# GeetCode Hub

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.

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.
```

Example 4:

```Input: s = "(a)(b)", knowledge = [["a","b"],["b","a"]]
Output: "ba"```

Constraints:

• `1 <= s.length <= 105`
• `0 <= knowledge.length <= 105`
• `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.

class Solution { public String evaluate(String s, List<List<String>> knowledge) { } }