Merge branch 'main' into feat/Crokily/mdEditor

Author: Crokily

.github/workflows/sync-uuid.yml

@@ -22,7 +22,7 @@ concurrency:
 jobs:
   backfill:
     # 防止 fork、限定 main、并避免机器人循环
-    if: github.repository == 'InvolutionHell/involutionhell.github.io' &&
+    if:
       (github.ref == 'refs/heads/main' || github.ref == 'refs/heads/feat/contributor') &&
       github.actor != 'github-actions[bot]'
     runs-on: ubuntu-latest

app/docs/CommunityShare/Leetcode/1004_translated.md

@@ -0,0 +1,69 @@
+---
+title: 1004.Maximum continuity1Number III Maximum continuity1Number III
+date: "2022.12.07-01:15"
+tags:
+  - - Python
+  - - solved
+    - answer
+abbrlink: ed19b576
+docId: ytg2bds2dnhzw37nrb3vassy
+---
+
+Today's daily question is too difficult，So find the problem by yourself。Today's question is a hash table+Sliding window algorithm，虽然感觉只用了Sliding window algorithm。
+
+```python
+
+class Solution:
+    def longestOnes(self, nums: List[int], k: int) -> int:
+        k_mean = k
+        # Used to record how many arrays there are now
+        flag = 0
+        # Used to record the maximum land number
+        max_flag = 0
+        for start in range(len(nums)):
+            tail = start
+            while k >= 0 and tail <= len(nums) - 1:
+                if nums[tail] == 1:
+                    tail += 1
+                    flag += 1
+                elif nums[tail] == 0 and k > 0:
+                    tail += 1
+                    k -= 1
+                    flag += 1
+                elif nums[tail] == 0 and k == 0:
+                    k = k_mean
+                    max_flag = max(max_flag, flag)
+                    flag = 0
+                    break
+                if tail == len(nums):
+                    max_flag = max(max_flag, flag)
+                    flag = 0
+                    break
+        return max_flag
+```
+
+This is my approach at the beginning，Although the double pointer is used，But there is no flexibility，Very empty feeling，Very dry slide。
+the following[@Lincoln](/u/lincoln)@Lincoln Big practice，Just as one record of yourself，不作为我的answer发表
+
+```
+class Solution:
+    def longestOnes(self, nums: List[int], k: int) -> int:
+        """
+        Thinking：1. k=0It can be understood as the problem of the maximum duplication sub -string
+             2. If the current window value-In the window1Number <= k: Then expand the window(right+1)
+                If the current window value-In the window1Number > k: Swipe the window to the right(left+1)
+        method：Hash table + Sliding window
+        """
+        n = len(nums)
+        o_res = 0
+        left = right = 0
+        while right < n:
+            if nums[right]== 1: o_res += 1
+            if right-left+1- o_res > k:
+                if nums[left]== 1: o_res -= 1
+                left += 1
+            right += 1
+        return right - left
+```
+
+Look atLincolnBigThinking，very clearly，remember

app/docs/CommunityShare/Leetcode/1234. 替换子串得到平衡字符串_translated.md

@@ -0,0 +1,73 @@
+---
+title: 1234. Replace the sub -string to get a balanced string One question daily
+date: "2024.01.01 0:00"
+tags:
+  - - Python
+  - - answer
+abbrlink: 56d97dcf
+docId: p8igr19xfxnuyo2lpngnr6fg
+---
+
+# topic：
+
+[1234. Replace the sub -string to get a balanced string.md](https://leetcode.cn/problems/replace-the-substring-for-balanced-string/description/)
+
+# Thought：
+
+`all()`:if bool(x) For all the values ​​in iterative objects x All for True，Then return True。 if可迭代对象为空，Then return True。
+
+Tongxiang dual pointer，The solution of the spiritual god of this question。
+If in this string，These four characters happen just to appear n/4 Second-rate，Then it is one「Balanced string」。
+if在待替换子串之外的任意字符的出现Second-rate数都**Exceed** $m=\dfrac{n}{4}$
+，So no matter how you replace it，都无法make这个字符的出现Second-rate数等于m。
+on the other hand，if在待替换子串之外的任意字符的出现Second-rate数都**不Exceed** m，
+
+> So it can be replaced ，make s 为Balanced string，即每个字符的出现Second-rate数均为 m。
+> For this question，The left and right end points of the sub -string are left and right，enumerate right，
+> if子串外的任意字符的出现Second-rate数都不Exceedm，The explanation from left arrive
+> rightThis sub -string can be to replace the sub -string，Length right−left+1
+> Update the minimum value of the answer，Move to the right left，Sumid sub -string length。
+
+# Code：
+
+```python
+class Solution:
+    def balancedString(self, s: str) -> int:
+        s_c = Counter(s)
+        n = len(s)
+        if all(s_c[v] <= n//4 for v in s_c):
+            return 0
+        ans, left = inf, 0
+        # enumerate右端点
+        for i, j in enumerate(s):
+            s_c[j] -= 1
+            while all(s_c[v] <= n // 4 for v in s_c):
+                ans = min(ans, i - left + 1)
+                s_c[s[left]] += 1
+                left += 1
+        return ans
+
+```
+
+```go
+func balancedString(s string) int {
+	cnt, m := ['X']int{}, len(s)/4
+	for _, c := range s {
+		cnt[c]++
+	}
+	if cnt['Q'] == m && cnt['W'] == m && cnt['E'] == m && cnt['R'] == m {
+		return 0
+	}
+	ans, left := len(s), 0
+	for right, c := range s {
+		cnt[c]--
+		for cnt['Q'] <= m && cnt['W'] <= m && cnt['E'] <= m && cnt['R'] <= m {
+			ans = min(ans, right-left+1)
+			cnt[s[left]]++
+			left++
+		}
+	}
+	return ans
+}
+func min(a, b int) int { if a > b { return b }; return a }
+```

