feat: add solutions to lc problem: No.3431 (#3988)

No.3431.Minimum Unlocked Indices to Sort Nums

Author: yanglbme

solution/2000-2099/2011.Final Value of Variable After Performing Operations/README.md

@@ -50,7 +50,7 @@ X++：X 加 1 ，X =  0 + 1 =  1
 <pre>
 <strong>输入：</strong>operations = ["++X","++X","X++"]
 <strong>输出：</strong>3
-<strong>解释：</strong>操作按下述步骤执行：
+<strong>解释：</strong>操作按下述步骤执行： 
 最初，X = 0
 ++X：X 加 1 ，X = 0 + 1 = 1
 ++X：X 加 1 ，X = 1 + 1 = 2

solution/2100-2199/2151.Maximum Good People Based on Statements/README.md

@@ -86,7 +86,7 @@ tags:
             - <strong>在认为 0 是坏人但说真话的情况下，这组玩家中没有一个好人。</strong>
         - 说假话。在这种情况下，1 是好人。
             - <strong>在认为 0 是坏人且说假话的情况下，这组玩家中只有一个好人。</strong>
-在最佳情况下，至多有一个好人，所以返回 1 。
+在最佳情况下，至多有一个好人，所以返回 1 。 
 注意，能得到此结论的方法不止一种。
 </pre>
 

solution/2100-2199/2165.Smallest Value of the Rearranged Number/README_EN.md

@@ -31,7 +31,7 @@ tags:
 <pre>
 <strong>Input:</strong> num = 310
 <strong>Output:</strong> 103
-<strong>Explanation:</strong> The possible arrangements for the digits of 310 are 013, 031, 103, 130, 301, 310.
+<strong>Explanation:</strong> The possible arrangements for the digits of 310 are 013, 031, 103, 130, 301, 310. 
 The arrangement with the smallest value that does not contain any leading zeros is 103.
 </pre>
 

solution/2100-2199/2169.Count Operations to Obtain Zero/README_EN.md

@@ -35,7 +35,7 @@ tags:
 <pre>
 <strong>Input:</strong> num1 = 2, num2 = 3
 <strong>Output:</strong> 3
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 - Operation 1: num1 = 2, num2 = 3. Since num1 &lt; num2, we subtract num1 from num2 and get num1 = 2, num2 = 3 - 2 = 1.
 - Operation 2: num1 = 2, num2 = 1. Since num1 &gt; num2, we subtract num2 from num1.
 - Operation 3: num1 = 1, num2 = 1. Since num1 == num2, we subtract num2 from num1.
@@ -48,7 +48,7 @@ So the total number of operations required is 3.
 <pre>
 <strong>Input:</strong> num1 = 10, num2 = 10
 <strong>Output:</strong> 1
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 - Operation 1: num1 = 10, num2 = 10. Since num1 == num2, we subtract num2 from num1 and get num1 = 10 - 10 = 0.
 Now num1 = 0 and num2 = 10. Since num1 == 0, we are done.
 So the total number of operations required is 1.

solution/2100-2199/2178.Maximum Split of Positive Even Integers/README_EN.md

@@ -53,7 +53,7 @@ Thus, we return an empty array.
 <pre>
 <strong>Input:</strong> finalSum = 28
 <strong>Output:</strong> [6,8,2,12]
-<strong>Explanation:</strong> The following are valid splits: <code>(2 + 26)</code>, <code>(6 + 8 + 2 + 12)</code>, and <code>(4 + 24)</code>.
+<strong>Explanation:</strong> The following are valid splits: <code>(2 + 26)</code>, <code>(6 + 8 + 2 + 12)</code>, and <code>(4 + 24)</code>. 
 <code>(6 + 8 + 2 + 12)</code> has the maximum number of integers, which is 4. Thus, we return [6,8,2,12].
 Note that [10,2,4,12], [6,2,4,16], etc. are also accepted.
 </pre>

solution/2200-2299/2206.Divide Array Into Equal Pairs/README_EN.md

@@ -38,7 +38,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [3,2,3,2,2,2]
 <strong>Output:</strong> true
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 There are 6 elements in nums, so they should be divided into 6 / 2 = 3 pairs.
 If nums is divided into the pairs (2, 2), (3, 3), and (2, 2), it will satisfy all the conditions.
 </pre>
@@ -48,7 +48,7 @@ If nums is divided into the pairs (2, 2), (3, 3), and (2, 2), it will satisfy al
 <pre>
 <strong>Input:</strong> nums = [1,2,3,4]
 <strong>Output:</strong> false
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 There is no way to divide nums into 4 / 2 = 2 pairs such that the pairs satisfy every condition.
 </pre>
 

