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committedDec 4, 2022
feat: add solutions to lc problems: No.2490~2493
* No.2490.Circular Sentence * No.2491.Divide Players Into Teams of Equal Skill * No.2492.Minimum Score of a Path Between Two Cities * No.2493.Divide Nodes Into the Maximum Number of Groups
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‎solution/1300-1399/1381.Design a Stack With Increment Operation/README_EN.md

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## Description
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<p>Design a stack which supports the following operations.</p>
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<p>Design a stack that supports increment operations on its elements.</p>
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<p>Implement the <code>CustomStack</code> class:</p>
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<ul>
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<li><code>CustomStack(int maxSize)</code> Initializes the object with <code>maxSize</code> which is the maximum number of elements in the stack or do nothing if the stack reached the <code>maxSize</code>.</li>
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<li><code>void push(int x)</code>&nbsp;Adds <code>x</code> to the top of the stack if the stack hasn&#39;t reached the <code>maxSize</code>.</li>
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<li><code>int pop()</code>&nbsp;Pops and returns the top of stack or <strong>-1</strong> if the stack is empty.</li>
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<li><code>void inc(int k, int val)</code> Increments the bottom <code>k</code> elements of the stack by <code>val</code>. If there are less than <code>k</code> elements in the stack, just increment all the elements in the stack.</li>
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<li><code>CustomStack(int maxSize)</code> Initializes the object with <code>maxSize</code> which is the maximum number of elements in the stack.</li>
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<li><code>void push(int x)</code> Adds <code>x</code> to the top of the stack if the stack has not reached the <code>maxSize</code>.</li>
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<li><code>int pop()</code> Pops and returns the top of the stack or <code>-1</code> if the stack is empty.</li>
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<li><code>void inc(int k, int val)</code> Increments the bottom <code>k</code> elements of the stack by <code>val</code>. If there are less than <code>k</code> elements in the stack, increment all the elements in the stack.</li>
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</ul>
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<p>&nbsp;</p>
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<strong>Output</strong>
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[null,null,null,2,null,null,null,null,null,103,202,201,-1]
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<strong>Explanation</strong>
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CustomStack customStack = new CustomStack(3); // Stack is Empty []
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customStack.push(1); // stack becomes [1]
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customStack.push(2); // stack becomes [1, 2]
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customStack.pop(); // return 2 --&gt; Return top of the stack 2, stack becomes [1]
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customStack.push(2); // stack becomes [1, 2]
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customStack.push(3); // stack becomes [1, 2, 3]
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customStack.push(4); // stack still [1, 2, 3], Don&#39;t add another elements as size is 4
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customStack.increment(5, 100); // stack becomes [101, 102, 103]
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customStack.increment(2, 100); // stack becomes [201, 202, 103]
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customStack.pop(); // return 103 --&gt; Return top of the stack 103, stack becomes [201, 202]
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customStack.pop(); // return 202 --&gt; Return top of the stack 102, stack becomes [201]
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customStack.pop(); // return 201 --&gt; Return top of the stack 101, stack becomes []
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customStack.pop(); // return -1 --&gt; Stack is empty return -1.
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CustomStack stk = new CustomStack(3); // Stack is Empty []
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stk.push(1); // stack becomes [1]
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stk.push(2); // stack becomes [1, 2]
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stk.pop(); // return 2 --&gt; Return top of the stack 2, stack becomes [1]
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stk.push(2); // stack becomes [1, 2]
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stk.push(3); // stack becomes [1, 2, 3]
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stk.push(4); // stack still [1, 2, 3], Do not add another elements as size is 4
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stk.increment(5, 100); // stack becomes [101, 102, 103]
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stk.increment(2, 100); // stack becomes [201, 202, 103]
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stk.pop(); // return 103 --&gt; Return top of the stack 103, stack becomes [201, 202]
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stk.pop(); // return 202 --&gt; Return top of the stack 202, stack becomes [201]
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stk.pop(); // return 201 --&gt; Return top of the stack 201, stack becomes []
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stk.pop(); // return -1 --&gt; Stack is empty return -1.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= maxSize &lt;= 1000</code></li>
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<li><code>1 &lt;= x &lt;= 1000</code></li>
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<li><code>1 &lt;= k &lt;= 1000</code></li>
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<li><code>1 &lt;= maxSize, x, k &lt;= 1000</code></li>
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<li><code>0 &lt;= val &lt;= 100</code></li>
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<li>At most&nbsp;<code>1000</code>&nbsp;calls will be made to each method of <code>increment</code>, <code>push</code> and <code>pop</code> each separately.</li>
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<li>At most <code>1000</code> calls will be made to each method of <code>increment</code>, <code>push</code> and <code>pop</code> each separately.</li>
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</ul>
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## Solutions

