feat: add solutions to lc problem: No.3344 (doocs#3731)

No.3344.Maximum Sized Array

Author: yanglbme

lcp/LCP 03. 机器人大冒险/README.md

@@ -247,7 +247,7 @@ class Solution {
         var visited: Set<[Int]> = []
         var i = 0, j = 0
         visited.insert([i, j])
-        
+
         for c in command {
             if c == "U" {
                 j += 1
@@ -256,16 +256,16 @@ class Solution {
             }
             visited.insert([i, j])
         }
-        
+
         func canReach(_ targetX: Int, _ targetY: Int) -> Bool {
             let k = min(targetX / i, targetY / j)
             return visited.contains([targetX - k * i, targetY - k * j])
         }
-        
+
         if !canReach(x, y) {
             return false
         }
-        
+
         for obstacle in obstacles {
             let obstacleX = obstacle[0]
             let obstacleY = obstacle[1]
@@ -276,7 +276,7 @@ class Solution {
                 return false
             }
         }
-        
+
         return true
     }
 }

solution/0400-0499/0410.Split Array Largest Sum/README.md

@@ -20,9 +20,11 @@ tags:
 
 <!-- description:start -->
 
-<p>给定一个非负整数数组 <code>nums</code> 和一个整数&nbsp;<code>k</code> ，你需要将这个数组分成&nbsp;<code>k</code><em>&nbsp;</em>个非空的连续子数组。</p>
+<p>给定一个非负整数数组 <code>nums</code> 和一个整数&nbsp;<code>k</code> ，你需要将这个数组分成&nbsp;<code>k</code><em>&nbsp;</em>个非空的连续子数组，使得这&nbsp;<code>k</code><em>&nbsp;</em>个子数组各自和的最大值 <strong>最小</strong>。</p>
 
-<p>设计一个算法使得这&nbsp;<code>k</code><em>&nbsp;</em>个子数组各自和的最大值最小。</p>
+<p>返回分割后最小的和的最大值。</p>
+
+<p><strong>子数组</strong> 是数组中连续的部份。</p>
 
 <p>&nbsp;</p>
 

solution/0600-0699/0636.Exclusive Time of Functions/README.md

@@ -34,10 +34,10 @@ tags:
 <strong>输入：</strong>n = 2, logs = ["0:start:0","1:start:2","1:end:5","0:end:6"]
 <strong>输出：</strong>[3,4]
 <strong>解释：</strong>
-函数 0 在时间戳 0 的起始开始执行，执行 2 个单位时间，于时间戳 1 的末尾结束执行。
-函数 1 在时间戳 2 的起始开始执行，执行 4 个单位时间，于时间戳 5 的末尾结束执行。
-函数 0 在时间戳 6 的开始恢复执行，执行 1 个单位时间。
-所以函数 0 总共执行 2 + 1 = 3 个单位时间，函数 1 总共执行 4 个单位时间。
+函数 0 在时间戳 0 的起始开始执行，执行 2 个单位时间，于时间戳 1 的末尾结束执行。 
+函数 1 在时间戳 2 的起始开始执行，执行 4 个单位时间，于时间戳 5 的末尾结束执行。 
+函数 0 在时间戳 6 的开始恢复执行，执行 1 个单位时间。 
+所以函数 0 总共执行 2 + 1 = 3 个单位时间，函数 1 总共执行 4 个单位时间。 
 </pre>
 
 <p><strong>示例 2：</strong></p>

solution/2200-2299/2265.Count Nodes Equal to Average of Subtree/README_EN.md

@@ -35,7 +35,7 @@ tags:
 <pre>
 <strong>Input:</strong> root = [4,8,5,0,1,null,6]
 <strong>Output:</strong> 5
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 For the node with value 4: The average of its subtree is (4 + 8 + 5 + 0 + 1 + 6) / 6 = 24 / 6 = 4.
 For the node with value 5: The average of its subtree is (5 + 6) / 2 = 11 / 2 = 5.
 For the node with value 0: The average of its subtree is 0 / 1 = 0.

solution/2200-2299/2270.Number of Ways to Split Array/README_EN.md

@@ -36,7 +36,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [10,4,-8,7]
 <strong>Output:</strong> 2
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 There are three ways of splitting nums into two non-empty parts:
 - Split nums at index 0. Then, the first part is [10], and its sum is 10. The second part is [4,-8,7], and its sum is 3. Since 10 &gt;= 3, i = 0 is a valid split.
 - Split nums at index 1. Then, the first part is [10,4], and its sum is 14. The second part is [-8,7], and its sum is -1. Since 14 &gt;= -1, i = 1 is a valid split.
@@ -49,9 +49,9 @@ Thus, the number of valid splits in nums is 2.
 <pre>
 <strong>Input:</strong> nums = [2,3,1,0]
 <strong>Output:</strong> 2
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 There are two valid splits in nums:
-- Split nums at index 1. Then, the first part is [2,3], and its sum is 5. The second part is [1,0], and its sum is 1. Since 5 &gt;= 1, i = 1 is a valid split.
+- Split nums at index 1. Then, the first part is [2,3], and its sum is 5. The second part is [1,0], and its sum is 1. Since 5 &gt;= 1, i = 1 is a valid split. 
 - Split nums at index 2. Then, the first part is [2,3,1], and its sum is 6. The second part is [0], and its sum is 0. Since 6 &gt;= 0, i = 2 is a valid split.
 </pre>
 

