added solution for 1807

dsa-solutions/lc-solutions/1800-1899/1807-evaluate-the-bracket-pairs-of-a-string.md

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+---
+id: evaluate-the-bracket-pairs-of-a-string
+title: Evaluate  the Bracket Pairs of a String
+sidebar_label: 1807-Evaluate the Bracket Pairs of a String
+tags:
+  - String Manipulation
+  - Hash Table
+  - LeetCode
+  - Java
+  - Python
+  - C++
+description: "This is a solution to the Evaluate Bracket Pairs problem on LeetCode."
+sidebar_position: 30
+---
+
+## Problem Description
+
+You are given a string `s` that contains some bracket pairs, with each pair containing a non-empty key.
+
+For example, in the string `"(name)is(age)yearsold"`, there are two bracket pairs that contain the keys "name" and "age". You know the values of a wide range of keys. This is represented by a 2D string array `knowledge` where each `knowledge[i] = [keyi, valuei]` indicates that key `keyi` has a value of `valuei`.
+
+You are tasked to evaluate all of the bracket pairs. When you evaluate a bracket pair that contains some key `keyi`, you will:
+- Replace `keyi` and the bracket pair with the key's corresponding `valuei`.
+- If you do not know the value of the key, you will replace `keyi` and the bracket pair with a question mark "?" (without the quotation marks).
+
+Each key will appear at most once in your knowledge. There will not be any nested brackets in `s`.
+
+Return the resulting string after evaluating all of the bracket pairs.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: s = "(name)is(age)yearsold", knowledge = [["name","bob"],["age","two"]]
+Output: "bobistwoyearsold"
+Explanation:
+The key "name" has a value of "bob", so replace "(name)" with "bob".
+The key "age" has a value of "two", so replace "(age)" with "two".
+```
+
+**Example 2:**
+
+```
+Input: s = "hi(name)", knowledge = [["a","b"]]
+Output: "hi?"
+Explanation: As you do not know the value of the key "name", replace "(name)" with "?".
+```
+
+**Example 3:**
+
+```
+Input: s = "(a)(a)(a)aaa", knowledge = [["a","yes"]]
+Output: "yesyesyesaaa"
+Explanation: The same key can appear multiple times.
+The key "a" has a value of "yes", so replace all occurrences of "(a)" with "yes".
+Notice that the "a"s not in a bracket pair are not evaluated.
+```
+
+### Constraints
+
+- `1 <= s.length <= 10^5`
+- `0 <= knowledge.length <= 10^5`
+- `knowledge[i].length == 2`
+- `1 <= keyi.length, valuei.length <= 10`
+- `s` consists of lowercase English letters and round brackets '(' and ')'.
+- Every open bracket '(' in `s` will have a corresponding close bracket ')'.
+- The key in each bracket pair of `s` will be non-empty.
+- There will not be any nested bracket pairs in `s`.
+- `keyi` and `valuei` consist of lowercase English letters.
+- Each `keyi` in knowledge is unique.
+
+---
+
+## Solution for Evaluate Bracket Pairs Problem
+
+To solve this problem, we need to use a hash map to store the key-value pairs from the knowledge array. Then, we can iterate through the string `s` and build the resulting string by replacing the keys within brackets with their corresponding values from the hash map.
+
+### Approach
+
+1. Create a hash map from the knowledge array.
+2. Initialize a result string.
+3. Traverse the string `s`:
+   - If we encounter a '(', start collecting the key until we find a ')'.
+   - Replace the key with its value from the hash map, or '?' if the key is not found.
+4. Continue until the entire string is processed.
+
+### Code in Different Languages
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```cpp
+class Solution {
+public:
+    string evaluate(string s, vector<vector<string>>& knowledge) {
+        unordered_map<string, string> map;
+        for (auto& pair : knowledge) {
+            map[pair[0]] = pair[1];
+        }
+
+        string result;
+        int i = 0;
+        while (i < s.size()) {
+            if (s[i] == '(') {
+                int j = i + 1;
+                while (s[j] != ')') j++;
+                string key = s.substr(i + 1, j - i - 1);
+                result += map.count(key) ? map[key] : "?";
+                i = j + 1;
+            } else {
+                result += s[i++];
+            }
+        }
+        return result;
+    }
+};
+```
+
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```java
+class Solution {
+    public String evaluate(String s, List<List<String>> knowledge) {
+        Map<String, String> map = new HashMap<>();
+        for (List<String> pair : knowledge) {
+            map.put(pair.get(0), pair.get(1));
+        }
+
+        StringBuilder result = new StringBuilder();
+        int i = 0;
+        while (i < s.length()) {
+            if (s.charAt(i) == '(') {
+                int j = i + 1;
+                while (s.charAt(j) != ')') j++;
+                String key = s.substring(i + 1, j);
+                result.append(map.getOrDefault(key, "?"));
+                i = j + 1;
+            } else {
+                result.append(s.charAt(i++));
+            }
+        }
+        return result.toString();
+    }
+}
+```
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```python
+class Solution:
+    def evaluate(self, s: str, knowledge: List[List[str]]) -> str:
+        knowledge_dict = {key: value for key, value in knowledge}
+        result = []
+        i = 0
+        while i < len(s):
+            if s[i] == '(':
+                j = i + 1
+                while s[j] != ')':
+                    j += 1
+                key = s[i + 1:j]
+                result.append(knowledge_dict.get(key, "?"))
+                i = j + 1
+            else:
+                result.append(s[i])
+                i += 1
+        return ''.join(result)
+```
+
+</TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- **Time Complexity**: $O(n + m)$, where `n` is the length of the string `s` and `m` is the number of key-value pairs in `knowledge`.
+- **Space Complexity**: $O(m)$, for storing the hash map of knowledge.
+
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['ImmidiSivani'].map(username => (
+ <Author key={username} username={username} />
+))}
+</div>
+```