chore(docs): sync doc metadata [skip ci]

Author: longsizhuo

app/docs/CommunityShare/Leetcode/2293_translated.md

@@ -1,65 +1,39 @@
 ---
 title: One question daily 2293. Great mini game
-date: '2024.01.01 0:00'
+date: "2024.01.01 0:00"
 tags:
   - - Python
   - - solved
     - answer
 abbrlink: 9df6242c
-category:
+category: null
+docId: mssz5wgh368yp55qcvs1op5e
 ---
-Although it is a simple question，But there are recursive ideas。      
+
+Although it is a simple question，But there are recursive ideas。
 
 Constantly convert the one -dimensional list into a two -dimensional list for operation（This can be avoidedindexChaos caused by changes）
 
-Then useflagCount，Comparison of maximum and minimum values。    
+Then useflagCount，Comparison of maximum and minimum values。
 
 Just return to the end，Although it must be returned to a number，Still‘nums[0]’To avoid warning。
 
-
 ```html
-2293. Great mini game
-Simple
-39
-company
-Google Google
-Give you a bidding 0 Started integer array nums ，Its length is 2 Power。
-
-right nums Execute the following algorithm：
-
-set up n equal nums length，if n == 1 ，termination Algorithm。otherwise，create A new integer array newNums ，The length of the new array is n / 2 ，Bidding from 0 start。
-right于满足 0 <= i < n / 2 Every even Bidding i ，Will newNums[i] Assignment for min(nums[2 * i], nums[2 * i + 1]) 。
-right于满足 0 <= i < n / 2 Every odd number Bidding i ，Will newNums[i] Assignment for max(nums[2 * i], nums[2 * i + 1]) 。
-use newNums replace nums 。
-From steps 1 start repeat the whole process。
-After executing the algorithm，return nums The remaining numbers。
-
- 
-
-Exemplary example 1：
-
-
-
-enter：nums = [1,3,5,2,4,8,2,2]
-Output：1
-explain：repeat执行算法会得到下述数组。
-first round：nums = [1,5,4,2]
-second round：nums = [1,4]
-Third round：nums = [1]
-1 Is the last number left，return 1 。
-Exemplary example 2：
-
-enter：nums = [3]
-Output：3
-explain：3 Is the last number left，return 3 。
- 
-
-hint：
-
-1 <= nums.length <= 1024
-1 <= nums[i] <= 109
-nums.length yes 2 Power
-
+2293. Great mini game Simple 39 company Google Google Give you a bidding 0
+Started integer array nums ，Its length is 2 Power。 right nums Execute the
+following algorithm： set up n equal nums length，if n == 1 ，termination
+Algorithm。otherwise，create A new integer array newNums ，The length of the new
+array is n / 2 ，Bidding from 0 start。 right于满足 0 <= i < n / 2 Every even
+Bidding i ，Will newNums[i] Assignment for min(nums[2 * i], nums[2 * i + 1]) 。
+right于满足 0 <= i < n / 2 Every odd number Bidding i ，Will newNums[i]
+Assignment for max(nums[2 * i], nums[2 * i + 1]) 。 use newNums replace nums 。
+From steps 1 start repeat the whole process。 After executing the
+algorithm，return nums The remaining numbers。 Exemplary example 1： enter：nums
+= [1,3,5,2,4,8,2,2] Output：1 explain：repeat执行算法会得到下述数组。 first
+round：nums = [1,5,4,2] second round：nums = [1,4] Third round：nums = [1] 1 Is
+the last number left，return 1 。 Exemplary example 2： enter：nums = [3]
+Output：3 explain：3 Is the last number left，return 3 。 hint： 1 <=
+nums.length <= 1024 1 <= nums[i] <= 109 nums.length yes 2 Power
 ```
 
 ```python

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