solution/2500-2599/2526.Find Consecutive Integers from a Data Stream/README.md

@@ -43,7 +43,7 @@ tags:
 [null, false, false, true, false]
 
 <strong>解释：</strong>
-DataStream dataStream = new DataStream(4, 3); // value = 4, k = 3
+DataStream dataStream = new DataStream(4, 3); // value = 4, k = 3 
 dataStream.consec(4); // 数据流中只有 1 个整数，所以返回 False 。
 dataStream.consec(4); // 数据流中只有 2 个整数
                       // 由于 2 小于 k ，返回 False 。

solution/2500-2599/2526.Find Consecutive Integers from a Data Stream/README_EN.md

@@ -42,11 +42,11 @@ tags:
 [null, false, false, true, false]
 
 <strong>Explanation</strong>
-DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3
-dataStream.consec(4); // Only 1 integer is parsed, so returns False.
+DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3 
+dataStream.consec(4); // Only 1 integer is parsed, so returns False. 
 dataStream.consec(4); // Only 2 integers are parsed.
-                      // Since 2 is less than k, returns False.
-dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True.
+                      // Since 2 is less than k, returns False. 
+dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True. 
 dataStream.consec(3); // The last k integers parsed in the stream are [4,4,3].
                       // Since 3 is not equal to value, it returns False.
 </pre>

solution/2500-2599/2528.Maximize the Minimum Powered City/README_EN.md

@@ -45,8 +45,8 @@ tags:
 <pre>
 <strong>Input:</strong> stations = [1,2,4,5,0], r = 1, k = 2
 <strong>Output:</strong> 5
-<strong>Explanation:</strong>
-One of the optimal ways is to install both the power stations at city 1.
+<strong>Explanation:</strong> 
+One of the optimal ways is to install both the power stations at city 1. 
 So stations will become [1,4,4,5,0].
 - City 0 is provided by 1 + 4 = 5 power stations.
 - City 1 is provided by 1 + 4 + 4 = 9 power stations.
@@ -62,7 +62,7 @@ Since it is not possible to obtain a larger power, we return 5.
 <pre>
 <strong>Input:</strong> stations = [4,4,4,4], r = 0, k = 3
 <strong>Output:</strong> 4
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 It can be proved that we cannot make the minimum power of a city greater than 4.
 </pre>
 

solution/2500-2599/2530.Maximal Score After Applying K Operations/README_EN.md

@@ -51,7 +51,7 @@ tags:
 <strong>Explanation: </strong>You can do the following operations:
 Operation 1: Select i = 1, so nums becomes [1,<strong><u>4</u></strong>,3,3,3]. Your score increases by 10.
 Operation 2: Select i = 1, so nums becomes [1,<strong><u>2</u></strong>,3,3,3]. Your score increases by 4.
-Operation 3: Select i = 2, so nums becomes [1,1,<u><strong>1</strong></u>,3,3]. Your score increases by 3.
+Operation 3: Select i = 2, so nums becomes [1,2,<u><strong>1</strong></u>,3,3]. Your score increases by 3.
 The final score is 10 + 4 + 3 = 17.
 </pre>
 

solution/3200-3299/3222.Find the Winning Player in Coin Game/README.md

@@ -22,7 +22,7 @@ tags:
 
 <p>给你两个 <strong>正</strong>&nbsp;整数&nbsp;<code>x</code>&nbsp;和&nbsp;<code>y</code>&nbsp;，分别表示价值为 75 和 10 的硬币的数目。</p>
 
-<p>Alice 和 Bob 正在玩一个游戏。每一轮中，Alice&nbsp;先进行操作，Bob 后操作。每次操作中，玩家需要拿出价值 <b>总和</b>&nbsp;为 115 的硬币。如果一名玩家无法执行此操作，那么这名玩家 <strong>输掉</strong>&nbsp;游戏。</p>
+<p>Alice 和 Bob 正在玩一个游戏。每一轮中，Alice&nbsp;先进行操作，Bob 后操作。每次操作中，玩家需要拿走价值 <b>总和</b>&nbsp;为 115 的硬币。如果一名玩家无法执行此操作，那么这名玩家 <strong>输掉</strong>&nbsp;游戏。</p>
 
 <p>两名玩家都采取 <strong>最优</strong>&nbsp;策略，请你返回游戏的赢家。</p>