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feat: add solutions to lc problem: No.3481 (#4150)
No.3481.Apply Substitutions
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‎solution/3400-3499/3481.Apply Substitutions/README.md

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@@ -76,32 +76,153 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3481.Ap
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<!-- solution:start -->
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### 方法一
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### 方法一：哈希表 + 递归
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我们用一个哈希表 $\textit{d}$ 来存储替换的映射关系，然后定义一个函数 $\textit{dfs}$ 来递归地替换字符串中的占位符。
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函数 $\textit{dfs}$ 的执行逻辑如下：
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1. 在字符串 $\textit{s}$ 中查找第一个占位符的起始位置 $i$，如果找不到，则返回 $\textit{s}$；
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2. 在字符串 $\textit{s}$ 中查找第一个占位符的结束位置 $j$，如果找不到，则返回 $\textit{s}$；
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3. 截取占位符的键值 $key$，然后递归地替换占位符的值 $d[key]$；
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4. 返回替换后的字符串。
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在主函数中，我们调用 $\textit{dfs}$ 函数，传入文本字符串 $\textit{text}$，并返回结果。
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时间复杂度 $O(m + n \times L)$，空间复杂度 $O(m + n \times L)$。其中 $m$ 为替换映射的长度，而 $n$ 和 $L$ 分别为文本字符串的长度和占位符的平均长度。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def applySubstitutions(self, replacements: List[List[str]], text: str) -> str:
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def dfs(s: str) -> str:
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i = s.find("%")
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if i == -1:
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return s
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j = s.find("%", i + 1)
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if j == -1:
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return s
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key = s[i + 1 : j]
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replacement = dfs(d[key])
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return s[:i] + replacement + dfs(s[j + 1 :])
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d = {s: t for s, t in replacements}
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return dfs(text)
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```
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#### Java
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```java
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class Solution {
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private final Map<String, String> d = new HashMap<>();
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public String applySubstitutions(List<List<String>> replacements, String text) {
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for (List<String> e : replacements) {
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d.put(e.get(0), e.get(1));
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}
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return dfs(text);
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}
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private String dfs(String s) {
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int i = s.indexOf("%");
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if (i == -1) {
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return s;
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}
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int j = s.indexOf("%", i + 1);
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if (j == -1) {
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return s;
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}
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String key = s.substring(i + 1, j);
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String replacement = dfs(d.getOrDefault(key, ""));
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return s.substring(0, i) + replacement + dfs(s.substring(j + 1));
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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string applySubstitutions(vector<vector<string>>& replacements, string text) {
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unordered_map<string, string> d;
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for (const auto& e : replacements) {
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d[e[0]] = e[1];
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}
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auto dfs = [&](this auto&& dfs, const string& s) -> string {
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size_t i = s.find('%');
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if (i == string::npos) {
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return s;
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}
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size_t j = s.find('%', i + 1);
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if (j == string::npos) {
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return s;
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}
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string key = s.substr(i + 1, j - i - 1);
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string replacement = dfs(d[key]);
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return s.substr(0, i) + replacement + dfs(s.substr(j + 1));
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};
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return dfs(text);
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}
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};
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```
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#### Go
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```go
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func applySubstitutions(replacements [][]string, text string) string {
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d := make(map[string]string)
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for _, e := range replacements {
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d[e[0]] = e[1]
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}
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var dfs func(string) string
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dfs = func(s string) string {
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i := strings.Index(s, "%")
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if i == -1 {
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return s
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}
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j := strings.Index(s[i+1:], "%")
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if j == -1 {
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return s
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}
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j += i + 1
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key := s[i+1 : j]
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replacement := dfs(d[key])
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return s[:i] + replacement + dfs(s[j+1:])
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}
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return dfs(text)
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}
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```
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#### TypeScript
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```ts
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function applySubstitutions(replacements: string[][], text: string): string {
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const d: Record<string, string> = {};
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for (const [key, value] of replacements) {
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d[key] = value;
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}
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const dfs = (s: string): string => {
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const i = s.indexOf('%');
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if (i === -1) {
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return s;
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}
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const j = s.indexOf('%', i + 1);
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if (j === -1) {
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return s;
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}
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const key = s.slice(i + 1, j);
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const replacement = dfs(d[key] ?? '');
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return s.slice(0, i) + replacement + dfs(s.slice(j + 1));
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};
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return dfs(text);
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}
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```
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<!-- tabs:end -->

‎solution/3400-3499/3481.Apply Substitutions/README_EN.md

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Hash Table + Recursion
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We use a hash table $\textit{d}$ to store the substitution mapping, and then define a function $\textit{dfs}$ to recursively replace the placeholders in the string.
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The execution logic of the function $\textit{dfs}$ is as follows:
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1. Find the starting position $i$ of the first placeholder in the string $\textit{s}$. If not found, return $\textit{s}$;
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2. Find the ending position $j$ of the first placeholder in the string $\textit{s}$. If not found, return $\textit{s}$;
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3. Extract the key of the placeholder, and then recursively replace the value of the placeholder $d[key]$;
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4. Return the replaced string.
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In the main function, we call the $\textit{dfs}$ function, pass in the text string $\textit{text}$, and return the result.
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The time complexity is $O(m + n \times L)$, and the space complexity is $O(m + n \times L)$. Where $m$ is the length of the substitution mapping, and $n$ and $L$ are the length of the text string and the average length of the placeholders, respectively.
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def applySubstitutions(self, replacements: List[List[str]], text: str) -> str:
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def dfs(s: str) -> str:
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i = s.find("%")
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if i == -1:
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return s
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j = s.find("%", i + 1)
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if j == -1:
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return s
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key = s[i + 1 : j]
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replacement = dfs(d[key])
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return s[:i] + replacement + dfs(s[j + 1 :])
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d = {s: t for s, t in replacements}
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return dfs(text)
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```
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#### Java
90117

