feat: add solutions to lc problems: No.3018~3020

solution/3000-3099/3018.Maximum Number of Removal Queries That Can Be Processed I/README.md

@@ -273,6 +273,50 @@ function maximumProcessableQueries(nums: number[], queries: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn maximum_processable_queries(nums: Vec<i32>, queries: Vec<i32>) -> i32 {
+        let n = nums.len();
+        let m = queries.len();
+        let mut f = vec![vec![0; n]; n];
+
+        for i in 0..n {
+            for j in (i..n).rev() {
+                if i > 0 {
+                    let idx = f[i - 1][j] as usize;
+                    if idx < m {
+                        f[i][j] = f[i][j].max(f[i - 1][j] + if nums[i - 1] >= queries[idx] { 1 } else { 0 });
+                    }
+                }
+                if j + 1 < n {
+                    let idx = f[i][j + 1] as usize;
+                    if idx < m {
+                        f[i][j] = f[i][j].max(f[i][j + 1] + if nums[j + 1] >= queries[idx] { 1 } else { 0 });
+                    }
+                }
+                if f[i][j] as usize == m {
+                    return m as i32;
+                }
+            }
+        }
+
+        let mut ans = 0;
+        for i in 0..n {
+            let idx = f[i][i] as usize;
+            if idx < m {
+                ans = ans.max(f[i][i] + if nums[i] >= queries[idx] { 1 } else { 0 });
+            } else {
+                ans = ans.max(f[i][i]);
+            }
+        }
+
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3000-3099/3018.Maximum Number of Removal Queries That Can Be Processed I/README_EN.md

@@ -271,6 +271,50 @@ function maximumProcessableQueries(nums: number[], queries: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn maximum_processable_queries(nums: Vec<i32>, queries: Vec<i32>) -> i32 {
+        let n = nums.len();
+        let m = queries.len();
+        let mut f = vec![vec![0; n]; n];
+
+        for i in 0..n {
+            for j in (i..n).rev() {
+                if i > 0 {
+                    let idx = f[i - 1][j] as usize;
+                    if idx < m {
+                        f[i][j] = f[i][j].max(f[i - 1][j] + if nums[i - 1] >= queries[idx] { 1 } else { 0 });
+                    }
+                }
+                if j + 1 < n {
+                    let idx = f[i][j + 1] as usize;
+                    if idx < m {
+                        f[i][j] = f[i][j].max(f[i][j + 1] + if nums[j + 1] >= queries[idx] { 1 } else { 0 });
+                    }
+                }
+                if f[i][j] as usize == m {
+                    return m as i32;
+                }
+            }
+        }
+
+        let mut ans = 0;
+        for i in 0..n {
+            let idx = f[i][i] as usize;
+            if idx < m {
+                ans = ans.max(f[i][i] + if nums[i] >= queries[idx] { 1 } else { 0 });
+            } else {
+                ans = ans.max(f[i][i]);
+            }
+        }
+
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3000-3099/3018.Maximum Number of Removal Queries That Can Be Processed I/Solution.rs

@@ -0,0 +1,41 @@
+impl Solution {
+    pub fn maximum_processable_queries(nums: Vec<i32>, queries: Vec<i32>) -> i32 {
+        let n = nums.len();
+        let m = queries.len();
+        let mut f = vec![vec![0; n]; n];
+
+        for i in 0..n {
+            for j in (i..n).rev() {
+                if i > 0 {
+                    let idx = f[i - 1][j] as usize;
+                    if idx < m {
+                        f[i][j] = f[i][j]
+                            .max(f[i - 1][j] + if nums[i - 1] >= queries[idx] { 1 } else { 0 });
+                    }
+                }
+                if j + 1 < n {
+                    let idx = f[i][j + 1] as usize;
+                    if idx < m {
+                        f[i][j] = f[i][j]
+                            .max(f[i][j + 1] + if nums[j + 1] >= queries[idx] { 1 } else { 0 });
+                    }
+                }
+                if f[i][j] as usize == m {
+                    return m as i32;
+                }
+            }
+        }
+
+        let mut ans = 0;
+        for i in 0..n {
+            let idx = f[i][i] as usize;
+            if idx < m {
+                ans = ans.max(f[i][i] + if nums[i] >= queries[idx] { 1 } else { 0 });
+            } else {
+                ans = ans.max(f[i][i]);
+            }
+        }
+
+        ans
+    }
+}

solution/3000-3099/3019.Number of Changing Keys/README.md

@@ -31,7 +31,7 @@ tags:
 <pre>
 <strong>输入：</strong>s = "aAbBcC"
 <strong>输出：</strong>2
-<strong>解释：</strong> 
+<strong>解释：</strong>
 从 s[0] = 'a' 到 s[1] = 'A'，不存在按键变更，因为不计入 caps lock 或 shift 。
 从 s[1] = 'A' 到 s[2] = 'b'，按键变更。
 从 s[2] = 'b' 到 s[3] = 'B'，不存在按键变更，因为不计入 caps lock 或 shift 。
@@ -139,6 +139,24 @@ function countKeyChanges(s: string): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn count_key_changes(s: String) -> i32 {
+        let s = s.to_lowercase();
+        let bytes = s.as_bytes();
+        let mut ans = 0;
+        for i in 1..bytes.len() {
+            if bytes[i] != bytes[i - 1] {
+                ans += 1;
+            }
+        }
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3000-3099/3019.Number of Changing Keys/README_EN.md

