feat: add solutions to lc problems: No.2574~2577

* No.2574.Left and Right Sum Differences
* No.2575.Find the Divisibility Array of a String
* No.2576.Find the Maximum Number of Marked Indices
* No.2577.Minimum Time to Visit a Cell In a Grid

Author: yanglbme

solution/1800-1899/1836.Remove Duplicates From an Unsorted Linked List/README_EN.md

@@ -57,8 +57,11 @@
 <p><strong>Constraints:</strong></p>
 
 <ul>
+
     <li>The number of nodes in the list is in the range&nbsp;<code>[1, 10<sup>5</sup>]</code></li>
+
     <li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
+
 </ul>
 
 ## Solutions

solution/2000-2099/2013.Detect Squares/README_EN.md

@@ -98,12 +98,12 @@ class DetectSquares {
     public DetectSquares() {
 
     }
-    
+
     public void add(int[] point) {
         int x = point[0], y = point[1];
         cnt.computeIfAbsent(x, k -> new HashMap<>()).merge(y, 1, Integer::sum);
     }
-    
+
     public int count(int[] point) {
         int x1 = point[0], y1 = point[1];
         if (!cnt.containsKey(x1)) {
@@ -140,12 +140,12 @@ public:
     DetectSquares() {
 
     }
-    
+
     void add(vector<int> point) {
         int x = point[0], y = point[1];
         ++cnt[x][y];
     }
-    
+
     int count(vector<int> point) {
         int x1 = point[0], y1 = point[1];
         if (!cnt.count(x1)) {

solution/2500-2599/2574.Left and Right Sum Differences/README.md

@@ -0,0 +1,152 @@
+# [2574. 左右元素和的差值](https://leetcode.cn/problems/left-and-right-sum-differences)
+
+[English Version](/solution/2500-2599/2574.Left%20and%20Right%20Sum%20Differences/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> ，请你找出一个下标从 <strong>0</strong> 开始的整数数组 <code>answer</code> ，其中：</p>
+
+<ul>
+	<li><code>answer.length == nums.length</code></li>
+	<li><code>answer[i] = |leftSum[i] - rightSum[i]|</code></li>
+</ul>
+
+<p>其中：</p>
+
+<ul>
+	<li><code>leftSum[i]</code> 是数组 <code>nums</code> 中下标 <code>i</code> 左侧元素之和。如果不存在对应的元素，<code>leftSum[i] = 0</code> 。</li>
+	<li><code>rightSum[i]</code> 是数组 <code>nums</code> 中下标 <code>i</code> 右侧元素之和。如果不存在对应的元素，<code>rightSum[i] = 0</code> 。</li>
+</ul>
+
+<p>返回数组 <code>answer</code> 。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><strong>输入：</strong>nums = [10,4,8,3]
+<strong>输出：</strong>[15,1,11,22]
+<strong>解释：</strong>数组 leftSum 为 [0,10,14,22] 且数组 rightSum 为 [15,11,3,0] 。
+数组 answer 为 [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22] 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><strong>输入：</strong>nums = [1]
+<strong>输出：</strong>[0]
+<strong>解释：</strong>数组 leftSum 为 [0] 且数组 rightSum 为 [0] 。
+数组 answer 为 [|0 - 0|] = [0] 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：前缀和**
+
+我们定义变量 $left$ 表示数组 `nums` 中下标 $i$ 左侧元素之和，变量 $right$ 表示数组 `nums` 中下标 $i$ 右侧元素之和。初始时 $left = 0$, $right = \sum_{i = 0}^{n - 1} nums[i]$。
+
+遍历数组 `nums`，对于当前遍历到的数字 $x$，我们更新 $right = right - x$，此时 $left$ 和 $right$ 分别表示数组 `nums` 中下标 $i$ 左侧元素之和和右侧元素之和。我们将 $left$ 和 $right$ 的差的绝对值加入答案数组 `ans` 中，然后更新 $left = left + x$。
+
+遍历完成后，返回答案数组 `ans` 即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `nums` 的长度。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def leftRigthDifference(self, nums: List[int]) -> List[int]:
+        left, right = 0, sum(nums)
+        ans = []
+        for x in nums:
+            right -= x
+            ans.append(abs(left - right))
+            left += x
+        return ans
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int[] leftRigthDifference(int[] nums) {
+        int left = 0, right = Arrays.stream(nums).sum();
+        int n = nums.length;
+        int[] ans = new int[n];
+        for (int i = 0; i < n; ++i) {
+            right -= nums[i];
+            ans[i] = Math.abs(left - right);
+            left += nums[i];
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> leftRigthDifference(vector<int>& nums) {
+        int left = 0, right = accumulate(nums.begin(), nums.end(), 0);
+        vector<int> ans;
+        for (int& x : nums) {
+            right -= x;
+            ans.push_back(abs(left - right));
+            left += x;
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func leftRigthDifference(nums []int) (ans []int) {
+	var left, right int
+	for _, x := range nums {
+		right += x
+	}
+	for _, x := range nums {
+		right -= x
+		ans = append(ans, abs(left-right))
+		left += x
+	}
+	return
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2500-2599/2574.Left and Right Sum Differences/README_EN.md

