feat: add solutions to lc problem: No.2227

No.2227.Encrypt and Decrypt Strings

Author: yanglbme

solution/2200-2299/2227.Encrypt and Decrypt Strings/README.md

@@ -61,11 +61,11 @@ tags:
 
 <strong>解释：</strong>
 Encrypter encrypter = new Encrypter([['a', 'b', 'c', 'd'], ["ei", "zf", "ei", "am"], ["abcd", "acbd", "adbc", "badc", "dacb", "cadb", "cbda", "abad"]);
-encrypter.encrypt("abcd"); // 返回 "eizfeiam"。 
+encrypter.encrypt("abcd"); // 返回 "eizfeiam"。
 &nbsp;                          // 'a' 映射为 "ei"，'b' 映射为 "zf"，'c' 映射为 "ei"，'d' 映射为 "am"。
-encrypter.decrypt("eizfeiam"); // return 2. 
-                              // "ei" 可以映射为 'a' 或 'c'，"zf" 映射为 'b'，"am" 映射为 'd'。 
-                              // 因此，解密后可以得到的字符串是 "abad"，"cbad"，"abcd" 和 "cbcd"。 
+encrypter.decrypt("eizfeiam"); // return 2.
+                              // "ei" 可以映射为 'a' 或 'c'，"zf" 映射为 'b'，"am" 映射为 'd'。
+                              // 因此，解密后可以得到的字符串是 "abad"，"cbad"，"abcd" 和 "cbcd"。
                               // 其中 2 个字符串，"abad" 和 "abcd"，在 dictionary 中出现，所以答案是 2 。
 </pre>
 
@@ -95,6 +95,16 @@ encrypter.decrypt("eizfeiam"); // return 2.
 
 ### 方法一：哈希表
 
+我们用一个哈希表 $\textit{mp}$ 记录每个字符的加密结果，用另一个哈希表 $\textit{cnt}$ 记录每个加密结果出现的次数。
+
+在构造函数中，我们遍历 $\textit{keys}$ 和 $\textit{values}$，将每个字符和其对应的加密结果存入 $\textit{mp}$ 中。然后遍历 $\textit{dictionary}$，统计每个加密结果出现的次数。时间复杂度 $(n + m)$，其中 $n$ 和 $m$ 分别是 $\textit{keys}$ 和 $\textit{dictionary}$ 的长度。
+
+在加密函数中，我们遍历输入字符串 $\textit{word1}$ 的每个字符，查找其加密结果并拼接起来。如果某个字符没有对应的加密结果，说明无法加密，返回空字符串。时间复杂度 $O(k)$，其中 $k$ 是 $\textit{word1}$ 的长度。
+
+在解密函数中，我们直接返回 $\textit{cnt}$ 中 $\textit{word2}$ 对应的次数。时间复杂度 $O(1)$。
+
+空间复杂度 $O(n + m)$。
+
 <!-- tabs:start -->
 
 #### Python3
@@ -135,8 +145,7 @@ class Encrypter {
             mp.put(keys[i], values[i]);
         }
         for (String w : dictionary) {
-            w = encrypt(w);
-            cnt.put(w, cnt.getOrDefault(w, 0) + 1);
+            cnt.merge(encrypt(w), 1, Integer::sum);
         }
     }
 
@@ -173,14 +182,20 @@ public:
     unordered_map<char, string> mp;
 
     Encrypter(vector<char>& keys, vector<string>& values, vector<string>& dictionary) {
-        for (int i = 0; i < keys.size(); ++i) mp[keys[i]] = values[i];
-        for (auto v : dictionary) cnt[encrypt(v)]++;
+        for (int i = 0; i < keys.size(); ++i) {
+            mp[keys[i]] = values[i];
+        }
+        for (auto v : dictionary) {
+            cnt[encrypt(v)]++;
+        }
     }
 
