feat: add solutions to lc problem: No.1884

solution/1800-1899/1884.Egg Drop With 2 Eggs and N Floors/README.md

@@ -65,32 +65,107 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：动态规划
+
+我们定义 $f[i]$ 表示有两枚鸡蛋，在 $i$ 层楼中确定 $f$ 的最小操作次数。初始时 $f[0] = 0$，其余 $f[i] = +\infty$。答案为 $f[n]$。
+
+考虑 $f[i]$，我们可以枚举第一枚鸡蛋从第 $j$ 层楼扔下，其中 $1 \leq j \leq i$，此时有两种情况：
+
+-   鸡蛋碎了，此时我们剩余一枚鸡蛋，需要在 $j - 1$ 层楼中确定 $f$，这需要 $j - 1$ 次操作，因此总操作次数为 $1 + (j - 1)$；
+-   鸡蛋没碎，此时我们剩余两枚鸡蛋，需要在 $i - j$ 层楼中确定 $f$，这需要 $f[i - j]$ 次操作，因此总操作次数为 $1 + f[i - j]$。
+
+综上，我们可以得到状态转移方程：
+
+$$
+f[i] = \min_{1 \leq j \leq i} \{1 + \max(j - 1, f[i - j])\}
+$$
+
+最后，我们返回 $f[n]$ 即可。
+
+时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 为楼层数。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def twoEggDrop(self, n: int) -> int:
+        f = [0] + [inf] * n
+        for i in range(1, n + 1):
+            for j in range(1, i + 1):
+                f[i] = min(f[i], 1 + max(j - 1, f[i - j]))
+        return f[n]
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int twoEggDrop(int n) {
+        int[] f = new int[n + 1];
+        Arrays.fill(f, 1 << 29);
+        f[0] = 0;
+        for (int i = 1; i <= n; i++) {
+            for (int j = 1; j <= i; j++) {
+                f[i] = Math.min(f[i], 1 + Math.max(j - 1, f[i - j]));
+            }
+        }
+        return f[n];
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    int twoEggDrop(int n) {
+        int f[n + 1];
+        memset(f, 0x3f, sizeof(f));
+        f[0] = 0;
+        for (int i = 1; i <= n; i++) {
+            for (int j = 1; j <= i; j++) {
+                f[i] = min(f[i], 1 + max(j - 1, f[i - j]));
+            }
+        }
+        return f[n];
+    }
+};
 ```
 
 #### Go
 
 ```go
+func twoEggDrop(n int) int {
+	f := make([]int, n+1)
+	for i := range f {
+		f[i] = 1 << 29
+	}
+	f[0] = 0
+	for i := 1; i <= n; i++ {
+		for j := 1; j <= i; j++ {
+			f[i] = min(f[i], 1+max(j-1, f[i-j]))
+		}
+	}
+	return f[n]
+}
+```
 