‎solution/2400-2499/2490.Circular Sentence/README.md

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# [2490. 回环句](https://leetcode.cn/problems/circular-sentence)
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[English Version](/solution/2400-2499/2490.Circular%20Sentence/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p><strong>句子</strong> 是由单个空格分隔的一组单词，且不含前导或尾随空格。</p>
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<ul>
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<li>例如，<code>"Hello World"</code>、<code>"HELLO"</code>、<code>"hello world hello world"</code> 都是符合要求的句子。</li>
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</ul>
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<p>单词 <strong>仅</strong> 由大写和小写英文字母组成。且大写和小写字母会视作不同字符。</p>
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<p>如果句子满足下述全部条件，则认为它是一个 <strong>回环句</strong> ：</p>
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<ul>
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<li>单词的最后一个字符和下一个单词的第一个字符相等。</li>
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<li>最后一个单词的最后一个字符和第一个单词的第一个字符相等。</li>
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</ul>
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<p>例如，<code>"leetcode exercises sound delightful"</code>、<code>"eetcode"</code>、<code>"leetcode eats soul"</code> 都是回环句。然而，<code>"Leetcode is cool"</code>、<code>"happy Leetcode"</code>、<code>"Leetcode"</code> 和 <code>"I like Leetcode"</code> 都 <strong>不</strong> 是回环句。</p>
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<p>给你一个字符串 <code>sentence</code> ，请你判断它是不是一个回环句。如果是，返回 <code>true</code><em> </em>；否则，返回 <code>false</code> 。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<strong>输入：</strong>sentence = "leetcode exercises sound delightful"
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<strong>输出：</strong>true
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<strong>解释：</strong>句子中的单词是 ["leetcode", "exercises", "sound", "delightful"] 。
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- leetcod<strong><em>e</em></strong> 的最后一个字符和 <strong><em>e</em></strong>xercises 的第一个字符相等。
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- exercise<em><strong>s</strong></em> 的最后一个字符和 <em><strong>s</strong></em>ound 的第一个字符相等。
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- <em><strong>s</strong></em>ound 的最后一个字符和 delightfu<em><strong>l</strong></em> 的第一个字符相等。
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- delightfu<em><strong>l</strong></em> 的最后一个字符和 <em><strong>l</strong></em>eetcode 的第一个字符相等。
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这个句子是回环句。</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>sentence = "eetcode"
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<strong>输出：</strong>true
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<strong>解释：</strong>句子中的单词是 ["eetcode"] 。
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- eetcod<em><strong>e</strong></em> 的最后一个字符和 eetcod<em><strong>e</strong></em> 的第一个字符相等。
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这个句子是回环句。</pre>
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<p><strong>示例 3：</strong></p>
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<pre>
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<strong>输入：</strong>sentence = "Leetcode is cool"
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<strong>输出：</strong>false
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<strong>解释：</strong>句子中的单词是 ["Leetcode", "is", "cool"] 。
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- Leetcod<em><strong>e</strong></em>&nbsp;的最后一个字符和 i<strong><em>s</em></strong> 的第一个字符 <strong>不</strong> 相等。
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这个句子 <strong>不</strong> 是回环句。</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= sentence.length &lt;= 500</code></li>
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<li><code>sentence</code> 仅由大小写英文字母和空格组成</li>
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<li><code>sentence</code> 中的单词由单个空格进行分隔</li>
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<li>不含任何前导或尾随空格</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：模拟**
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根据题意模拟即可。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是字符串的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def isCircularSentence(self, sentence: str) -> bool:
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sentence = sentence.split()
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return all(s[0] == sentence[i - 1][-1] for i, s in enumerate(sentence))
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public boolean isCircularSentence(String sentence) {
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if (sentence.charAt(0) != sentence.charAt(sentence.length() - 1)) {
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return false;
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}
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String[] ss = sentence.split(" ");
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for (int i = 1; i < ss.length; ++i) {
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if (ss[i].charAt(0) != ss[i - 1].charAt(ss[i - 1].length() - 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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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool isCircularSentence(string sentence) {
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if (sentence[0] != sentence[sentence.size() - 1]) return false;
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istringstream is(sentence);
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vector<string> ss;
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string s;
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while (is >> s) ss.emplace_back(s);
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for (int i = 1; i < ss.size(); ++i) {
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if (ss[i][0] != ss[i - 1][ss[i - 1].size() - 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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};
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```
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### **Go**
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```go
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func isCircularSentence(sentence string) bool {
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if sentence[0] != sentence[len(sentence)-1] {
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return false
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}
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ss := strings.Split(sentence, " ")
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for i := 1; i < len(ss); i++ {
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if ss[i][0] != ss[i-1][len(ss[i-1])-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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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2400-2499/2490.Circular Sentence/README_EN.md

