feat: add solutions for lc No.1545,3666 (#5054)

Author: yanglbme

solution/1500-1599/1545.Find Kth Bit in Nth Binary String/README.md

@@ -211,27 +211,55 @@ function findKthBit(n: number, k: number): string {
 }
 ```
 
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
-<!-- solution:start -->
-
-### 方法二：位运算
-
-<!-- tabs:start -->
+#### Rust
 
-#### TypeScript
+```rust
+impl Solution {
+    pub fn find_kth_bit(n: i32, k: i32) -> char {
+        fn dfs(n: i32, k: i32) -> i32 {
+            if k == 1 {
+                return 0;
+            }
+            if (k & (k - 1)) == 0 {
+                return 1;
+            }
+            let m: i32 = 1 << n;
+            if k * 2 < m - 1 {
+                dfs(n - 1, k)
+            } else {
+                dfs(n - 1, m - k) ^ 1
+            }
+        }
 
-```ts
-const findKthBit = (n: number, k: number): string =>
-    String((((k / (k & -k)) >> 1) & 1) ^ (k & 1) ^ 1);
+        if dfs(n, k) == 0 { '0' } else { '1' }
+    }
+}
 ```
 
 #### JavaScript
 
 ```js
-const findKthBit = (n, k) => String((((k / (k & -k)) >> 1) & 1) ^ (k & 1) ^ 1);
+/**
+ * @param {number} n
+ * @param {number} k
+ * @return {character}
+ */
+var findKthBit = function (n, k) {
+    const dfs = function (n, k) {
+        if (k === 1) {
+            return 0;
+        }
+        if ((k & (k - 1)) === 0) {
+            return 1;
+        }
+        const m = 1 << n;
+        if (k * 2 < m - 1) {
+            return dfs(n - 1, k);
+        }
+        return dfs(n - 1, m - k) ^ 1;
+    };
+    return dfs(n, k).toString();
+};
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1545.Find Kth Bit in Nth Binary String/README_EN.md

@@ -197,27 +197,55 @@ function findKthBit(n: number, k: number): string {
 }
 ```
 
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
-<!-- solution:start -->
-
-### Solution 2: Bit Manipulation
-
-<!-- tabs:start -->
+#### Rust
 
-#### TypeScript
+```rust
+impl Solution {
+    pub fn find_kth_bit(n: i32, k: i32) -> char {
+        fn dfs(n: i32, k: i32) -> i32 {
+            if k == 1 {
+                return 0;
+            }
+            if (k & (k - 1)) == 0 {
+                return 1;
+            }
+            let m: i32 = 1 << n;
+            if k * 2 < m - 1 {
+                dfs(n - 1, k)
+            } else {
+                dfs(n - 1, m - k) ^ 1
+            }
+        }
 
-```ts
-const findKthBit = (n: number, k: number): string =>
-    String((((k / (k & -k)) >> 1) & 1) ^ (k & 1) ^ 1);
+        if dfs(n, k) == 0 { '0' } else { '1' }
+    }
+}
 ```
 
 #### JavaScript
 
 ```js
-const findKthBit = (n, k) => String((((k / (k & -k)) >> 1) & 1) ^ (k & 1) ^ 1);
+/**
+ * @param {number} n
+ * @param {number} k
+ * @return {character}
+ */
+var findKthBit = function (n, k) {
+    const dfs = function (n, k) {
+        if (k === 1) {
+            return 0;
+        }
+        if ((k & (k - 1)) === 0) {
+            return 1;
+        }
+        const m = 1 << n;
+        if (k * 2 < m - 1) {
+            return dfs(n - 1, k);
+        }
+        return dfs(n - 1, m - k) ^ 1;
+    };
+    return dfs(n, k).toString();
+};
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.js

@@ -0,0 +1,21 @@
+/**
+ * @param {number} n
+ * @param {number} k
+ * @return {character}
+ */
+var findKthBit = function (n, k) {
+    const dfs = function (n, k) {
+        if (k === 1) {
+            return 0;
+        }
+        if ((k & (k - 1)) === 0) {
+            return 1;
+        }
+        const m = 1 << n;
+        if (k * 2 < m - 1) {
+            return dfs(n - 1, k);
+        }
+        return dfs(n - 1, m - k) ^ 1;
+    };
+    return dfs(n, k).toString();
+};

solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.rs

@@ -0,0 +1,20 @@
+impl Solution {
+    pub fn find_kth_bit(n: i32, k: i32) -> char {
+        fn dfs(n: i32, k: i32) -> i32 {
+            if k == 1 {
+                return 0;
+            }
+            if (k & (k - 1)) == 0 {
+                return 1;
+            }
+            let m: i32 = 1 << n;
+            if k * 2 < m - 1 {
+                dfs(n - 1, k)
+            } else {
+                dfs(n - 1, m - k) ^ 1
+            }
+        }
+
+        if dfs(n, k) == 0 { '0' } else { '1' }
+    }
+}

solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution2.js

@@ -132,7 +132,7 @@ $$
 
 接下来，我们确定二分查找的左边界 $l = 1$，由于最少有一个工作，且每个工人的工作时间不超过 $10^6$，要想使山的高度降低到 $0$，最少需要 $(1 + \textit{mountainHeight}) \cdot \textit{mountainHeight} / 2 \cdot \textit{workerTimes}[i] \leq 10^{16}$ 秒，因此我们可以确定二分查找的右边界 $r = 10^{16}$。然后我们不断地将区间 $[l, r]$ 二分，直到 $l = r$，此时 $l$ 即为答案。
 
