chore: update lc problems

solution/0100-0199/0184.Department Highest Salary/Solution.py

@@ -17,4 +17,4 @@ def department_highest_salary(
     result = top_earners[['name_y', 'name_x', 'salary']].copy()
     result.columns = ['Department', 'Employee', 'Salary']
 
-    return result
+    return result

solution/0400-0499/0416.Partition Equal Subset Sum/README.md

@@ -202,14 +202,14 @@ impl Solution {
         let n = nums.len();
         let mut f = vec![vec![false; m + 1]; n + 1];
         f[0][0] = true;
-        
+
         for i in 1..=n {
             let x = nums[i - 1] as usize;
             for j in 0..=m {
                 f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
             }
         }
-        
+
         f[n][m]
     }
 }

solution/0400-0499/0416.Partition Equal Subset Sum/README_EN.md

@@ -201,14 +201,14 @@ impl Solution {
         let n = nums.len();
         let mut f = vec![vec![false; m + 1]; n + 1];
         f[0][0] = true;
-        
+
         for i in 1..=n {
             let x = nums[i - 1] as usize;
             for j in 0..=m {
                 f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
             }
         }
-        
+
         f[n][m]
     }
 }

solution/0400-0499/0416.Partition Equal Subset Sum/Solution.rs

@@ -8,14 +8,14 @@ impl Solution {
         let n = nums.len();
         let mut f = vec![vec![false; m + 1]; n + 1];
         f[0][0] = true;
-        
+
         for i in 1..=n {
             let x = nums[i - 1] as usize;
             for j in 0..=m {
                 f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
             }
         }
-        
+
         f[n][m]
     }
 }

solution/0700-0799/0752.Open the Lock/Solution.py

@@ -17,7 +17,7 @@ def next(s):
         s = set(deadends)
         if '0000' in s:
             return -1
-        q = deque([('0000')])
+        q = deque(['0000'])
         s.add('0000')
         ans = 0
         while q:

solution/0700-0799/0752.Open the Lock/Solution2.py

@@ -27,7 +27,7 @@ def extend(m1, m2, q):
 
         def bfs():
             m1, m2 = {"0000": 0}, {target: 0}
-            q1, q2 = deque([('0000')]), deque([(target)])
+            q1, q2 = deque(['0000']), deque([target])
             while q1 and q2:
                 t = extend(m1, m2, q1) if len(q1) <= len(q2) else extend(m2, m1, q2)
                 if t != -1:

solution/0700-0799/0773.Sliding Puzzle/Solution.py

@@ -29,7 +29,7 @@ def f():
         if start == end:
             return 0
         vis = {start}
-        q = deque([(start)])
+        q = deque([start])
         ans = 0
         while q:
             ans += 1

solution/1200-1299/1237.Find Positive Integer Solution for a Given Equation/Solution.py

@@ -1,11 +1,11 @@
 """
-   This is the custom function interface.
-   You should not implement it, or speculate about its implementation
-   class CustomFunction:
-       # Returns f(x, y) for any given positive integers x and y.
-       # Note that f(x, y) is increasing with respect to both x and y.
-       # i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
-       def f(self, x, y):
+This is the custom function interface.
+You should not implement it, or speculate about its implementation
+class CustomFunction:
+    # Returns f(x, y) for any given positive integers x and y.
+    # Note that f(x, y) is increasing with respect to both x and y.
+    # i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
+    def f(self, x, y):
 
 """
 

solution/1200-1299/1237.Find Positive Integer Solution for a Given Equation/Solution2.py

@@ -1,11 +1,11 @@
 """
-   This is the custom function interface.
-   You should not implement it, or speculate about its implementation
-   class CustomFunction:
-       # Returns f(x, y) for any given positive integers x and y.
-       # Note that f(x, y) is increasing with respect to both x and y.
-       # i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
-       def f(self, x, y):
+This is the custom function interface.
+You should not implement it, or speculate about its implementation
+class CustomFunction:
+    # Returns f(x, y) for any given positive integers x and y.
+    # Note that f(x, y) is increasing with respect to both x and y.
+    # i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
+    def f(self, x, y):
 
 """
 

solution/3300-3399/3396.Minimum Number of Operations to Make Elements in Array Distinct/Solution.rs