     string encrypt(string word1) {
         string res = "";
         for (char c : word1) {
-            if (!mp.count(c)) return "";
+            if (!mp.count(c)) {
+                return "";
+            }
             res += mp[c];
         }
         return res;
@@ -244,6 +259,49 @@ func (this *Encrypter) Decrypt(word2 string) int {
  */
 ```
 
+#### TypeScript
+
+```ts
+class Encrypter {
+    private mp: Map<string, string> = new Map();
+    private cnt: Map<string, number> = new Map();
+
+    constructor(keys: string[], values: string[], dictionary: string[]) {
+        for (let i = 0; i < keys.length; i++) {
+            this.mp.set(keys[i], values[i]);
+        }
+        for (const w of dictionary) {
+            const encrypted = this.encrypt(w);
+            if (encrypted !== '') {
+                this.cnt.set(encrypted, (this.cnt.get(encrypted) || 0) + 1);
+            }
+        }
+    }
+
+    encrypt(word: string): string {
+        let res = '';
+        for (const c of word) {
+            if (!this.mp.has(c)) {
+                return '';
+            }
+            res += this.mp.get(c);
+        }
+        return res;
+    }
+
+    decrypt(word: string): number {
+        return this.cnt.get(word) || 0;
+    }
+}
+
+/**
+ * Your Encrypter object will be instantiated and called as such:
+ * const obj = new Encrypter(keys, values, dictionary);
+ * const param_1 = obj.encrypt(word1);
+ * const param_2 = obj.decrypt(word2);
+ */
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2200-2299/2227.Encrypt and Decrypt Strings/README_EN.md

@@ -60,11 +60,11 @@ tags:
 
 <strong>Explanation</strong>
 Encrypter encrypter = new Encrypter([[&#39;a&#39;, &#39;b&#39;, &#39;c&#39;, &#39;d&#39;], [&quot;ei&quot;, &quot;zf&quot;, &quot;ei&quot;, &quot;am&quot;], [&quot;abcd&quot;, &quot;acbd&quot;, &quot;adbc&quot;, &quot;badc&quot;, &quot;dacb&quot;, &quot;cadb&quot;, &quot;cbda&quot;, &quot;abad&quot;]);
-encrypter.encrypt(&quot;abcd&quot;); // return &quot;eizfeiam&quot;. 
+encrypter.encrypt(&quot;abcd&quot;); // return &quot;eizfeiam&quot;.
 &nbsp;                          // &#39;a&#39; maps to &quot;ei&quot;, &#39;b&#39; maps to &quot;zf&quot;, &#39;c&#39; maps to &quot;ei&quot;, and &#39;d&#39; maps to &quot;am&quot;.
-encrypter.decrypt(&quot;eizfeiam&quot;); // return 2. 
-                              // &quot;ei&quot; can map to &#39;a&#39; or &#39;c&#39;, &quot;zf&quot; maps to &#39;b&#39;, and &quot;am&quot; maps to &#39;d&#39;. 
-                              // Thus, the possible strings after decryption are &quot;abad&quot;, &quot;cbad&quot;, &quot;abcd&quot;, and &quot;cbcd&quot;. 
+encrypter.decrypt(&quot;eizfeiam&quot;); // return 2.
+                              // &quot;ei&quot; can map to &#39;a&#39; or &#39;c&#39;, &quot;zf&quot; maps to &#39;b&#39;, and &quot;am&quot; maps to &#39;d&#39;.
+                              // Thus, the possible strings after decryption are &quot;abad&quot;, &quot;cbad&quot;, &quot;abcd&quot;, and &quot;cbcd&quot;.
                               // 2 of those strings, &quot;abad&quot; and &quot;abcd&quot;, appear in dictionary, so the answer is 2.
 </pre>
 
@@ -91,7 +91,17 @@ encrypter.decrypt(&quot;eizfeiam&quot;); // return 2.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Hash Table
+
+We use a hash table $\textit{mp}$ to record the encryption result of each character, and another hash table $\textit{cnt}$ to record the number of occurrences of each encryption result.
+
+In the constructor, we traverse $\textit{keys}$ and $\textit{values}$, storing each character and its corresponding encryption result in $\textit{mp}$. Then, we traverse $\textit{dictionary}$ to count the occurrences of each encryption result. The time complexity is $O(n + m)$, where $n$ and $m$ are the lengths of $\textit{keys}$ and $\textit{dictionary}$, respectively.
+
+In the encryption function, we traverse each character of the input string $\textit{word1}$, look up its encryption result, and concatenate them. If a character does not have a corresponding encryption result, it means encryption is not possible, and we return an empty string. The time complexity is $O(k)$, where $k$ is the length of $\textit{word1}$.
+
+In the decryption function, we directly return the count of $\textit{word2}$ in $\textit{cnt}$. The time complexity is $O(1)$.
+
+The space complexity is $O(n + m)$.
 
 <!-- tabs:start -->
 
@@ -133,8 +143,7 @@ class Encrypter {
             mp.put(keys[i], values[i]);
         }
         for (String w : dictionary) {
-            w = encrypt(w);
-            cnt.put(w, cnt.getOrDefault(w, 0) + 1);
+            cnt.merge(encrypt(w), 1, Integer::sum);
         }
     }
 
@@ -171,14 +180,20 @@ public:
     unordered_map<char, string> mp;
 
     Encrypter(vector<char>& keys, vector<string>& values, vector<string>& dictionary) {
-        for (int i = 0; i < keys.size(); ++i) mp[keys[i]] = values[i];
-        for (auto v : dictionary) cnt[encrypt(v)]++;
+        for (int i = 0; i < keys.size(); ++i) {
+            mp[keys[i]] = values[i];
+        }
+        for (auto v : dictionary) {
+            cnt[encrypt(v)]++;
+        }
     }
 
     string encrypt(string word1) {
         string res = "";
         for (char c : word1) {
-            if (!mp.count(c)) return "";
+            if (!mp.count(c)) {
+                return "";
+            }
             res += mp[c];
         }
         return res;
@@ -242,6 +257,49 @@ func (this *Encrypter) Decrypt(word2 string) int {
  */
 ```
 