+#### TypeScript
+
+```ts
+function twoEggDrop(n: number): number {
+    const f: number[] = Array(n + 1).fill(Infinity);
+    f[0] = 0;
+    for (let i = 1; i <= n; ++i) {
+        for (let j = 1; j <= i; ++j) {
+            f[i] = Math.min(f[i], 1 + Math.max(j - 1, f[i - j]));
+        }
+    }
+    return f[n];
+}
 ```
 
 <!-- tabs:end -->

solution/1800-1899/1884.Egg Drop With 2 Eggs and N Floors/README_EN.md

@@ -62,32 +62,107 @@ Regardless of the outcome, it takes at most 14 drops to determine f.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Dynamic Programming
+
+We define $f[i]$ to represent the minimum number of operations to determine $f$ in $i$ floors with two eggs. Initially, $f[0] = 0$, and the rest $f[i] = +\infty$. The answer is $f[n]$.
+
+Considering $f[i]$, we can enumerate the first egg thrown from the $j$-th floor, where $1 \leq j \leq i$. At this point, there are two cases:
+
+-   The egg breaks. At this time, we have one egg left and need to determine $f$ in $j - 1$ floors, which requires $j - 1$ operations. Therefore, the total number of operations is $1 + (j - 1)$;
+-   The egg does not break. At this time, we have two eggs left and need to determine $f$ in $i - j$ floors, which requires $f[i - j]$ operations. Therefore, the total number of operations is $1 + f[i - j]$.
+
+In summary, we can obtain the state transition equation:
+
+$$
+f[i] = \min_{1 \leq j \leq i} \{1 + \max(j - 1, f[i - j])\}
+$$
+
+Finally, we return $f[n]$.
+
+The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Where $n$ is the number of floors.
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def twoEggDrop(self, n: int) -> int:
+        f = [0] + [inf] * n
+        for i in range(1, n + 1):
+            for j in range(1, i + 1):
+                f[i] = min(f[i], 1 + max(j - 1, f[i - j]))
+        return f[n]
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int twoEggDrop(int n) {
+        int[] f = new int[n + 1];
+        Arrays.fill(f, 1 << 29);
+        f[0] = 0;
+        for (int i = 1; i <= n; i++) {
+            for (int j = 1; j <= i; j++) {
+                f[i] = Math.min(f[i], 1 + Math.max(j - 1, f[i - j]));
+            }
+        }
+        return f[n];
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    int twoEggDrop(int n) {
+        int f[n + 1];
+        memset(f, 0x3f, sizeof(f));
+        f[0] = 0;
+        for (int i = 1; i <= n; i++) {
+            for (int j = 1; j <= i; j++) {
+                f[i] = min(f[i], 1 + max(j - 1, f[i - j]));
+            }
+        }
+        return f[n];
+    }
+};
 ```
 
 #### Go
 
 ```go
+func twoEggDrop(n int) int {
+	f := make([]int, n+1)
+	for i := range f {
+		f[i] = 1 << 29
+	}
+	f[0] = 0
+	for i := 1; i <= n; i++ {
+		for j := 1; j <= i; j++ {
+			f[i] = min(f[i], 1+max(j-1, f[i-j]))
+		}
+	}
+	return f[n]
+}
+```
 
+#### TypeScript
+
+```ts
+function twoEggDrop(n: number): number {
+    const f: number[] = Array(n + 1).fill(Infinity);
+    f[0] = 0;
+    for (let i = 1; i <= n; ++i) {
+        for (let j = 1; j <= i; ++j) {
+            f[i] = Math.min(f[i], 1 + Math.max(j - 1, f[i - j]));
+        }
+    }
+    return f[n];
+}
 ```
 
 <!-- tabs:end -->

solution/1800-1899/1884.Egg Drop With 2 Eggs and N Floors/Solution.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+    int twoEggDrop(int n) {
+        int f[n + 1];
+        memset(f, 0x3f, sizeof(f));
+        f[0] = 0;
+        for (int i = 1; i <= n; i++) {
+            for (int j = 1; j <= i; j++) {
+                f[i] = min(f[i], 1 + max(j - 1, f[i - j]));
+            }
+        }
+        return f[n];
+    }
+};

solution/1800-1899/1884.Egg Drop With 2 Eggs and N Floors/Solution.go

@@ -0,0 +1,13 @@
+func twoEggDrop(n int) int {
+	f := make([]int, n+1)
+	for i := range f {
+		f[i] = 1 << 29
+	}
+	f[0] = 0
+	for i := 1; i <= n; i++ {
+		for j := 1; j <= i; j++ {
+			f[i] = min(f[i], 1+max(j-1, f[i-j]))
+		}
+	}
+	return f[n]
+}

