see image for instructionsstarter code for Main.java import java.util.*;public class Main { public static boolean checkParen(String s) { //implement this method to return true if grouping symbols match //each other, otherwise return false. /* Type your code here. */ }} 11.14 NVCC Lab: Parentheses MatchA fairly common algorithmic task is to process some data set in reverse order. Typically you put some data in temporary storage, then takeit out, always in a Last-In, First-Out (LIFO) order. A stack is the data structure that was invented to help manage this process. A real-worldexample of a stack is the dispenser of the trays in the lunch room. You always take the top tray from the top, never from the middle or thebottom. Similarly, the lunch workers always put trays on the top, never at the middle or bottom. For this problem, you are going to determinewhether the grouping symbols--parentheses, brackets, curly braces, etc.--in an arithmetic expression, such as [(5+7)*3], match each other.Here are all matching group symbols: ().I.), and <>. For this problem, you can ignore all digits and operands. Study the following cases toexamine how matching symbols are evaluated.5+7(5+7))5+7(((5+7)*3)<{5+7}*3>[(5+7)*]3(5+7)*35+(7*3)((5+7)*3[(5+7]*3)[(5+7)*3])([(5+7)*3]return false.}ParenthesesPlease implement the following method:public static boolean checkParen (String s) {Matchtruetruefalsetruetruetruetruetruefalsefalsefalsefalse<50 +(37- (3*2}} + [ 19 + {8*9}]> true//implement this method to return true if grouping symbols match each other, otherwise

This particular question belongs to Computer Science. Computer Science is an umbrella term for subjects such as Programming Languages, Data Structures, Algorithms, Object Oriented Programming, Databases, Computer Networks, Software Engineering, C Language, C++ Language, Java Language etc. The solution for this question consists of writing Java Code to implement checkParen method.// Implementation of checkParen method. import java.util.*; public class Main { public static boolean checkParen(String s) { //implement this method to return true if grouping symbols match //each other, otherwise return false. // create the stack to store symbols. Stack<Character> st = new Stack<>(); // Traverse the input string s. for (int i=0; i < s.length(); i++) { char ch = s.charAt(i); // skip all digits and operators if (ch != '(' || ch != ')' || ch != '[' || ch != ']' || ch != '{' || ch != '}' || ch != '<' || ch != '>') continue; // if character is a starting symbol, then push it onto stack. if (ch == '(' || ch == '[' || ch == '{' || ch == '<') { st.push(ch); continue; } // character is a closing symbol. stack should not be empty. // if stack is empty then return false. if (st.isEmpty()) return false; // pop character from the stack. if the closing symbol matches // the popped character, then it is okay. otherwise not a match. char popch; switch (ch) { case ')': popch = st.pop(); if (popch == '{' || popch =='[' || popch == '<') return false; break; case ']': popch = st.pop(); if (popch == '(' || popch == '{' || popch == '<') return false; break; case '}': popch = st.pop(); if (popch == '(' || popch == '[' || popch == '<') return false; break; case '>': popch = st.pop(); if (popch == '(' || popch == '[' || popch == '{') return false; break; } // switch } // for loop // At the end of traversal of input string s, stack should be // empty. if stack is empty then it is a perfect match. otherwise // it is not a perfect match. if (st.isEmpty()) return true; // perfect match. } // end of checkParen method} // end of Main. Conclusion: Assumption: Here the digits, operators and operands are ignored.