feat: add solutions to lc problem: No.3224

No.3324.Find the Sequence of Strings Appeared on the Screen

Author: yanglbme

solution/3300-3399/3324.Find the Sequence of Strings Appeared on the Screen/README.md

@@ -73,32 +73,99 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3324.Fi
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：模拟
+
+我们可以模拟 Alice 按键的过程，从空字符串开始，每次按键后更新字符串，直到得到目标字符串。
+
+时间复杂度 $O(n^2 \times |\Sigma|)$，其中 $n$ 是目标字符串的长度，而 $\Sigma$ 是字符集，这里是小写字母集合，因此 $|\Sigma| = 26$。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def stringSequence(self, target: str) -> List[str]:
+        ans = []
+        for c in target:
+            s = ans[-1] if ans else ""
+            for a in ascii_lowercase:
+                t = s + a
+                ans.append(t)
+                if a == c:
+                    break
+        return ans
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public List<String> stringSequence(String target) {
+        List<String> ans = new ArrayList<>();
+        for (char c : target.toCharArray()) {
+            String s = ans.isEmpty() ? "" : ans.get(ans.size() - 1);
+            for (char a = 'a'; a <= c; ++a) {
+                String t = s + a;
+                ans.add(t);
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    vector<string> stringSequence(string target) {
+        vector<string> ans;
+        for (char c : target) {
+            string s = ans.empty() ? "" : ans.back();
+            for (char a = 'a'; a <= c; ++a) {
+                string t = s + a;
+                ans.push_back(t);
+            }
+        }
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func stringSequence(target string) (ans []string) {
+	for _, c := range target {
+		s := ""
+		if len(ans) > 0 {
+			s = ans[len(ans)-1]
+		}
+		for a := 'a'; a <= c; a++ {
+			t := s + string(a)
+			ans = append(ans, t)
+		}
+	}
+	return
+}
+```
 
+#### TypeScript
+
+```ts
+function stringSequence(target: string): string[] {
+    const ans: string[] = [];
+    for (const c of target) {
+        let s = ans.length > 0 ? ans[ans.length - 1] : '';
+        for (let a = 'a'.charCodeAt(0); a <= c.charCodeAt(0); a++) {
+            const t = s + String.fromCharCode(a);
+            ans.push(t);
+        }
+    }
+    return ans;
+}
 ```
 
 <!-- tabs:end -->

solution/3300-3399/3324.Find the Sequence of Strings Appeared on the Screen/README_EN.md

@@ -71,32 +71,99 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3324.Fi
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+We can simulate Alice's typing process, starting from an empty string and updating the string after each keystroke until the target string is obtained.
+
+The time complexity is $O(n^2 \times |\Sigma|)$, where $n$ is the length of the target string and $\Sigma$ is the character set, which in this case is the set of lowercase letters, so $|\Sigma| = 26$.
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def stringSequence(self, target: str) -> List[str]:
+        ans = []
+        for c in target:
+            s = ans[-1] if ans else ""
+            for a in ascii_lowercase:
+                t = s + a
+                ans.append(t)
+                if a == c:
+                    break
+        return ans
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public List<String> stringSequence(String target) {
+        List<String> ans = new ArrayList<>();
+        for (char c : target.toCharArray()) {
+            String s = ans.isEmpty() ? "" : ans.get(ans.size() - 1);
+            for (char a = 'a'; a <= c; ++a) {
+                String t = s + a;
+                ans.add(t);
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    vector<string> stringSequence(string target) {
+        vector<string> ans;
+        for (char c : target) {
+            string s = ans.empty() ? "" : ans.back();
+            for (char a = 'a'; a <= c; ++a) {
+                string t = s + a;
+                ans.push_back(t);
+            }
+        }
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func stringSequence(target string) (ans []string) {
+	for _, c := range target {
+		s := ""
+		if len(ans) > 0 {
+			s = ans[len(ans)-1]
+		}
+		for a := 'a'; a <= c; a++ {
+			t := s + string(a)
+			ans = append(ans, t)
+		}
+	}
+	return
+}
+```
 
+#### TypeScript
+
+```ts
+function stringSequence(target: string): string[] {
+    const ans: string[] = [];
+    for (const c of target) {
+        let s = ans.length > 0 ? ans[ans.length - 1] : '';
+        for (let a = 'a'.charCodeAt(0); a <= c.charCodeAt(0); a++) {
+            const t = s + String.fromCharCode(a);
+            ans.push(t);
+        }
+    }
+    return ans;
+}
 ```
 
 <!-- tabs:end -->

solution/3300-3399/3324.Find the Sequence of Strings Appeared on the Screen/Solution.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+    vector<string> stringSequence(string target) {
+        vector<string> ans;
+        for (char c : target) {
+            string s = ans.empty() ? "" : ans.back();
+            for (char a = 'a'; a <= c; ++a) {
+                string t = s + a;
+                ans.push_back(t);
+            }
+        }
+        return ans;
+    }
+};

solution/3300-3399/3324.Find the Sequence of Strings Appeared on the Screen/Solution.go

@@ -0,0 +1,13 @@
+func stringSequence(target string) (ans []string) {
+	for _, c := range target {
+		s := ""
+		if len(ans) > 0 {
+			s = ans[len(ans)-1]
+		}
+		for a := 'a'; a <= c; a++ {
+			t := s + string(a)
+			ans = append(ans, t)
+		}
+	}
+	return
+}

solution/3300-3399/3324.Find the Sequence of Strings Appeared on the Screen/Solution.java

@@ -0,0 +1,13 @@
+class Solution {
+    public List<String> stringSequence(String target) {
+        List<String> ans = new ArrayList<>();
+        for (char c : target.toCharArray()) {
+            String s = ans.isEmpty() ? "" : ans.get(ans.size() - 1);
+            for (char a = 'a'; a <= c; ++a) {
+                String t = s + a;
+                ans.add(t);
+            }
+        }
+        return ans;
+    }
+}

solution/3300-3399/3324.Find the Sequence of Strings Appeared on the Screen/Solution.py

@@ -0,0 +1,11 @@
+class Solution:
+    def stringSequence(self, target: str) -> List[str]:
+        ans = []
+        for c in target:
+            s = ans[-1] if ans else ""
+            for a in ascii_lowercase:
+                t = s + a
+                ans.append(t)
+                if a == c:
+                    break
+        return ans

solution/3300-3399/3324.Find the Sequence of Strings Appeared on the Screen/Solution.ts

@@ -0,0 +1,11 @@
+function stringSequence(target: string): string[] {
+    const ans: string[] = [];
+    for (const c of target) {
+        let s = ans.length > 0 ? ans[ans.length - 1] : '';
+        for (let a = 'a'.charCodeAt(0); a <= c.charCodeAt(0); a++) {
+            const t = s + String.fromCharCode(a);
+            ans.push(t);
+        }
+    }
+    return ans;
+}