@@ -1,13 +1,13 @@
-use std::collections::HashSet;
-
-impl Solution {
-    pub fn minimum_operations(nums: Vec<i32>) -> i32 {
-        let mut s = HashSet::new();
-        for i in (0..nums.len()).rev() {
-            if !s.insert(nums[i]) {
-                return (i / 3) as i32 + 1;
-            }
-        }
-        0
-    }
-}
+use std::collections::HashSet;
+
+impl Solution {
+    pub fn minimum_operations(nums: Vec<i32>) -> i32 {
+        let mut s = HashSet::new();
+        for i in (0..nums.len()).rev() {
+            if !s.insert(nums[i]) {
+                return (i / 3) as i32 + 1;
+            }
+        }
+        0
+    }
+}

solution/3500-3599/3506.Find Time Required to Eliminate Bacterial Strains/README.md

@@ -6,7 +6,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3506.Fi
 
 <!-- problem:start -->
 
-# [3506. 查找消除细菌菌株所需时间 II 🔒](https://leetcode.cn/problems/find-time-required-to-eliminate-bacterial-strains)
+# [3506. 查找消除细菌菌株所需时间 🔒](https://leetcode.cn/problems/find-time-required-to-eliminate-bacterial-strains)
 
 [English Version](/solution/3500-3599/3506.Find%20Time%20Required%20to%20Eliminate%20Bacterial%20Strains/README_EN.md)
 

solution/3500-3599/3511.Make a Positive Array/README.md

@@ -0,0 +1,147 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3511.Make%20a%20Positive%20Array/README.md
+---
+
+<!-- problem:start -->
+
+# [3511. 构造正数组 🔒](https://leetcode.cn/problems/make-a-positive-array)
+
+[English Version](/solution/3500-3599/3511.Make%20a%20Positive%20Array/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给定一个数组&nbsp;<code>nums</code>。一个数组被认为是 <strong>正</strong> 的，如果每个包含 <strong>超过两个</strong> 元素的 <strong><span data-keyword="subarray">子数组</span></strong> 的所有数字之和都是正数。</p>
+
+<p>您可以执行以下操作任意多次：</p>
+
+<ul>
+	<li>用 -10<sup>18</sup> 和 10<sup>18&nbsp;</sup>之间的任意整数替换&nbsp;<code>nums</code>&nbsp;中的 <strong>一个</strong>&nbsp;元素。</li>
+</ul>
+
+<p>找到使 <code>nums</code> 变为正数组所需的最小操作数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [-10,15,-12]</span></p>
+
+<p><span class="example-io"><b>输出：</b>1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>唯一有超过 2 个元素的子数组是这个数组本身。所有元素的和是&nbsp;<code>(-10) + 15 + (-12) = -7</code>。通过将&nbsp;<code>nums[0]</code>&nbsp;替换为 0，新的和变为&nbsp;<code>0 + 15 + (-12) = 3</code>。因此，现在数组是正的。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [-1,-2,3,-1,2,6]</span></p>
+
+<p><strong>输出：</strong><span class="example-io">1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>具有 2 个以上元素且和非正的子数组是：</p>
+
+<table style="border: 1px solid black;">
+	<tbody>
+		<tr>
+			<th style="border: 1px solid black;">子数组下标</th>
+			<th style="border: 1px solid black;">子数组</th>
+			<th style="border: 1px solid black;">和</th>
+			<th style="border: 1px solid black;">替换后的子数组（令 nums[1] = 1）</th>
+			<th style="border: 1px solid black;">新的和</th>
+		</tr>
+		<tr>
+			<td style="border: 1px solid black;">nums[0...2]</td>
+			<td style="border: 1px solid black;">[-1, -2, 3]</td>
+			<td style="border: 1px solid black;">0</td>
+			<td style="border: 1px solid black;">[-1, 1, 3]</td>
+			<td style="border: 1px solid black;">3</td>
+		</tr>
+		<tr>
+			<td style="border: 1px solid black;">nums[0...3]</td>
+			<td style="border: 1px solid black;">[-1, -2, 3, -1]</td>
+			<td style="border: 1px solid black;">-1</td>
+			<td style="border: 1px solid black;">[-1, 1, 3, -1]</td>
+			<td style="border: 1px solid black;">2</td>
+		</tr>
+		<tr>
+			<td style="border: 1px solid black;">nums[1...3]</td>
+			<td style="border: 1px solid black;">[-2, 3, -1]</td>
+			<td style="border: 1px solid black;">0</td>
+			<td style="border: 1px solid black;">[1, 3, -1]</td>
+			<td style="border: 1px solid black;">3</td>
+		</tr>
+	</tbody>
+</table>
+
+<p>因此，<code>nums</code>&nbsp;在一次操作后是正的。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [1,2,3]</span></p>
+
+<p><span class="example-io"><b>输出：</b>0</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>数组已经是正的，所以不需要操作。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->