+#### TypeScript
+
+```ts
+class Encrypter {
+    private mp: Map<string, string> = new Map();
+    private cnt: Map<string, number> = new Map();
+
+    constructor(keys: string[], values: string[], dictionary: string[]) {
+        for (let i = 0; i < keys.length; i++) {
+            this.mp.set(keys[i], values[i]);
+        }
+        for (const w of dictionary) {
+            const encrypted = this.encrypt(w);
+            if (encrypted !== '') {
+                this.cnt.set(encrypted, (this.cnt.get(encrypted) || 0) + 1);
+            }
+        }
+    }
+
+    encrypt(word: string): string {
+        let res = '';
+        for (const c of word) {
+            if (!this.mp.has(c)) {
+                return '';
+            }
+            res += this.mp.get(c);
+        }
+        return res;
+    }
+
+    decrypt(word: string): number {
+        return this.cnt.get(word) || 0;
+    }
+}
+
+/**
+ * Your Encrypter object will be instantiated and called as such:
+ * const obj = new Encrypter(keys, values, dictionary);
+ * const param_1 = obj.encrypt(word1);
+ * const param_2 = obj.decrypt(word2);
+ */
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2200-2299/2227.Encrypt and Decrypt Strings/Solution.cpp

@@ -4,14 +4,20 @@ class Encrypter {
     unordered_map<char, string> mp;
 
     Encrypter(vector<char>& keys, vector<string>& values, vector<string>& dictionary) {
-        for (int i = 0; i < keys.size(); ++i) mp[keys[i]] = values[i];
-        for (auto v : dictionary) cnt[encrypt(v)]++;
+        for (int i = 0; i < keys.size(); ++i) {
+            mp[keys[i]] = values[i];
+        }
+        for (auto v : dictionary) {
+            cnt[encrypt(v)]++;
+        }
     }
 
     string encrypt(string word1) {
         string res = "";
         for (char c : word1) {
-            if (!mp.count(c)) return "";
+            if (!mp.count(c)) {
+                return "";
+            }
             res += mp[c];
         }
         return res;
@@ -27,4 +33,4 @@ class Encrypter {
  * Encrypter* obj = new Encrypter(keys, values, dictionary);
  * string param_1 = obj->encrypt(word1);
  * int param_2 = obj->decrypt(word2);
- */
+ */

solution/2200-2299/2227.Encrypt and Decrypt Strings/Solution.java

@@ -7,8 +7,7 @@ public Encrypter(char[] keys, String[] values, String[] dictionary) {
             mp.put(keys[i], values[i]);
         }
         for (String w : dictionary) {
-            w = encrypt(w);
-            cnt.put(w, cnt.getOrDefault(w, 0) + 1);
+            cnt.merge(encrypt(w), 1, Integer::sum);
         }
     }
 
@@ -33,4 +32,4 @@ public int decrypt(String word2) {
  * Encrypter obj = new Encrypter(keys, values, dictionary);
  * String param_1 = obj.encrypt(word1);
  * int param_2 = obj.decrypt(word2);
- */
+ */

solution/2200-2299/2227.Encrypt and Decrypt Strings/Solution.ts

@@ -0,0 +1,38 @@
+class Encrypter {
+    private mp: Map<string, string> = new Map();
+    private cnt: Map<string, number> = new Map();
+
+    constructor(keys: string[], values: string[], dictionary: string[]) {
+        for (let i = 0; i < keys.length; i++) {
+            this.mp.set(keys[i], values[i]);
+        }
+        for (const w of dictionary) {
+            const encrypted = this.encrypt(w);
+            if (encrypted !== '') {
+                this.cnt.set(encrypted, (this.cnt.get(encrypted) || 0) + 1);
+            }
+        }
+    }
+
+    encrypt(word: string): string {
+        let res = '';
+        for (const c of word) {
+            if (!this.mp.has(c)) {
+                return '';
+            }
+            res += this.mp.get(c);
+        }
+        return res;
+    }
+
+    decrypt(word: string): number {
+        return this.cnt.get(word) || 0;
+    }
+}
+
+/**
+ * Your Encrypter object will be instantiated and called as such:
+ * const obj = new Encrypter(keys, values, dictionary);
+ * const param_1 = obj.encrypt(word1);
+ * const param_2 = obj.decrypt(word2);
+ */