New Problem Solution -"Evaluate the Bracket Pairs of a String"

Author: haoel

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1807|[Evaluate the Bracket Pairs of a String](https://leetcode.com/problems/evaluate-the-bracket-pairs-of-a-string/) | [C++](./algorithms/cpp/evaluateTheBracketPairsOfAString/EvaluateTheBracketPairsOfAString.cpp)|Medium|
 |1806|[Minimum Number of Operations to Reinitialize a Permutation](https://leetcode.com/problems/minimum-number-of-operations-to-reinitialize-a-permutation/) | [C++](./algorithms/cpp/minimumNumberOfOperationsToReinitializeAPermutation/MinimumNumberOfOperationsToReinitializeAPermutation.cpp)|Medium|
 |1805|[Number of Different Integers in a String](https://leetcode.com/problems/number-of-different-integers-in-a-string/) | [C++](./algorithms/cpp/numberOfDifferentIntegersInAString/NumberOfDifferentIntegersInAString.cpp)|Easy|
 |1803|[Count Pairs With XOR in a Range](https://leetcode.com/problems/count-pairs-with-xor-in-a-range/) | [C++](./algorithms/cpp/countPairsWithXorInARange/CountPairsWithXorInARange.cpp)|Hard|

algorithms/cpp/evaluateTheBracketPairsOfAString/EvaluateTheBracketPairsOfAString.cpp

@@ -0,0 +1,107 @@
+// Source : https://leetcode.com/problems/evaluate-the-bracket-pairs-of-a-string/
+// Author : Hao Chen
+// Date   : 2021-03-28
+
+/***************************************************************************************************** 
+ *
+ * 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 <= 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.
+ ******************************************************************************************************/
+
+class Solution {
+private:
+    bool isBracket(char c) {
+        return c=='(' || c == ')';
+    }
+public:
+    string evaluate(string s, vector<vector<string>>& knowledge) {
+        unordered_map<string, string> dict;
+        for(auto& k : knowledge) {
+            dict[k[0]] = k[1];
+        }
+        
+        string result;
+        string key;
+        bool meetLeftBracket = false;
+        for(auto& c : s) {
+           
+            if (c == '(') {
+                meetLeftBracket = true;
+            } else if (c == ')') {
+                meetLeftBracket = false;
+                //cout << key << endl;
+                if (dict.find(key) != dict.end()) {
+                    result += dict[key];
+                }else {
+                    result += '?';
+                }
+                key = "";
+            } else {
+                if (meetLeftBracket) {
+                    key += c;
+                }else{
+                    result +=c;
+                }
+            }
+        }
+        
+        return result;
+    }
+};