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

Author: longsizhuo

app/docs/CommunityShare/Leetcode/3138. Minimum Length of Anagram Concatenation.md

@@ -8,6 +8,7 @@ tags:
   - String
   - Counting
 abbrlink: d1339d55
+docId: o3knuvbpnki6isfjv3g5ohau
 ---
 
 # Description:
@@ -22,24 +23,25 @@ An anagram is formed by rearranging the letters of a string. For example, "aab",
 #### Example 1:
 
     Input: s = "abba"
-    
+
     Output: 2
-    
+
     Explanation:
-    
+
     One possible string t could be "ba".
 
 #### Example 2:
 
     Input: s = "cdef"
-    
+
     Output: 4
-    
+
     Explanation:
-    
+
     One possible string t could be "cdef", notice that t can be equal to s.
 
 # Thinking:
+
 Since the problem naturally suggests using a counting method (`Counter`), we need to find the minimum substring for each string. For example, for `abba`, the result is `ab`; for `cdef`, it's `cdef`.
 We iterate from length `1` (a single character) onwards, slicing the string to get the current substring.
 
@@ -50,6 +52,7 @@ Next, we only need to check if the count of each character in the current substr
 因为最开始我们对原始字符串进行了Counter，得到了字符数量和字符对应的字典。接下来我们只需要判断当前子串的Counter到的某个字符的数值乘以`n/k`是否等于原始字符串的Counter的值即可（即当前子串乘以x倍是否等于源字符串）。
 
 # Code:
+
 ```python
 import collections
 class Solution:
@@ -73,4 +76,4 @@ class Solution:
         for i in range(1, n+1):
             if n % i == 0 and check(i):
                 return i
-```
+```

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