solution/2200-2299/2229.Check if an Array Is Consecutive/README.md

@@ -41,9 +41,9 @@ tags:
 <strong>输入：</strong>nums = [1,3]
 <strong>输出：</strong>false
 <strong>解释：
-</strong>最小值是 1 ，数组长度为 2 。
-范围 [x, x + n - 1] 中的所有值没有都出现在 nums 中：[1, 1 + 2 - 1] = [1, 2] = (1, 2) 。
-因此，nums 不是一个连贯数组。
+</strong>最小值是 1 ，数组长度为 2 。 
+范围 [x, x + n - 1] 中的所有值没有都出现在 nums 中：[1, 1 + 2 - 1] = [1, 2] = (1, 2) 。 
+因此，nums 不是一个连贯数组。 
 </pre>
 
 <p><strong>示例 3：</strong></p>

solution/2200-2299/2243.Calculate Digit Sum of a String/README.md

@@ -39,11 +39,11 @@ tags:
 <strong>输出：</strong>"135"
 <strong>解释：</strong>
 - 第一轮，将 s 分成："111"、"112"、"222" 和 "23" 。
-  接着，计算每一组的数字和：1 + 1 + 1 = 3、1 + 1 + 2 = 4、2 + 2 + 2 = 6 和 2 + 3 = 5 。
+  接着，计算每一组的数字和：1 + 1 + 1 = 3、1 + 1 + 2 = 4、2 + 2 + 2 = 6 和 2 + 3 = 5 。 
 &nbsp; 这样，s 在第一轮之后变成 "3" + "4" + "6" + "5" = "3465" 。
 - 第二轮，将 s 分成："346" 和 "5" 。
 &nbsp; 接着，计算每一组的数字和：3 + 4 + 6 = 13 、5 = 5 。
-&nbsp; 这样，s 在第二轮之后变成 "13" + "5" = "135" 。
+&nbsp; 这样，s 在第二轮之后变成 "13" + "5" = "135" 。 
 现在，s.length &lt;= k ，所以返回 "135" 作为答案。
 </pre>
 
@@ -53,7 +53,7 @@ tags:
 <strong>输出：</strong>"000"
 <strong>解释：</strong>
 将 "000", "000", and "00".
-接着，计算每一组的数字和：0 + 0 + 0 = 0 、0 + 0 + 0 = 0 和 0 + 0 = 0 。
+接着，计算每一组的数字和：0 + 0 + 0 = 0 、0 + 0 + 0 = 0 和 0 + 0 = 0 。 
 s 变为 "0" + "0" + "0" = "000" ，其长度等于 k ，所以返回 "000" 。
 </pre>
 

solution/2200-2299/2243.Calculate Digit Sum of a String/README_EN.md

@@ -37,13 +37,13 @@ tags:
 <pre>
 <strong>Input:</strong> s = &quot;11111222223&quot;, k = 3
 <strong>Output:</strong> &quot;135&quot;
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 - For the first round, we divide s into groups of size 3: &quot;111&quot;, &quot;112&quot;, &quot;222&quot;, and &quot;23&quot;.
-  ​​​​​Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5.
+  ​​​​​Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5. 
 &nbsp; So, s becomes &quot;3&quot; + &quot;4&quot; + &quot;6&quot; + &quot;5&quot; = &quot;3465&quot; after the first round.
 - For the second round, we divide s into &quot;346&quot; and &quot;5&quot;.
-&nbsp; Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5.
-&nbsp; So, s becomes &quot;13&quot; + &quot;5&quot; = &quot;135&quot; after second round.
+&nbsp; Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5. 
+&nbsp; So, s becomes &quot;13&quot; + &quot;5&quot; = &quot;135&quot; after second round. 
 Now, s.length &lt;= k, so we return &quot;135&quot; as the answer.
 </pre>
 
@@ -52,9 +52,9 @@ Now, s.length &lt;= k, so we return &quot;135&quot; as the answer.
 <pre>
 <strong>Input:</strong> s = &quot;00000000&quot;, k = 3
 <strong>Output:</strong> &quot;000&quot;
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 We divide s into &quot;000&quot;, &quot;000&quot;, and &quot;00&quot;.
-Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0.
+Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0. 
 s becomes &quot;0&quot; + &quot;0&quot; + &quot;0&quot; = &quot;000&quot;, whose length is equal to k, so we return &quot;000&quot;.
 </pre>
 

solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/README.md

@@ -52,7 +52,7 @@ tags:
 <strong>输入：</strong>nums = [1,2,3,4]
 <strong>输出：</strong>1
 <strong>解释：</strong>
-子数组按位与运算的最大值是 4 。
+子数组按位与运算的最大值是 4 。 
 能得到此结果的最长子数组是 [4]，所以返回 1 。
 </pre>