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feat: add solutions to lc problems: No.2395~2398
* No.2395.Find Subarrays With Equal Sum * No.2396.Strictly Palindromic Number * No.2397.Maximum Rows Covered by Columns * No.2398.Maximum Number of Robots Within Budget
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‎solution/2300-2399/2395.Find Subarrays With Equal Sum/README.md

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# [2395. 和相等的子数组](https://leetcode.cn/problems/find-subarrays-with-equal-sum)
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[English Version](/solution/2300-2399/2395.Find%20Subarrays%20With%20Equal%20Sum/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>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;，判断是否存在&nbsp;<strong>两个</strong>&nbsp;长度为&nbsp;<code>2</code>&nbsp;的子数组且它们的&nbsp;<strong>和</strong>&nbsp;相等。注意，这两个子数组起始位置的下标必须&nbsp;<strong>不相同</strong>&nbsp;。</p>
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<p>如果这样的子数组存在，请返回&nbsp;<code>true</code>，否则返回&nbsp;<code>false</code><em>&nbsp;</em>。</p>
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<p><strong>子数组</strong> 是一个数组中一段连续非空的元素组成的序列。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>nums = [4,2,4]
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<b>输出：</b>true
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<b>解释：</b>元素为 [4,2] 和 [2,4] 的子数组有相同的和 6 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>nums = [1,2,3,4,5]
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<b>输出：</b>false
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<b>解释：</b>没有长度为 2 的两个子数组和相等。
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</pre>
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<p><strong>示例 3：</strong></p>
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<pre><b>输入：</b>nums = [0,0,0]
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<b>输出：</b>true
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<b>解释：</b>子数组 [nums[0],nums[1]] 和 [nums[1],nums[2]] 的和相等，都为 0 。
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注意即使子数组的元素相同，这两个子数组也视为不相同的子数组，因为它们在原数组中的起始位置不同。
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</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>2 &lt;= nums.length &lt;= 1000</code></li>
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<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></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：哈希表**
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用哈希表 `s` 记录数组相邻两元素的和。
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遍历数组 `nums`，若 `s` 中存在 `nums[i] + nums[i + 1]`，则返回 `true`；否则将 `nums[i] + nums[i + 1]` 加入 `s` 中。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$，其中 $n$ 为数组 `nums` 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 findSubarrays(self, nums: List[int]) -> bool:
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s = set()
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for a, b in pairwise(nums):
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if (v := a + b) in s:
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return True
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s.add(v)
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return False
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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 findSubarrays(int[] nums) {
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Set<Integer> s = new HashSet<>();
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for (int i = 0; i < nums.length - 1; ++i) {
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int v = nums[i] + nums[i + 1];
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if (!s.add(v)) {
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return true;
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}
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}
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return false;
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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 findSubarrays(vector<int>& nums) {
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unordered_set<int> s;
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for (int i = 0; i < nums.size() - 1; ++i) {
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int v = nums[i] + nums[i + 1];
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if (s.count(v)) return true;
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s.insert(v);
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}
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return false;
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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 findSubarrays(nums []int) bool {
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s := map[int]bool{}
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for i := 0; i < len(nums)-1; i++ {
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v := nums[i] + nums[i+1]
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if s[v] {
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return true
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}
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s[v] = true
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}
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return false
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}
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2300-2399/2395.Find Subarrays With Equal Sum/README_EN.md

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# [2395. Find Subarrays With Equal Sum](https://leetcode.com/problems/find-subarrays-with-equal-sum)
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[中文文档](/solution/2300-2399/2395.Find%20Subarrays%20With%20Equal%20Sum/README.md)
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## Description
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<p>Given a <strong>0-indexed</strong> integer array <code>nums</code>, determine whether there exist <strong>two</strong> subarrays of length <code>2</code> with <strong>equal</strong> sum. Note that the two subarrays must begin at <strong>different</strong> indices.</p>
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<p>Return <code>true</code><em> if these subarrays exist, and </em><code>false</code><em> otherwise.</em></p>
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<p>A <b>subarray</b> is a contiguous non-empty sequence of elements within an array.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [4,2,4]
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<strong>Output:</strong> true
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<strong>Explanation:</strong> The subarrays with elements [4,2] and [2,4] have the same sum of 6.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,3,4,5]
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<strong>Output:</strong> false
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<strong>Explanation:</strong> No two subarrays of size 2 have the same sum.
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [0,0,0]
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<strong>Output:</strong> true
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<strong>Explanation:</strong> The subarrays [nums[0],nums[1]] and [nums[1],nums[2]] have the same sum of 0.
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Note that even though the subarrays have the same content, the two subarrays are considered different because they are in different positions in the original array.
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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>2 &lt;= nums.length &lt;= 1000</code></li>
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<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></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 findSubarrays(self, nums: List[int]) -> bool:
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s = set()
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for a, b in pairwise(nums):
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if (v := a + b) in s:
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return True
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s.add(v)
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return False
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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 findSubarrays(int[] nums) {
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Set<Integer> s = new HashSet<>();
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for (int i = 0; i < nums.length - 1; ++i) {
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int v = nums[i] + nums[i + 1];
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if (!s.add(v)) {
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return true;
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}
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}
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return false;
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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 findSubarrays(vector<int>& nums) {
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unordered_set<int> s;
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for (int i = 0; i < nums.size() - 1; ++i) {
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int v = nums[i] + nums[i + 1];
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if (s.count(v)) return true;
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s.insert(v);
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}
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return false;
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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 findSubarrays(nums []int) bool {
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s := map[int]bool{}
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for i := 0; i < len(nums)-1; i++ {
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v := nums[i] + nums[i+1]
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if s[v] {
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return true
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}
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s[v] = true
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}
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return false
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}
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2300-2399/2395.Find Subarrays With Equal Sum/Solution.cpp

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class Solution {
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public:
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bool findSubarrays(vector<int>& nums) {
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unordered_set<int> s;
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for (int i = 0; i < nums.size() - 1; ++i) {
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int v = nums[i] + nums[i + 1];
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if (s.count(v)) return true;
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s.insert(v);
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}
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return false;
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}
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};

‎solution/2300-2399/2395.Find Subarrays With Equal Sum/Solution.go

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func findSubarrays(nums []int) bool {
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s := map[int]bool{}
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for i := 0; i < len(nums)-1; i++ {
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v := nums[i] + nums[i+1]
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if s[v] {
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return true
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}
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s[v] = true
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}
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return false
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}

‎solution/2300-2399/2395.Find Subarrays With Equal Sum/Solution.java

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class Solution {
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public boolean findSubarrays(int[] nums) {
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Set<Integer> s = new HashSet<>();
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for (int i = 0; i < nums.length - 1; ++i) {
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int v = nums[i] + nums[i + 1];
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if (!s.add(v)) {
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return true;
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}
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}
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return false;
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}
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}

‎solution/2300-2399/2395.Find Subarrays With Equal Sum/Solution.py

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class Solution:
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def findSubarrays(self, nums: List[int]) -> bool:
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s = set()
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for a, b in pairwise(nums):
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if (v := a + b) in s:
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return True
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s.add(v)
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return False

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