feat: add biweekly contest 170

solution/0700-0799/0757.Set Intersection Size At Least Two/README.md

@@ -45,7 +45,7 @@ tags:
 <strong>输入：</strong>intervals = [[1,3],[1,4],[2,5],[3,5]]
 <strong>输出：</strong>3
 <strong>解释：</strong>nums = [2, 3, 4].
-可以证明不存在元素数量为 2 的包含集合。
+可以证明不存在元素数量为 2 的包含集合。 
 </pre>
 
 <p><strong class="example">示例 3：</strong></p>
@@ -54,7 +54,7 @@ tags:
 <strong>输入：</strong>intervals = [[1,2],[2,3],[2,4],[4,5]]
 <strong>输出：</strong>5
 <strong>解释：</strong>nums = [1, 2, 3, 4, 5].
-可以证明不存在元素数量为 4 的包含集合。
+可以证明不存在元素数量为 4 的包含集合。 
 </pre>
 
 <p>&nbsp;</p>

solution/2100-2199/2154.Keep Multiplying Found Values by Two/README.md

@@ -40,7 +40,7 @@ tags:
 <pre>
 <strong>输入：</strong>nums = [5,3,6,1,12], original = 3
 <strong>输出：</strong>24
-<strong>解释：</strong>
+<strong>解释：</strong> 
 - 3 能在 nums 中找到。3 * 2 = 6 。
 - 6 能在 nums 中找到。6 * 2 = 12 。
 - 12 能在 nums 中找到。12 * 2 = 24 。

solution/2100-2199/2154.Keep Multiplying Found Values by Two/README_EN.md

@@ -39,7 +39,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [5,3,6,1,12], original = 3
 <strong>Output:</strong> 24
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 - 3 is found in nums. 3 is multiplied by 2 to obtain 6.
 - 6 is found in nums. 6 is multiplied by 2 to obtain 12.
 - 12 is found in nums. 12 is multiplied by 2 to obtain 24.

solution/3700-3799/3745.Maximize Expression of Three Elements/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 简单
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3745.Maximize%20Expression%20of%20Three%20Elements/README.md
+rating: 1218
+source: 第 476 场周赛 Q1
 ---
 
 <!-- problem:start -->

solution/3700-3799/3745.Maximize Expression of Three Elements/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Easy
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3745.Maximize%20Expression%20of%20Three%20Elements/README_EN.md
+rating: 1218
+source: Weekly Contest 476 Q1
 ---
 
 <!-- problem:start -->

solution/3700-3799/3746.Minimum String Length After Balanced Removals/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3746.Minimum%20String%20Length%20After%20Balanced%20Removals/README.md
+rating: 1326
+source: 第 476 场周赛 Q2
 ---
 
 <!-- problem:start -->

solution/3700-3799/3746.Minimum String Length After Balanced Removals/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3746.Minimum%20String%20Length%20After%20Balanced%20Removals/README_EN.md
+rating: 1326
+source: Weekly Contest 476 Q2
 ---
 
 <!-- problem:start -->

solution/3700-3799/3747.Count Distinct Integers After Removing Zeros/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3747.Count%20Distinct%20Integers%20After%20Removing%20Zeros/README.md
+rating: 1848
+source: 第 476 场周赛 Q3
 ---
 
 <!-- problem:start -->

solution/3700-3799/3747.Count Distinct Integers After Removing Zeros/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3747.Count%20Distinct%20Integers%20After%20Removing%20Zeros/README_EN.md
+rating: 1848
+source: Weekly Contest 476 Q3
 ---
 
 <!-- problem:start -->

solution/3700-3799/3748.Count Stable Subarrays/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3748.Count%20Stable%20Subarrays/README.md
+rating: 2209
+source: 第 476 场周赛 Q4
 ---
 
 <!-- problem:start -->

solution/3700-3799/3748.Count Stable Subarrays/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3748.Count%20Stable%20Subarrays/README_EN.md
+rating: 2209
+source: Weekly Contest 476 Q4
 ---
 