@@ -0,0 +1,134 @@
+# [2574. Left and Right Sum Differences](https://leetcode.com/problems/left-and-right-sum-differences)
+
+[中文文档](/solution/2500-2599/2574.Left%20and%20Right%20Sum%20Differences/README.md)
+
+## Description
+
+<p>Given a <strong>0-indexed</strong> integer array <code>nums</code>, find a <strong>0-indexed </strong>integer array <code>answer</code> where:</p>
+
+<ul>
+    <li><code>answer.length == nums.length</code>.</li>
+    <li><code>answer[i] = |leftSum[i] - rightSum[i]|</code>.</li>
+</ul>
+
+<p>Where:</p>
+
+<ul>
+    <li><code>leftSum[i]</code> is the sum of elements to the left of the index <code>i</code> in the array <code>nums</code>. If there is no such element, <code>leftSum[i] = 0</code>.</li>
+    <li><code>rightSum[i]</code> is the sum of elements to the right of the index <code>i</code> in the array <code>nums</code>. If there is no such element, <code>rightSum[i] = 0</code>.</li>
+</ul>
+
+<p>Return <em>the array</em> <code>answer</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [10,4,8,3]
+<strong>Output:</strong> [15,1,11,22]
+<strong>Explanation:</strong> The array leftSum is [0,10,14,22] and the array rightSum is [15,11,3,0].
+The array answer is [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1]
+<strong>Output:</strong> [0]
+<strong>Explanation:</strong> The array leftSum is [0] and the array rightSum is [0].
+The array answer is [|0 - 0|] = [0].
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def leftRigthDifference(self, nums: List[int]) -> List[int]:
+        left, right = 0, sum(nums)
+        ans = []
+        for x in nums:
+            right -= x
+            ans.append(abs(left - right))
+            left += x
+        return ans
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int[] leftRigthDifference(int[] nums) {
+        int left = 0, right = Arrays.stream(nums).sum();
+        int n = nums.length;
+        int[] ans = new int[n];
+        for (int i = 0; i < n; ++i) {
+            right -= nums[i];
+            ans[i] = Math.abs(left - right);
+            left += nums[i];
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> leftRigthDifference(vector<int>& nums) {
+        int left = 0, right = accumulate(nums.begin(), nums.end(), 0);
+        vector<int> ans;
+        for (int& x : nums) {
+            right -= x;
+            ans.push_back(abs(left - right));
+            left += x;
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func leftRigthDifference(nums []int) (ans []int) {
+	var left, right int
+	for _, x := range nums {
+		right += x
+	}
+	for _, x := range nums {
+		right -= x
+		ans = append(ans, abs(left-right))
+		left += x
+	}
+	return
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2500-2599/2574.Left and Right Sum Differences/Solution.cpp

@@ -0,0 +1,13 @@
+class Solution {
+public:
+    vector<int> leftRigthDifference(vector<int>& nums) {
+        int left = 0, right = accumulate(nums.begin(), nums.end(), 0);
+        vector<int> ans;
+        for (int& x : nums) {
+            right -= x;
+            ans.push_back(abs(left - right));
+            left += x;
+        }
+        return ans;
+    }
+};

solution/2500-2599/2574.Left and Right Sum Differences/Solution.go

@@ -0,0 +1,19 @@
+func leftRigthDifference(nums []int) (ans []int) {
+	var left, right int
+	for _, x := range nums {
+		right += x
+	}
+	for _, x := range nums {
+		right -= x
+		ans = append(ans, abs(left-right))
+		left += x
+	}
+	return
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}

solution/2500-2599/2574.Left and Right Sum Differences/Solution.java

@@ -0,0 +1,13 @@
+class Solution {
+    public int[] leftRigthDifference(int[] nums) {
+        int left = 0, right = Arrays.stream(nums).sum();
+        int n = nums.length;
+        int[] ans = new int[n];
+        for (int i = 0; i < n; ++i) {
+            right -= nums[i];
+            ans[i] = Math.abs(left - right);
+            left += nums[i];
+        }
+        return ans;
+    }
+}

solution/2500-2599/2574.Left and Right Sum Differences/Solution.py

@@ -0,0 +1,9 @@
+class Solution:
+    def leftRigthDifference(self, nums: List[int]) -> List[int]:
+        left, right = 0, sum(nums)
+        ans = []
+        for x in nums:
+            right -= x
+            ans.append(abs(left - right))
+            left += x
+        return ans