Added solution for leet code 1190.

anxious-coder-lhs · anxious-coder-lhs · commit 05b9a5ddbc66 · 2020-10-12T00:26:47.000+05:30
diff --git a/README.md b/README.md
@@ -50,6 +50,7 @@ Happy to accept any contributions to this in any langugage.
 15. [Leet Code 362 Design Hit Counter](https://leetcode.com/problems/design-hit-counter/) | [Solution 1](./level_medium/LeetCode_362_Hit_Counter_1.ts) | [Solution 2](./level_medium/LeetCode_362_Hit_Counter_Medium.ts)
 16. [LeetCode Web Crawler 1242](https://leetcode.com/problems/web-crawler-multithreaded/) | [Solution 1 ](./level_medium/LeetCode_WebCrawler_MultiThreaded_1242.java) | [Solution 2](level_medium/LeetCode_WebCrawler_MultiThreaded_1242_1.java)
 17. [Leet Code 554 Brick Wall](https://leetcode.com/problems/brick-wall/) | [Solution 1](./level_medium/LeetCode_554_Brick_Wall_Medium.ts)
+18. [Leet Code 1190 Reverse Substrings Between Each Pair of Paranthesis](https://leetcode.com/problems/reverse-substrings-between-each-pair-of-parentheses/) | [Solution 1](level_medium/LeetCode_1190_Reverse_Substrings.ts)
 
 
 ## Hard
diff --git a/level_medium/LeetCode_1190_Reverse_Substrings.ts b/level_medium/LeetCode_1190_Reverse_Substrings.ts
@@ -0,0 +1,36 @@
+/**
+ * Problem: Leet Code 1190. Reverse Substrings between each pair of paranthesis.
+ * 
+ * Solution: Uses the stack approach of reversing the elements of the array. Also the
+ * same approach as used in the problems requiring backtracking where paranthesis is
+ * involved. The approach uses the stack as the final value which will not be reversed
+ * in the last. This allows for storing additional chars in the array without using
+ * additional auxilliary array.
+ * 
+ * The elements of the internal brackets are added after reversing them back into the
+ * stack for further processing.
+ * 
+ * Complexity: Time O(n), Space O(n)
+ * 
+ * @param s 
+ */
+function reverseParentheses(s: string): string {
+    const stack: string[] = []
+    for (let idx = 0; idx < s.length; idx++) {
+        const char: string = s[idx];
+        if (char === ')') {
+            // Start pulling from the stack and adding to result.
+            let elem = stack.pop();
+            const reversed: string[] = [];
+            while(elem !== '(') {
+                if (elem !== undefined) reversed.push(elem);
+                elem = stack.pop();
+            }
+            stack.push(...reversed);
+        } else {
+            stack.push(char);    
+        }
+    }
+    
+    return stack.join("");
+};