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# [2490. Circular Sentence](https://leetcode.com/problems/circular-sentence)
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[中文文档](/solution/2400-2499/2490.Circular%20Sentence/README.md)
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## Description
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<p>A <strong>sentence</strong> is a list of words that are separated by a<strong> single</strong> space with no leading or trailing spaces.</p>
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<ul>
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<li>For example, <code>&quot;Hello World&quot;</code>, <code>&quot;HELLO&quot;</code>, <code>&quot;hello world hello world&quot;</code> are all sentences.</li>
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</ul>
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<p>Words consist of <strong>only</strong> uppercase and lowercase English letters. Uppercase and lowercase English letters are considered different.</p>
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<p>A sentence is <strong>circular </strong>if:</p>
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<ul>
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<li>The last character of a word is equal to the first character of the next word.</li>
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<li>The last character of the last word is equal to the first character of the first word.</li>
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</ul>
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<p>For example, <code>&quot;leetcode exercises sound delightful&quot;</code>, <code>&quot;eetcode&quot;</code>, <code>&quot;leetcode eats soul&quot; </code>are all circular sentences. However, <code>&quot;Leetcode is cool&quot;</code>, <code>&quot;happy Leetcode&quot;</code>, <code>&quot;Leetcode&quot;</code> and <code>&quot;I like Leetcode&quot;</code> are <strong>not</strong> circular sentences.</p>
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<p>Given a string <code>sentence</code>, return <code>true</code><em> if it is circular</em>. Otherwise, return <code>false</code>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> sentence = &quot;leetcode exercises sound delightful&quot;
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<strong>Output:</strong> true
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<strong>Explanation:</strong> The words in sentence are [&quot;leetcode&quot;, &quot;exercises&quot;, &quot;sound&quot;, &quot;delightful&quot;].
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- leetcod<u>e</u>&#39;s&nbsp;last character is equal to <u>e</u>xercises&#39;s first character.
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- exercise<u>s</u>&#39;s&nbsp;last character is equal to <u>s</u>ound&#39;s first character.
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- soun<u>d</u>&#39;s&nbsp;last character is equal to <u>d</u>elightful&#39;s first character.
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- delightfu<u>l</u>&#39;s&nbsp;last character is equal to <u>l</u>eetcode&#39;s first character.
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The sentence is circular.</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> sentence = &quot;eetcode&quot;
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<strong>Output:</strong> true
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<strong>Explanation:</strong> The words in sentence are [&quot;eetcode&quot;].
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- eetcod<u>e</u>&#39;s&nbsp;last character is equal to <u>e</u>etcode&#39;s first character.
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The sentence is circular.</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> sentence = &quot;Leetcode is cool&quot;
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<strong>Output:</strong> false
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<strong>Explanation:</strong> The words in sentence are [&quot;Leetcode&quot;, &quot;is&quot;, &quot;cool&quot;].
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- Leetcod<u>e</u>&#39;s&nbsp;last character is <strong>not</strong> equal to <u>i</u>s&#39;s first character.
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The sentence is <strong>not</strong> circular.</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= sentence.length &lt;= 500</code></li>
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<li><code>sentence</code> consist of only lowercase and uppercase English letters and spaces.</li>
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<li>The words in <code>sentence</code> are separated by a single space.</li>
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<li>There are no leading or trailing spaces.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def isCircularSentence(self, sentence: str) -> bool:
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sentence = sentence.split()
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return all(s[0] == sentence[i - 1][-1] for i, s in enumerate(sentence))
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```
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### **Java**
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```java
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class Solution {
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public boolean isCircularSentence(String sentence) {
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if (sentence.charAt(0) != sentence.charAt(sentence.length() - 1)) {
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return false;
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}
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String[] ss = sentence.split(" ");
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for (int i = 1; i < ss.length; ++i) {
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if (ss[i].charAt(0) != ss[i - 1].charAt(ss[i - 1].length() - 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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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool isCircularSentence(string sentence) {
105+
if (sentence[0] != sentence[sentence.size() - 1]) return false;
106+
istringstream is(sentence);
107+
vector<string> ss;
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string s;
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while (is >> s) ss.emplace_back(s);
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for (int i = 1; i < ss.size(); ++i) {
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if (ss[i][0] != ss[i - 1][ss[i - 1].size() - 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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};
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```
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### **Go**
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```go
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func isCircularSentence(sentence string) bool {
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if sentence[0] != sentence[len(sentence)-1] {
125+
return false
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}
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ss := strings.Split(sentence, " ")
128+
for i := 1; i < len(ss); i++ {
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if ss[i][0] != ss[i-1][len(ss[i-1])-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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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2400-2499/2490.Circular Sentence/Solution.cpp

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class Solution {
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public:
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bool isCircularSentence(string sentence) {
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if (sentence[0] != sentence[sentence.size() - 1]) return false;
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istringstream is(sentence);
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vector<string> ss;
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string s;
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while (is >> s) ss.emplace_back(s);
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for (int i = 1; i < ss.size(); ++i) {
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if (ss[i][0] != ss[i - 1][ss[i - 1].size() - 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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};

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