app/docs/CommunityShare/Leetcode/1653. 使字符串平衡的最少删除次数_translated.md

@@ -0,0 +1,39 @@
+---
+title: 1653. The minimum number of times to balance the string balance.md
+date: "2024.01.01 0:00"
+tags:
+  - - Python
+  - - answer
+abbrlink: cac21f27
+docId: bsf0yz1zrmlz7masrdmq8fq6
+---
+
+# topic：
+
+[1653. The minimum number of times to balance the string balance.md](https://leetcode.cn/problems/minimum-deletions-to-make-string-balanced/description/)
+
+# Thought：
+
+ask：Why if-else Write (c - 'a') \* 2 - 1 It will be much faster？
+
+answer：CPU I encounter a branch（Condition jump instruction）Which branch of the code will be predicted when the code is executed，If the prediction is correct，
+CPU It will continue to execute the program in accordance with the predicted path。But if the prediction fails，CPU You need to roll back the previous instructions and load the correct instructions，To ensure the correctness of the program execution。
+
+For the data of this question，character ‘a’ and ‘b’ It can be considered to appear random，In this case, the branch prediction will be available 50% Probability failure。
+失败导致的回滚and加载操作需要消耗额外的 CPU cycle，If the branch can be removed at a smaller price，The situation of this question can inevitably bring efficiency improvement。
+
+Notice：This optimization method often reduces readability，It's best not to use in business code。
+
+# Code：
+
+```python
+class Solution:
+    def minimumDeletions(self, s: str) -> int:
+        ans = delete = s.count('a')
+        for c in s:
+            delete -= 1 if c == 'a' else -1
+            if delete < ans:  # Manually min It will be much faster
+                ans = delete
+        return ans
+
+```

app/docs/CommunityShare/Leetcode/2270. Number of Ways to Split Array.md

@@ -0,0 +1,41 @@
+---
+title: 2270. Number of Ways to Split Array.md
+date: "2025/1/14-9:31"
+tags:
+  - - Python
+  - - Answer
+abbrlink: c25bb550
+docId: a6inw303oslb7i5tcqj5xxx4
+---
+
+# QUESTION：
+
+[2270. Number of Ways to Split Array.md](https://leetcode.cn/problems/number-of-ways-to-split-array/description/)
+
+# My Think：
+
+`2 <= nums.length <= 105`, 因此我们可以直接获取到第一个数字, 初始状态就在指针于index0, 正要往index1走的时候.
+然后只需要一次For循环就可以搞定
+
+重点是第二个方法, 来自题解.
+
+# Code：
+
+```python
+class Solution:
+    def waysToSplitArray(self, nums: List[int]) -> int:
+        temp_sum = nums[0]
+        total_sum = sum(nums) - temp_sum
+        ans = 0
+        for i in range(1, len(nums)):
+            if temp_sum >= total_sum:
+                ans += 1
+            temp_sum += nums[i]
+            total_sum -= nums[i]
+        return ans
+```
+
+```python
+t = (sum(nums) + 1) // 2
+return sum(s >= t for s in accumulate(nums[:-1]))
+```