solution/1800-1899/1884.Egg Drop With 2 Eggs and N Floors/Solution.java

@@ -0,0 +1,13 @@
+class Solution {
+    public int twoEggDrop(int n) {
+        int[] f = new int[n + 1];
+        Arrays.fill(f, 1 << 29);
+        f[0] = 0;
+        for (int i = 1; i <= n; i++) {
+            for (int j = 1; j <= i; j++) {
+                f[i] = Math.min(f[i], 1 + Math.max(j - 1, f[i - j]));
+            }
+        }
+        return f[n];
+    }
+}

solution/1800-1899/1884.Egg Drop With 2 Eggs and N Floors/Solution.py

@@ -0,0 +1,7 @@
+class Solution:
+    def twoEggDrop(self, n: int) -> int:
+        f = [0] + [inf] * n
+        for i in range(1, n + 1):
+            for j in range(1, i + 1):
+                f[i] = min(f[i], 1 + max(j - 1, f[i - j]))
+        return f[n]

solution/1800-1899/1884.Egg Drop With 2 Eggs and N Floors/Solution.ts

@@ -0,0 +1,10 @@
+function twoEggDrop(n: number): number {
+    const f: number[] = Array(n + 1).fill(Infinity);
+    f[0] = 0;
+    for (let i = 1; i <= n; ++i) {
+        for (let j = 1; j <= i; ++j) {
+            f[i] = Math.min(f[i], 1 + Math.max(j - 1, f[i - j]));
+        }
+    }
+    return f[n];
+}

solution/3100-3199/3171.Find Subarray With Bitwise OR Closest to K/README.md

@@ -85,7 +85,7 @@ tags:
 
 根据题目描述，我们需要求出数组 $\textit{nums}$ 下标 $l$ 到 $r$ 的元素的按位或运算的结果，即 $\textit{nums}[l] \lor \textit{nums}[l + 1] \lor \cdots \lor \textit{nums}[r]$。其中 $\lor$ 表示按位或运算。
 
-如果我们每次固定右端点 $r$，那么左端点 $l$ 的范围是 $[0, r]$。每次移动右端点 $r$，按位或的结果只会变大，我们用一个变量 $s$ 记录当前的按位或的结果，如果 $s$ 大于 $k$，我们就将左端点 $l$ 向右移动，直到 $s$ 小于等于 $k$。在移动左端点 $l$ 的过程中，我们需要维护一个数组 $cnt$，记录当前区间内每个二进制位上 $0$ 的个数，当 $cnt[h]$ 为 $0$ 时，说明当前区间内的元素在第 $h$ 位上都为 $1$，我们就可以将 $s$ 的第 $h$ 位设置为 $0$。
+如果我们每次固定右端点 $r$，那么左端点 $l$ 的范围是 $[0, r]$。每次移动右端点 $r$，按位或的结果只会变大，我们用一个变量 $s$ 记录当前的按位或的结果，如果 $s$ 大于 $k$，我们就将左端点 $l$ 向右移动，直到 $s$ 小于等于 $k$。在移动左端点 $l$ 的过程中，我们需要维护一个数组 $cnt$，记录当前区间内每个二进制位上 $0$ 的个数，当 $cnt[h]$ 为 $0$ 时，说明当前区间内的元素在第 $h$ 位上都为 $0$，我们就可以将 $s$ 的第 $h$ 位设置为 $0$。
 
 时间复杂度 $O(n \times \log M)$，空间复杂度 $O(\log M)$。其中 $n$ 和 $M$ 分别是数组 $\textit{nums}$ 的长度和数组 $\textit{nums}$ 中元素的最大值。
 

solution/3100-3199/3171.Find Subarray With Bitwise OR Closest to K/README_EN.md

@@ -83,7 +83,7 @@ tags:
 
 According to the problem description, we need to calculate the result of the bitwise OR operation of elements from index $l$ to $r$ in the array $\textit{nums}$, that is, $\textit{nums}[l] \lor \textit{nums}[l + 1] \lor \cdots \lor \textit{nums}[r]$, where $\lor$ represents the bitwise OR operation.
 