-时间复杂度 $O(n \times \log M)$，其中 $n$ 为工人的数量，而 $M$ 为二分查找的右边界，本题中 $M = 10^{16}$。
+时间复杂度 $O(n \times \log M)$，其中 $n$ 为工人的数量，而 $M$ 为二分查找的右边界，本题中 $M = 10^{16}$。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 

solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution2.ts

@@ -128,7 +128,7 @@ We can sum up all the $h'$ values for the workers to get the total reduced heigh
 
 Next, we determine the left boundary of the binary search $l = 1$. Since there is at least one worker, and each worker's working time does not exceed $10^6$, to reduce the mountain height to $0$, it takes at least $(1 + \textit{mountainHeight}) \cdot \textit{mountainHeight} / 2 \cdot \textit{workerTimes}[i] \leq 10^{16}$ seconds. Therefore, we can set the right boundary of the binary search to $r = 10^{16}$. Then, we continuously halve the interval $[l, r]$ until $l = r$. At this point, $l$ is the answer.
 
-The time complexity is $O(n \times \log M)$, where $n$ is the number of workers, and $M$ is the right boundary of the binary search, which is $10^{16}$ in this problem.
+The time complexity is $O(n \times \log M)$, where $n$ is the number of workers, and $M$ is the right boundary of the binary search, which is $10^{16}$ in this problem. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 

solution/3200-3299/3296.Minimum Number of Seconds to Make Mountain Height Zero/README.md

@@ -402,6 +402,66 @@ impl Solution {
 }
 ```
 
+#### C#
+
+```cs
+public class Solution {
+    public int MinOperations(string s, int k) {
+        int n = s.Length;
+
+        var ts = new SortedSet<int>[2];
+        ts[0] = new SortedSet<int>();
+        ts[1] = new SortedSet<int>();
+
+        for (int i = 0; i <= n; i++) {
+            ts[i % 2].Add(i);
+        }
+
+        int cnt0 = 0;
+        foreach (char c in s) {
+            if (c == '0') {
+                cnt0++;
+            }
+        }
+
+        ts[cnt0 % 2].Remove(cnt0);
+
+        var q = new Queue<int>();
+        q.Enqueue(cnt0);
+
+        int ans = 0;
+
+        while (q.Count > 0) {
+            int size = q.Count;
+            for (int i = 0; i < size; i++) {
+                int cur = q.Dequeue();
+                if (cur == 0) {
+                    return ans;
+                }
+
+                int l = cur + k - 2 * Math.Min(cur, k);
+                int r = cur + k - 2 * Math.Max(k - n + cur, 0);
+
+                var t = ts[l % 2];
+
+                var toRemove = new List<int>();
+                foreach (int next in t.GetViewBetween(l, r)) {
+                    q.Enqueue(next);
+                    toRemove.Add(next);
+                }
+
+                foreach (int next in toRemove) {
+                    t.Remove(next);
+                }
+            }
+            ans++;
+        }
+
+        return -1;
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3200-3299/3296.Minimum Number of Seconds to Make Mountain Height Zero/README_EN.md

@@ -399,6 +399,66 @@ impl Solution {
 }
 ```
 
+#### C#
+
+```cs
+public class Solution {
+    public int MinOperations(string s, int k) {
+        int n = s.Length;
+
+        var ts = new SortedSet<int>[2];
+        ts[0] = new SortedSet<int>();
+        ts[1] = new SortedSet<int>();
+
+        for (int i = 0; i <= n; i++) {
+            ts[i % 2].Add(i);
+        }
+
+        int cnt0 = 0;
+        foreach (char c in s) {
+            if (c == '0') {
+                cnt0++;
+            }
+        }
+
+        ts[cnt0 % 2].Remove(cnt0);
+
+        var q = new Queue<int>();
+        q.Enqueue(cnt0);
+
+        int ans = 0;
+
+        while (q.Count > 0) {
+            int size = q.Count;
+            for (int i = 0; i < size; i++) {
+                int cur = q.Dequeue();
+                if (cur == 0) {
+                    return ans;
+                }
+
+                int l = cur + k - 2 * Math.Min(cur, k);
+                int r = cur + k - 2 * Math.Max(k - n + cur, 0);
+
+                var t = ts[l % 2];
+
+                var toRemove = new List<int>();
+                foreach (int next in t.GetViewBetween(l, r)) {
+                    q.Enqueue(next);
+                    toRemove.Add(next);
+                }
+
+                foreach (int next in toRemove) {
+                    t.Remove(next);
+                }
+            }
+            ans++;
+        }
+
+        return -1;
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->