app/docs/CommunityShare/Leetcode/2679.矩阵中的和_translated.md

@@ -0,0 +1,90 @@
+---
+title: 2679.In the matrix and the harmony.md
+date: "2024.01.01 0:00"
+tags:
+  - - Python
+  - - answer
+  - - Array
+  - - matrix
+  - - Sort
+  - - simulation
+  - - heap（Priority queue）
+abbrlink: "5277100"
+docId: clx9mmqqvxipdfamqciuo146
+---
+
+# topic：
+
+[2679.In the matrix and the harmony.md](https://leetcode.cn/problems/sum-in-a-matrix/)
+
+# Thought：
+
+**One -line**
+First of all, the meaning is to find the largest number of each sub -list，Thenpopgo out，Finally ask for peace。
+The effect of traversing again and again is too bad，So I thought of using itzip一次性Traversal多个子Array。
+于yes先对每个子ArraySort，Then usezipTraversal，Find the maximum。
+for example：
+`nums = [[7,2,1],[6,4,2],[6,5,3],[3,2,1]]`
+Sort后：
+`nums = [[1,2,7],[2,4,6],[3,5,6],[1,2,3]]`
+Then usezipTraversal得到：
+`[(1,2,3,1),(2,4,5,2),(7,6,6,3)]`
+Find the maximum：
+`[3,5,7]`
+Context：
+`15`
+
+# Code：
+
+```python
+class Solution:
+    def matrixSum(self, nums: List[List[int]]) -> int:
+        return sum(max(i) for i in \
+        zip(*(sorted(sublist) for sublist in nums)))
+```
+
+### \*The role and the rolezip()explain
+
+```python
+nums = [[1,2,7],[2,4,6],[3,5,6],[1,2,3]]
+
+for i in range(len(nums[1])):
+    for j in range(len(nums)):
+        print(nums[j][i])
+# ans = 123124527663
+num1 = [1,2,7]
+num2 = [2,4,6]
+num3 = [3,5,6]
+num4 = [1,2,3]
+```
+
+`zip()`The function corresponds to one -to -one elements in multiple lists，Then return onezipObject，Can uselist()Function convert to list。
+
+```python
+for i in zip(num1, num2, num3, num4):
+    print(i)
+#(1, 2, 3, 1)
+#(2, 4, 5, 2)
+#(7, 6, 6, 3)
+```
+
+`*nums`The role ispythonNot a pointer，InsteadnumsEach element in the parameter is passed into the function。I understand here as a list。
+
+```python
+
+# Enumerate
+print(*nums)
+# [1, 2, 7] [2, 4, 6] [3, 5, 6] [1, 2, 3]
+```
+
+`zip(*nums)`WillnumsEach element is passed in as a parameterzip()In the function，Then return onezipObject，Can uselist()Function convert to list。
+`zip(*nums)`Equivalent to`zip(num1, num2, num3, num4)`，innum1, num2, num3, num4yesnumsElement。
+
+```python
+for i in zip(*nums):
+    print(i)
+# Equivalent
+#(1, 2, 3, 1)
+#(2, 4, 5, 2)
+#(7, 6, 6, 3)
+```

app/docs/CommunityShare/Leetcode/80_translated.md

@@ -0,0 +1,41 @@
+---
+title: Python beat98.40% collectionsofCounter method！
+date: "2024.01.01 0:00"
+tags:
+  - - Python
+  - - solved
+    - answer
+abbrlink: 73b5ce9c
+category: null
+docId: ryp6s59uwc10w2dywgs6f66h
+---
+
+Pyhton `collection` In the bag `Counter()` Be rightlist里出现of元素conduct计数，And the output is a dictionary。
+for example`nums=[1, 1, 1, 2, 2, 3]` ToCounter后of结果是`Counter({1: 3, 2: 2, 3: 1})`。
+
+1. So traverse a dictionary，when`value>3`of时候`value=2`，Can be greater than2indivualof元素计数become2indivual。
+1. So we will`Counter({1: 3, 2: 2, 3: 1})`become`Counter({1: 2, 2: 2, 3: 1})`后再Toelementsconductlist操作就可以得到改变后of列表了。
+
+---
+
+Due to the meaning of the question“Input the array「Quote」方式传递of”，So we willnumsJust fill in and fill in
+
+```python
+from collections import Counter # Imported package
+class Solution:
+    def removeDuplicates(self, nums: List[int]) -> List[int]:
+        dict1 = Counter(nums)
+        for i in dict1:
+            if dict1[i] > 2:
+                dict1[i] = 2
+        list1 = list(dict1.elements())
+        nums.clear() # clear the list
+        nums.extend(list1) # Add the collection to the list
+        return len(nums)
+```
+
+Complexity analysis
+
+time complexity：`O(n)`，in `n` 是数组of长度。We have a maximum of this array once。
+
+Spatial complexity：`O(1)`。我们只需要常数of空间存储若干变量。