Add solution to the 'Reverse in parenthesis' challenge

Add java and js solutions

Reverse_In_Parenthesis/Reverse_in_parenthesis.java

@@ -0,0 +1,24 @@
+import java.util.Arrays;
+import java.util.function.UnaryOperator;
+public class Reverse_in_parenthesis{
+	public static void main (String[] args){
+            String[] tests = {
+                "(bar)",
+                "foo(bar(baz))blim",
+                "(abc)d(efg)"
+            };
+            for(String test : tests){
+                System.out.format(" Input: '%s'%n Solution: '%s' %n%n", test, solution(test));
+            }
+	}
+
+	static String solution(String input){
+            UnaryOperator<String> mostNestedOperator1 = x -> {
+                int start = x.lastIndexOf('(');
+                return (start != -1) ?x.substring(start, x.indexOf(')', start) + 1) :null; // returns most nested with parenthesis
+            };
+            for(String mostNested; (mostNested = mostNestedOperator1.apply(input)) != null; )
+                input = input.replace(mostNested, new StringBuilder(mostNested.substring(1, mostNested.length() - 1)).reverse().toString());
+            return input;
+        }
+}

Reverse_In_Parenthesis/Reverse_in_parenthesis.js

@@ -0,0 +1,21 @@
+function solution(input) {
+  for(let mostNested; (mostNested=findMostNested(input))!=null;){
+    input = input.replace(mostNested, mostNested.substring(1, mostNested.length -1).split('').reverse().join(''))
+  } 
+  return input;
+}
+
+function findMostNested(string){
+  const start = string.lastIndexOf("(");
+  return start !== -1 ? string.substring(start, string.indexOf(")", start) + 1) :null;
+}
+
+const tests = [
+	"(bar)",
+        "foo(bar(baz))blim",
+        "(abc)d(efg)"
+];
+
+for (const test of tests) {
+  console.log(` Input: ${test} \n Inverted: ${solution(test)}\n\n`)
+}