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Commit ca1ddbe

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feat: add solutions to lc problem: No.3344 (#3731)
No.3344.Maximum Sized Array
1 parent 4524ae8 commit ca1ddbe

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20 files changed

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-26
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20 files changed

+661
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‎lcp/LCP 03. 机器人大冒险/README.md‎

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

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

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<!-- description:start -->
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<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>
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<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>
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<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>
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<p>返回分割后最小的和的最大值。</p>
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<p><strong>子数组</strong> 是数组中连续的部份。</p>
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<p>&nbsp;</p>
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‎solution/0600-0699/0636.Exclusive Time of Functions/README.md‎

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@@ -34,10 +34,10 @@ tags:
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<strong>输入:</strong>n = 2, logs = ["0:start:0","1:start:2","1:end:5","0:end:6"]
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<strong>输出:</strong>[3,4]
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<strong>解释:</strong>
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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。
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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。
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函数 0 在时间戳 6 的开始恢复执行,执行 1 个单位时间。
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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。
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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。
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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。
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函数 0 在时间戳 6 的开始恢复执行,执行 1 个单位时间。
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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。
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</pre>
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<p><strong>示例 2:</strong></p>

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

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

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<pre>
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<strong>Input:</strong> nums = [10,4,-8,7]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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There are three ways of splitting nums into two non-empty parts:
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- 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.
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- 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.
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<pre>
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<strong>Input:</strong> nums = [2,3,1,0]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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There are two valid splits in nums:
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- 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.
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- 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.
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- 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.
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</pre>
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‎solution/2500-2599/2526.Find Consecutive Integers from a Data Stream/README.md‎

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

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

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[null, false, false, true, false]
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<strong>Explanation</strong>
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DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3
46-
dataStream.consec(4); // Only 1 integer is parsed, so returns False.
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DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3
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dataStream.consec(4); // Only 1 integer is parsed, so returns False.
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dataStream.consec(4); // Only 2 integers are parsed.
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// Since 2 is less than k, returns False.
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dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True.
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// Since 2 is less than k, returns False.
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dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True.
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dataStream.consec(3); // The last k integers parsed in the stream are [4,4,3].
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// Since 3 is not equal to value, it returns False.
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</pre>

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

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<pre>
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<strong>Input:</strong> stations = [1,2,4,5,0], r = 1, k = 2
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<strong>Output:</strong> 5
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<strong>Explanation:</strong>
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One of the optimal ways is to install both the power stations at city 1.
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<strong>Explanation:</strong>
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One of the optimal ways is to install both the power stations at city 1.
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So stations will become [1,4,4,5,0].
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- City 0 is provided by 1 + 4 = 5 power stations.
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- City 1 is provided by 1 + 4 + 4 = 9 power stations.
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<pre>
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<strong>Input:</strong> stations = [4,4,4,4], r = 0, k = 3
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<strong>Output:</strong> 4
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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It can be proved that we cannot make the minimum power of a city greater than 4.
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</pre>
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‎solution/2500-2599/2530.Maximal Score After Applying K Operations/README_EN.md‎

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<strong>Explanation: </strong>You can do the following operations:
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Operation 1: Select i = 1, so nums becomes [1,<strong><u>4</u></strong>,3,3,3]. Your score increases by 10.
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Operation 2: Select i = 1, so nums becomes [1,<strong><u>2</u></strong>,3,3,3]. Your score increases by 4.
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Operation 3: Select i = 2, so nums becomes [1,1,<u><strong>1</strong></u>,3,3]. Your score increases by 3.
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Operation 3: Select i = 2, so nums becomes [1,2,<u><strong>1</strong></u>,3,3]. Your score increases by 3.
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The final score is 10 + 4 + 3 = 17.
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</pre>
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‎solution/3200-3299/3222.Find the Winning Player in Coin Game/README.md‎

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<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>
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<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>
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<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>
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<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>
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