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```java
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class Solution {
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private final Map<String, String> d = new HashMap<>();
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public String applySubstitutions(List<List<String>> replacements, String text) {
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for (List<String> e : replacements) {
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d.put(e.get(0), e.get(1));
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}
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return dfs(text);
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}
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private String dfs(String s) {
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int i = s.indexOf("%");
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if (i == -1) {
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return s;
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}
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int j = s.indexOf("%", i + 1);
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if (j == -1) {
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return s;
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}
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String key = s.substring(i + 1, j);
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String replacement = dfs(d.getOrDefault(key, ""));
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return s.substring(0, i) + replacement + dfs(s.substring(j + 1));
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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string applySubstitutions(vector<vector<string>>& replacements, string text) {
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unordered_map<string, string> d;
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for (const auto& e : replacements) {
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d[e[0]] = e[1];
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}
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auto dfs = [&](this auto&& dfs, const string& s) -> string {
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size_t i = s.find('%');
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if (i == string::npos) {
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return s;
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}
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size_t j = s.find('%', i + 1);
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if (j == string::npos) {
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return s;
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}
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string key = s.substr(i + 1, j - i - 1);
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string replacement = dfs(d[key]);
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return s.substr(0, i) + replacement + dfs(s.substr(j + 1));
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};
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return dfs(text);
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}
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};
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```
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#### Go
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```go
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func applySubstitutions(replacements [][]string, text string) string {
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d := make(map[string]string)
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for _, e := range replacements {
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d[e[0]] = e[1]
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}
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var dfs func(string) string
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dfs = func(s string) string {
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i := strings.Index(s, "%")
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if i == -1 {
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return s
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}
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j := strings.Index(s[i+1:], "%")
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if j == -1 {
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return s
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}
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j += i + 1
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key := s[i+1 : j]
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replacement := dfs(d[key])
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return s[:i] + replacement + dfs(s[j+1:])
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}
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return dfs(text)
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}
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```
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#### TypeScript
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```ts
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function applySubstitutions(replacements: string[][], text: string): string {
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const d: Record<string, string> = {};
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for (const [key, value] of replacements) {
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d[key] = value;
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}
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const dfs = (s: string): string => {
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const i = s.indexOf('%');
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if (i === -1) {
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return s;
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}
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const j = s.indexOf('%', i + 1);
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if (j === -1) {
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return s;
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}
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const key = s.slice(i + 1, j);
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const replacement = dfs(d[key] ?? '');
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return s.slice(0, i) + replacement + dfs(s.slice(j + 1));
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};
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return dfs(text);
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}
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```
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<!-- tabs:end -->

‎solution/3400-3499/3481.Apply Substitutions/Solution.cpp

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class Solution {
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public:
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string applySubstitutions(vector<vector<string>>& replacements, string text) {
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unordered_map<string, string> d;
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for (const auto& e : replacements) {
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d[e[0]] = e[1];
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}
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auto dfs = [&](this auto&& dfs, const string& s) -> string {
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size_t i = s.find('%');
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if (i == string::npos) {
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return s;
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}
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size_t j = s.find('%', i + 1);
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if (j == string::npos) {
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return s;
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}
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string key = s.substr(i + 1, j - i - 1);
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string replacement = dfs(d[key]);
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return s.substr(0, i) + replacement + dfs(s.substr(j + 1));
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};
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return dfs(text);
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}
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};

‎solution/3400-3499/3481.Apply Substitutions/Solution.go

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func applySubstitutions(replacements [][]string, text string) string {
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d := make(map[string]string)
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for _, e := range replacements {
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d[e[0]] = e[1]
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}
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var dfs func(string) string
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dfs = func(s string) string {
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i := strings.Index(s, "%")
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if i == -1 {
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return s
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}
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j := strings.Index(s[i+1:], "%")
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if j == -1 {
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return s
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}
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j += i + 1
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key := s[i+1 : j]
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replacement := dfs(d[key])
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return s[:i] + replacement + dfs(s[j+1:])
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}
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return dfs(text)
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}

‎solution/3400-3499/3481.Apply Substitutions/Solution.java

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class Solution {
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private final Map<String, String> d = new HashMap<>();
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public String applySubstitutions(List<List<String>> replacements, String text) {
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for (List<String> e : replacements) {
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d.put(e.get(0), e.get(1));
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}
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return dfs(text);
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}
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private String dfs(String s) {
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int i = s.indexOf("%");
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if (i == -1) {
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return s;
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}
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int j = s.indexOf("%", i + 1);
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if (j == -1) {
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return s;
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}
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String key = s.substring(i + 1, j);
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String replacement = dfs(d.getOrDefault(key, ""));
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return s.substring(0, i) + replacement + dfs(s.substring(j + 1));
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}
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}

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