-If we fix the right endpoint $r$, then the range of the left endpoint $l$ is $[0, r]$. Each time we move the right endpoint $r$, the result of the bitwise OR operation will only increase. We use a variable $s$ to record the current result of the bitwise OR operation. If $s$ is greater than $k$, we move the left endpoint $l$ to the right until $s$ is less than or equal to $k$. During the process of moving the left endpoint $l$, we need to maintain an array $cnt$ to record the number of $0$s on each binary digit in the current interval. When $cnt[h] = 0$, it means that all elements in the current interval have a $1$ on the $h^{th}$ bit, and we can set the $h^{th}$ bit of $s$ to $0$.
+If we fix the right endpoint $r$, then the range of the left endpoint $l$ is $[0, r]$. Each time we move the right endpoint $r$, the result of the bitwise OR operation will only increase. We use a variable $s$ to record the current result of the bitwise OR operation. If $s$ is greater than $k$, we move the left endpoint $l$ to the right until $s$ is less than or equal to $k$. During the process of moving the left endpoint $l$, we need to maintain an array $cnt$ to record the number of $0$s on each binary digit in the current interval. When $cnt[h] = 0$, it means that all elements in the current interval have a $0$ on the $h^{th}$ bit, and we can set the $h^{th}$ bit of $s$ to $0$.
 
 The time complexity is $O(n \times \log M)$, and the space complexity is $O(\log M)$. Here, $n$ and $M$ respectively represent the length of the array $\textit{nums}$ and the maximum value in the array $\textit{nums}$.
 

solution/3100-3199/3194.Minimum Average of Smallest and Largest Elements/README.md

@@ -189,7 +189,7 @@ class Solution:
     def minimumAverage(self, nums: List[int]) -> float:
         nums.sort()
         n = len(nums)
-        return min(nums[i] + nums[n - i - 1] for i in range(n // 2)) / 2
+        return min(nums[i] + nums[-i - 1] for i in range(n // 2)) / 2
 ```
 
 #### Java
@@ -252,6 +252,19 @@ function minimumAverage(nums: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn minimum_average(mut nums: Vec<i32>) -> f64 {
+        nums.sort();
+        let n = nums.len();
+        let ans = (0..n / 2).map(|i| nums[i] + nums[n - i - 1]).min().unwrap();
+        ans as f64 / 2.0
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3100-3199/3194.Minimum Average of Smallest and Largest Elements/README_EN.md

@@ -187,7 +187,7 @@ class Solution:
     def minimumAverage(self, nums: List[int]) -> float:
         nums.sort()
         n = len(nums)
-        return min(nums[i] + nums[n - i - 1] for i in range(n // 2)) / 2
+        return min(nums[i] + nums[-i - 1] for i in range(n // 2)) / 2
 ```
 
 #### Java
@@ -250,6 +250,19 @@ function minimumAverage(nums: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn minimum_average(mut nums: Vec<i32>) -> f64 {
+        nums.sort();
+        let n = nums.len();
+        let ans = (0..n / 2).map(|i| nums[i] + nums[n - i - 1]).min().unwrap();
+        ans as f64 / 2.0
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3100-3199/3194.Minimum Average of Smallest and Largest Elements/Solution.py

@@ -2,4 +2,4 @@ class Solution:
     def minimumAverage(self, nums: List[int]) -> float:
         nums.sort()
         n = len(nums)
-        return min(nums[i] + nums[n - i - 1] for i in range(n // 2)) / 2
+        return min(nums[i] + nums[-i - 1] for i in range(n // 2)) / 2

solution/3100-3199/3194.Minimum Average of Smallest and Largest Elements/Solution.rs

@@ -0,0 +1,8 @@
+impl Solution {
+    pub fn minimum_average(mut nums: Vec<i32>) -> f64 {
+        nums.sort();
+        let n = nums.len();
+        let ans = (0..n / 2).map(|i| nums[i] + nums[n - i - 1]).min().unwrap();
+        ans as f64 / 2.0
+    }
+}