Hello Programmers, In this post, you will know how to solve the Java Stack HackerRank Solution. This problem is a part of the HackerRank Java Programming Series.Java Stack HackerRank SolutionsOne more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.Java Stack HackerRank SolutionProblemIn computer science, a stack or LIFO (last in, first out) is an abstract data type that serves as a collection of elements, with two principal operations: push, which adds an element to the collection, and pop, which removes the last element that was added.(Wikipedia) A string containing only parentheses is balanced if the following is true: 1. if it is an empty string 2. if A and B are correct, AB is correct, 3. if A is correct, (A) and {A} and [A] are also correct.Examples of some correctly balanced strings are: “{}()”, “[{()}]”, “({()})”Examples of some unbalanced strings are: “{}(“, “({)}”, “[[“, “}{” etc.Given a string, determine if it is balanced or not.Input FormatThere will be multiple lines in the input file, each having a single non-empty string. You should read input till end-of-file.The part of the code that handles input operation is already provided in the editor.Output FormatFor each case, print ‘true’ if the string is balanced, ‘false’ otherwise.Sample Input{}() ({()}) {}( [] Sample Outputtrue true false trueJava Stack HackerRank Solutionsimport java.util.Scanner; import java.util.Stack; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); while (in.hasNext()) { System.out.println(isBalanced(in.next())); } in.close(); } static boolean isBalanced(String parentheses) { Stack<Character> stack = new Stack<Character>(); for (int i = 0; i < parentheses.length(); i++) { char ch = parentheses.charAt(i); if (ch == '(' || ch == '[' || ch == '{') { stack.push(ch); } else if (stack.empty()) { return false; } else { char top = stack.pop(); if ((top == '(' && ch != ')') || (top == '[' && ch != ']') || (top == '{' && ch != '}')) { return false; } } } return stack.empty(); } }Disclaimer: The above Problem (Java Stack) is generated by Hackerrank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: Java Generics HackerRank Solution Post navigationJava Map HackerRank Solution Java Generics HackerRank Solution