 <!-- problem:start -->

solution/3700-3799/3750.Minimum Number of Flips to Reverse Binary String/README.md

@@ -0,0 +1,163 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3750.Minimum%20Number%20of%20Flips%20to%20Reverse%20Binary%20String/README.md
+---
+
+<!-- problem:start -->
+
+# [3750. 最少反转次数得到翻转二进制字符串](https://leetcode.cn/problems/minimum-number-of-flips-to-reverse-binary-string)
+
+[English Version](/solution/3700-3799/3750.Minimum%20Number%20of%20Flips%20to%20Reverse%20Binary%20String/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个 <strong>正</strong> 整数 <code>n</code>。</p>
+
+<p>令 <code>s</code> 为 <code>n</code> 的 <strong>二进制表示</strong>（不含前导零）。</p>
+
+<p>二进制字符串 <code>s</code> 的 <strong>反转</strong> 是指将 <code>s</code> 中的字符按相反顺序排列得到的字符串。</p>
+
+<p>你可以翻转 <code>s</code> 中的任意一位（将 <code>0 → 1</code> 或 <code>1 → 0</code>）。每次翻转 <strong>仅</strong> 影响一位。</p>
+
+<p>请返回使 <code>s</code> 等于其原始形式的反转所需的 <strong>最少</strong> 翻转次数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">n = 7</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">0</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>7 的二进制表示为 <code>"111"</code>。其反转也是 <code>"111"</code>，两者相同。因此，不需要翻转。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">n = 10</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">4</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>10 的二进制表示为 <code>"1010"</code>。其反转为 <code>"0101"</code>。必须翻转所有四位才能使它们相等。因此，所需的最少翻转次数为 4。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：模拟
+
+我们首先将整数 $n$ 转换成二进制字符串 $s$，然后我们使用双指针分别从字符串的两端向中间遍历，统计对应位置字符不同的个数 $cnt$。由于每次翻转只能影响一位，因此总的翻转次数为 $cnt \times 2$。
+
+时间复杂度 $O(\log n)$，空间复杂度 $O(\log n)$，其中 $n$ 是输入的整数。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def minimumFlips(self, n: int) -> int:
+        s = bin(n)[2:]
+        m = len(s)
+        return sum(s[i] != s[m - i - 1] for i in range(m // 2)) * 2
+```
+
+#### Java
+
+```java
+class Solution {
+    public int minimumFlips(int n) {
+        String s = Integer.toBinaryString(n);
+        int m = s.length();
+        int cnt = 0;
+        for (int i = 0; i < m / 2; i++) {
+            if (s.charAt(i) != s.charAt(m - i - 1)) {
+                cnt++;
+            }
+        }
+        return cnt * 2;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int minimumFlips(int n) {
+        vector<int> s;
+        while (n > 0) {
+            s.push_back(n & 1);
+            n >>= 1;
+        }
+
+        int m = s.size();
+        int cnt = 0;
+        for (int i = 0; i < m / 2; i++) {
+            if (s[i] != s[m - i - 1]) {
+                cnt++;
+            }
+        }
+        return cnt * 2;
+    }
+};
+```
+
+#### Go
+
+```go
+func minimumFlips(n int) int {
+    s := strconv.FormatInt(int64(n), 2)
+    m := len(s)
+    cnt := 0
+    for i := 0; i < m/2; i++ {
+        if s[i] != s[m-i-1] {
+            cnt++
+        }
+    }
+    return cnt * 2
+}
+```
+
+#### TypeScript
+
+```ts
+function minimumFlips(n: number): number {
+    const s = n.toString(2);
+    const m = s.length;
+    let cnt = 0;
+    for (let i = 0; i < m / 2; i++) {
+        if (s[i] !== s[m - i - 1]) {
+            cnt++;
+        }
+    }
+    return cnt * 2;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->