '2024.01.01 0:00'
+---
+
+# topic：
+
+[2341. How much can the array be formed.md](https://leetcode.cn/problems/maximum-number-of-pairs-in-array/description/)
+
+# Thought：
+## I am：
+I don't know if I saw it“Simple”Two words，This question has the initiative to think about the optimal solution。Actually this timeylbBig guy's hash table method is still fast。
+Sort the list，Whether two or two are equal to whether。
+## Hash tableThought：
+CounterAfter counting，`a+=v//2`,`b+=v%2`For each number x，
+if x Number of times v more than the 1，You can choose two from the array x Form a number pair，we willv Divide 2 Take down，
+You can get the current number x Number pairs that can be formed，Then we accumulate this number to the variable s middle。
+
+# Code：
+
+```python   Ordinary count
+class Solution:
+    def numberOfPairs(self, nums: List[int]) -> List[int]:
+        nums.sort()
+        ans = [0, len(nums)]
+        for index in range(1, len(nums)):
+            if nums[index - 1] == nums[index]:
+                ans[0] += 1
+                ans[1] -= 2
+                nums[index - 1] = nums[index] = -1
+        return ans
+```
+```python   Hash table
+class Solution:
+    def numberOfPairs(self, nums: List[int]) -> List[int]:
+        x = Counter(nums)
+        a = 0
+        b = 0
+        for k,v in x.items():
+            a+=v//2
+            b+=v%2
+        return [a,b]
+```