@@ -30,7 +30,7 @@ tags:
 <pre>
 <strong>Input:</strong> s = &quot;aAbBcC&quot;
 <strong>Output:</strong> 2
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 From s[0] = &#39;a&#39; to s[1] = &#39;A&#39;, there is no change of key as caps lock or shift is not counted.
 From s[1] = &#39;A&#39; to s[2] = &#39;b&#39;, there is a change of key.
 From s[2] = &#39;b&#39; to s[3] = &#39;B&#39;, there is no change of key as caps lock or shift is not counted.
@@ -138,6 +138,24 @@ function countKeyChanges(s: string): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn count_key_changes(s: String) -> i32 {
+        let s = s.to_lowercase();
+        let bytes = s.as_bytes();
+        let mut ans = 0;
+        for i in 1..bytes.len() {
+            if bytes[i] != bytes[i - 1] {
+                ans += 1;
+            }
+        }
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3000-3099/3019.Number of Changing Keys/Solution.rs

@@ -0,0 +1,13 @@
+impl Solution {
+    pub fn count_key_changes(s: String) -> i32 {
+        let s = s.to_lowercase();
+        let bytes = s.as_bytes();
+        let mut ans = 0;
+        for i in 1..bytes.len() {
+            if bytes[i] != bytes[i - 1] {
+                ans += 1;
+            }
+        }
+        ans
+    }
+}

solution/3000-3099/3020.Find the Maximum Number of Elements in Subset/README.md

@@ -193,6 +193,39 @@ function maximumLength(nums: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn maximum_length(nums: Vec<i32>) -> i32 {
+        let mut cnt: HashMap<i64, i32> = HashMap::new();
+        for &x in &nums {
+            *cnt.entry(x as i64).or_insert(0) += 1;
+        }
+
+        let mut ans = 0;
+        if let Some(t) = cnt.remove(&1) {
+            ans = t - ((t % 2) ^ 1);
+        }
+
+        for &key in cnt.keys() {
+            let mut x = key;
+            let mut t = 0;
+            while *cnt.get(&x).unwrap_or(&0) > 1 {
+                x = x * x;
+                t += 2;
+            }
+            t += cnt.get(&x).unwrap_or(&-1);
+            ans = ans.max(t);
+        }
+
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3000-3099/3020.Find the Maximum Number of Elements in Subset/README_EN.md

@@ -44,7 +44,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [1,3,2,4]
 <strong>Output:</strong> 1
-<strong>Explanation:</strong> We can select the subset {1}, which can be placed in the array as [1] which follows the pattern. Hence the answer is 1. Note that we could have also selected the subsets {2}, {3}, or {4}, there may be multiple subsets which provide the same answer. 
+<strong>Explanation:</strong> We can select the subset {1}, which can be placed in the array as [1] which follows the pattern. Hence the answer is 1. Note that we could have also selected the subsets {2}, {3}, or {4}, there may be multiple subsets which provide the same answer.
 </pre>
 
 <p>&nbsp;</p>
@@ -191,6 +191,39 @@ function maximumLength(nums: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn maximum_length(nums: Vec<i32>) -> i32 {
+        let mut cnt: HashMap<i64, i32> = HashMap::new();
+        for &x in &nums {
+            *cnt.entry(x as i64).or_insert(0) += 1;
+        }
+
+        let mut ans = 0;
+        if let Some(t) = cnt.remove(&1) {
+            ans = t - ((t % 2) ^ 1);
+        }
+
+        for &key in cnt.keys() {
+            let mut x = key;
+            let mut t = 0;
+            while *cnt.get(&x).unwrap_or(&0) > 1 {
+                x = x * x;
+                t += 2;
+            }
+            t += cnt.get(&x).unwrap_or(&-1);
+            ans = ans.max(t);
+        }
+
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3000-3099/3020.Find the Maximum Number of Elements in Subset/Solution.rs

@@ -0,0 +1,28 @@
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn maximum_length(nums: Vec<i32>) -> i32 {
+        let mut cnt: HashMap<i64, i32> = HashMap::new();
+        for &x in &nums {
+            *cnt.entry(x as i64).or_insert(0) += 1;
+        }
+
+        let mut ans = 0;
+        if let Some(t) = cnt.remove(&1) {
+            ans = t - ((t % 2) ^ 1);
+        }
+
+        for &key in cnt.keys() {
+            let mut x = key;
+            let mut t = 0;
+            while *cnt.get(&x).unwrap_or(&0) > 1 {
+                x = x * x;
+                t += 2;
+            }
+            t += cnt.get(&x).unwrap_or(&-1);
+            ans = ans.max(t);
+        }
+
+        ans
+    }
+}