feat: add solutions to lc problems: No.2357,2358

solution/2300-2399/2357.Make Array Zero by Subtracting Equal Amounts/README.md

@@ -69,9 +69,9 @@ tags:
 
 ### 方法一：哈希表或数组
 
-我们观察到，每一次操作，都可以把数组 `nums` 中相同且非零的元素减少到 $0$，因此，我们只需要统计数组 `nums` 中有多少个不同的非零元素，即为最少操作数。统计不同的非零元素，可以使用哈希表或数组来实现。
+我们观察到，每一次操作，都可以把数组 $\textit{nums}$ 中相同且非零的元素减少到 $0$，因此，我们只需要统计数组 $\textit{nums}$ 中有多少个不同的非零元素，即为最少操作数。统计不同的非零元素，可以使用哈希表或数组来实现。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组长度。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -141,9 +141,9 @@ func minimumOperations(nums []int) (ans int) {
 
 ```ts
 function minimumOperations(nums: number[]): number {
-    const set = new Set(nums);
-    set.delete(0);
-    return set.size;
+    const s = new Set(nums);
+    s.delete(0);
+    return s.size;
 }
 ```
 
@@ -153,9 +153,9 @@ function minimumOperations(nums: number[]): number {
 use std::collections::HashSet;
 impl Solution {
     pub fn minimum_operations(nums: Vec<i32>) -> i32 {
-        let mut set = nums.iter().collect::<HashSet<&i32>>();
-        set.remove(&0);
-        set.len() as i32
+        let mut s = nums.iter().collect::<HashSet<&i32>>();
+        s.remove(&0);
+        s.len() as i32
     }
 }
 ```

solution/2300-2399/2357.Make Array Zero by Subtracting Equal Amounts/README_EN.md

@@ -66,7 +66,11 @@ In the third operation, choose x = 2. Now, nums = [0,0,0,0,0].
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Hash Table or Array
+
+We observe that in each operation, all identical nonzero elements in the array $\textit{nums}$ can be reduced to $0$. Therefore, we only need to count the number of distinct nonzero elements in $\textit{nums}$, which is the minimum number of operations required. To count the distinct nonzero elements, we can use a hash table or an array.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$.
 
 <!-- tabs:start -->
 
@@ -136,9 +140,9 @@ func minimumOperations(nums []int) (ans int) {
 
 ```ts
 function minimumOperations(nums: number[]): number {
-    const set = new Set(nums);
-    set.delete(0);
-    return set.size;
+    const s = new Set(nums);
+    s.delete(0);
+    return s.size;
 }
 ```
 
@@ -148,9 +152,9 @@ function minimumOperations(nums: number[]): number {
 use std::collections::HashSet;
 impl Solution {
     pub fn minimum_operations(nums: Vec<i32>) -> i32 {
-        let mut set = nums.iter().collect::<HashSet<&i32>>();
-        set.remove(&0);
-        set.len() as i32
+        let mut s = nums.iter().collect::<HashSet<&i32>>();
+        s.remove(&0);
+        s.len() as i32
     }
 }
 ```

solution/2300-2399/2357.Make Array Zero by Subtracting Equal Amounts/Solution.rs

@@ -1,8 +1,8 @@
 use std::collections::HashSet;
 impl Solution {
     pub fn minimum_operations(nums: Vec<i32>) -> i32 {
-        let mut set = nums.iter().collect::<HashSet<&i32>>();
-        set.remove(&0);
-        set.len() as i32
+        let mut s = nums.iter().collect::<HashSet<&i32>>();
+        s.remove(&0);
+        s.len() as i32
     }
 }

solution/2300-2399/2357.Make Array Zero by Subtracting Equal Amounts/Solution.ts

@@ -1,5 +1,5 @@
 function minimumOperations(nums: number[]): number {
-    const set = new Set(nums);
-    set.delete(0);
-    return set.size;
+    const s = new Set(nums);
+    s.delete(0);
+    return s.size;
 }

solution/2300-2399/2358.Maximum Number of Groups Entering a Competition/README.md

@@ -160,6 +160,26 @@ function maximumGroups(grades: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn maximum_groups(grades: Vec<i32>) -> i32 {
+        let n = grades.len() as i64;
+        let (mut l, mut r) = (0i64, n);
+        while l < r {
+            let mid = (l + r + 1) / 2;
+            if mid * mid + mid > 2 * n {
+                r = mid - 1;
+            } else {
+                l = mid;
+            }
+        }
+        l as i32
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2300-2399/2358.Maximum Number of Groups Entering a Competition/README_EN.md

@@ -65,7 +65,17 @@ It can be shown that it is not possible to form more than 3 groups.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy + Binary Search
+
+Observing the conditions in the problem, the number of students in the $i$-th group must be less than that in the $(i+1)$-th group, and the total score of students in the $i$-th group must be less than that in the $(i+1)$-th group. We only need to sort the students by their scores in ascending order, and then assign $1$, $2$, ..., $k$ students to each group in order. If the last group does not have enough students for $k$, we can distribute these students to the previous last group.
+
+Therefore, we need to find the largest $k$ such that $\frac{(1 + k) \times k}{2} \leq n$, where $n$ is the total number of students. We can use binary search to solve this.
+
+We define the left boundary of binary search as $l = 1$ and the right boundary as $r = n$. Each time, the midpoint is $mid = \lfloor \frac{l + r + 1}{2} \rfloor$. If $(1 + mid) \times mid \gt 2 \times n$, it means $mid$ is too large, so we shrink the right boundary to $mid - 1$; otherwise, we increase the left boundary to $mid$.
+
+Finally, we return $l$ as the answer.
+
+The time complexity is $O(\log n)$ and the space complexity is $O(1)$, where $n$ is the total number of students.
 
 <!-- tabs:start -->
 
@@ -150,6 +160,26 @@ function maximumGroups(grades: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn maximum_groups(grades: Vec<i32>) -> i32 {
+        let n = grades.len() as i64;
+        let (mut l, mut r) = (0i64, n);
+        while l < r {
+            let mid = (l + r + 1) / 2;
+            if mid * mid + mid > 2 * n {
+                r = mid - 1;
+            } else {
+                l = mid;
+            }
+        }
+        l as i32
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2300-2399/2358.Maximum Number of Groups Entering a Competition/Solution.rs

@@ -0,0 +1,15 @@
+impl Solution {
+    pub fn maximum_groups(grades: Vec<i32>) -> i32 {
+        let n = grades.len() as i64;
+        let (mut l, mut r) = (0i64, n);
+        while l < r {
+            let mid = (l + r + 1) / 2;
+            if mid * mid + mid > 2 * n {
+                r = mid - 1;
+            } else {
+                l = mid;
+            }
+        }
+        l as i32
+    }
+}