feat: add solutions to lc problem: No.2452

No.2452.Words Within Two Edits of Dictionary

Author: YangFong

solution/2400-2499/2452.Words Within Two Edits of Dictionary/README.md

@@ -149,7 +149,50 @@ func twoEditWords(queries []string, dictionary []string) (ans []string) {
 ### **TypeScript**
 
 ```ts
+function twoEditWords(queries: string[], dictionary: string[]): string[] {
+    const n = queries[0].length;
+    return queries.filter(querie => {
+        for (const s of dictionary) {
+            let diff = 0;
+            for (let i = 0; i < n; i++) {
+                if (querie[i] !== s[i] && ++diff > 2) {
+                    break;
+                }
+            }
+            if (diff <= 2) {
+                return true;
+            }
+        }
+        return false;
+    });
+}
+```
 
+### **Rust**
+
+```rust
+impl Solution {
+    pub fn two_edit_words(queries: Vec<String>, dictionary: Vec<String>) -> Vec<String> {
+        let n = queries[0].len();
+        queries
+            .into_iter()
+            .filter(|querie| {
+                for s in dictionary.iter() {
+                    let mut diff = 0;
+                    for i in 0..n {
+                        if querie.as_bytes()[i] != s.as_bytes()[i] {
+                            diff += 1;
+                        }
+                    }
+                    if diff <= 2 {
+                        return true;
+                    }
+                }
+                false
+            })
+            .collect()
+    }
+}
 ```
 
 ### **...**

solution/2400-2499/2452.Words Within Two Edits of Dictionary/README_EN.md

@@ -134,7 +134,50 @@ func twoEditWords(queries []string, dictionary []string) (ans []string) {
 ### **TypeScript**
 
 ```ts
+function twoEditWords(queries: string[], dictionary: string[]): string[] {
+    const n = queries[0].length;
+    return queries.filter(querie => {
+        for (const s of dictionary) {
+            let diff = 0;
+            for (let i = 0; i < n; i++) {
+                if (querie[i] !== s[i] && ++diff > 2) {
+                    break;
+                }
+            }
+            if (diff <= 2) {
+                return true;
+            }
+        }
+        return false;
+    });
+}
+```
 
+### **Rust**
+
+```rust
+impl Solution {
+    pub fn two_edit_words(queries: Vec<String>, dictionary: Vec<String>) -> Vec<String> {
+        let n = queries[0].len();
+        queries
+            .into_iter()
+            .filter(|querie| {
+                for s in dictionary.iter() {
+                    let mut diff = 0;
+                    for i in 0..n {
+                        if querie.as_bytes()[i] != s.as_bytes()[i] {
+                            diff += 1;
+                        }
+                    }
+                    if diff <= 2 {
+                        return true;
+                    }
+                }
+                false
+            })
+            .collect()
+    }
+}
 ```
 
 ### **...**

solution/2400-2499/2452.Words Within Two Edits of Dictionary/Solution.rs

@@ -0,0 +1,22 @@
+impl Solution {
+    pub fn two_edit_words(queries: Vec<String>, dictionary: Vec<String>) -> Vec<String> {
+        let n = queries[0].len();
+        queries
+            .into_iter()
+            .filter(|querie| {
+                for s in dictionary.iter() {
+                    let mut diff = 0;
+                    for i in 0..n {
+                        if querie.as_bytes()[i] != s.as_bytes()[i] {
+                            diff += 1;
+                        }
+                    }
+                    if diff <= 2 {
+                        return true;
+                    }
+                }
+                false
+            })
+            .collect()
+    }
+}

solution/2400-2499/2452.Words Within Two Edits of Dictionary/Solution.ts

@@ -0,0 +1,17 @@
+function twoEditWords(queries: string[], dictionary: string[]): string[] {
+    const n = queries[0].length;
+    return queries.filter(querie => {
+        for (const s of dictionary) {
+            let diff = 0;
+            for (let i = 0; i < n; i++) {
+                if (querie[i] !== s[i] && ++diff > 2) {
+                    break;
+                }
+            }
+            if (diff <= 2) {
+                return true;
+            }
+        }
+        return false;
+    });
+}