public class Solution
extends Object
1807 - Evaluate the Bracket Pairs of a String\.
Medium
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.
**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.