Java balanced expressions check {[()]}

Question

I am trying to create a program that takes a string as an argument into its constructor. I need a method that checks whether the string is a balanced parenthesized expressio

Answer 1

public class StackProb {

public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);
    List<Boolean> list = new ArrayList<>();

    while (sc.hasNextLine()) {
        String s=sc.nextLine();
        if(!s.isEmpty()) {
            list.add(isBalanced(s));
            //System.out.println(isBalanced(s));
        }else {
            sc.close();
            break;
        }
    }

    for (int i = 0; i < list.size(); i++) {
        System.out.println(list.get(i) + " ");
    }

}

private static boolean isBalanced(String s) {
    boolean res = false;
    Stack<Character> stack = new Stack();
    int countA = 0;
    int countB = 0;
    for (int i = 0; i < s.length(); i++) {

        if(s.charAt(i)=='{' || s.charAt(i)=='(' || s.charAt(i)=='[') {

            stack.push(s.charAt(i));
            countA++;
        }


        if(s.charAt(i)=='}' || s.charAt(i)==')' || s.charAt(i)==']') {

            stack.push(s.charAt(i));
            countB++;
        }

        if(stack.firstElement()=='}' || stack.firstElement()==')' || stack.firstElement()==']') {
            countB++;
        }


    }
    if(countA==countB) {
        return true;
    }
    return false;

}

}

Answer 2

This can be used. Passes all the tests.

static String isBalanced(String s) {

    if(null == s){
        return "";
    }

    Stack<Character> bracketStack = new Stack<>();


    int length = s.length();

    if(length < 2 || length > 1000){
        return "NO";
    }


    for(int i = 0; i < length; i++){
        Character c= s.charAt(i);
        if(c == '(' || c == '{' || c == '[' ){
            bracketStack.push(c);
        } else {
            if(!bracketStack.isEmpty()){
               char cPop = bracketStack.pop();

               if(c == ']' && cPop!= '['){
                  return "NO";
               }

               if(c == ')' && cPop!= '('){
                  return "NO";
               }

               if(c == '}' && cPop!= '{'){
                  return "NO";
               }
            } else{
                return "NO";
            }

        }
    }

    if(bracketStack.isEmpty()){
        return "YES";
    } else {
        return "NO";
    }

}

Answer 3

I hope this code can help:

import java.util.Stack;

public class BalancedParenthensies {

    public static void main(String args[]) {

        System.out.println(balancedParenthensies("{(a,b)}"));
        System.out.println(balancedParenthensies("{(a},b)"));
        System.out.println(balancedParenthensies("{)(a,b}"));
    }

    public static boolean balancedParenthensies(String s) {
        Stack<Character> stack  = new Stack<Character>();
        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if(c == '[' || c == '(' || c == '{' ) {     
                stack.push(c);
            } else if(c == ']') {
                if(stack.isEmpty() || stack.pop() != '[') {
                    return false;
                }
            } else if(c == ')') {
                if(stack.isEmpty() || stack.pop() != '(') {
                    return false;
                }           
            } else if(c == '}') {
                if(stack.isEmpty() || stack.pop() != '{') {
                    return false;
                }
            }

        }
        return stack.isEmpty();
    }
}

Answer 4

public static void main(String[] args) {
    System.out.println("is balanced : "+isBalanced("(){}[]<>"));
    System.out.println("is balanced : "+isBalanced("({})[]<>"));
    System.out.println("is balanced : "+isBalanced("({[]})<>"));
    System.out.println("is balanced : "+isBalanced("({[<>]})"));
    System.out.println("is balanced : "+isBalanced("({})[<>]"));


    System.out.println("is balanced : "+isBalanced("({[}])[<>]"));
    System.out.println("is balanced : "+isBalanced("([{})]"));
    System.out.println("is balanced : "+isBalanced("[({}])"));
    System.out.println("is balanced : "+isBalanced("[(<{>})]"));

    System.out.println("is balanced : "+isBalanced("["));
    System.out.println("is balanced : "+isBalanced("]"));

    System.out.println("is balanced : "+isBalanced("asdlsa"));
}

private static boolean isBalanced(String brackets){
    char[] bracketsArray = brackets.toCharArray();
    Stack<Character> stack = new Stack<Character>();
    Map<Character, Character> openingClosingMap = initOpeningClosingMap();

    for (char bracket : bracketsArray) {
        if(openingClosingMap.keySet().contains(bracket)){ 
            stack.push(bracket);
        }else if(openingClosingMap.values().contains(bracket)){
            if(stack.isEmpty() || openingClosingMap.get(stack.pop())!=bracket){
                return false;
            }
        }else{
            System.out.println("Only  < > ( ) { } [ ] brackets  are allowed .");
            return false;
        }
    }
    return stack.isEmpty();
}

private static Map<Character, Character> initOpeningClosingMap() {
    Map<Character, Character> openingClosingMap = new HashMap<Character, Character>();
    openingClosingMap.put(Character.valueOf('('), Character.valueOf(')'));
    openingClosingMap.put(Character.valueOf('{'), Character.valueOf('}'));
    openingClosingMap.put(Character.valueOf('['), Character.valueOf(']'));
    openingClosingMap.put(Character.valueOf('<'), Character.valueOf('>'));
    return openingClosingMap;
}

Simplifying and making readable. Using One Map only and minimum conditions to get desired result.

Answer 5

Code snippet for implementing matching parenthesis using java.util.Stack data structure -

    //map for storing matching parenthesis pairs
    private static final Map<Character, Character> matchingParenMap = new HashMap<>();

    //set for storing opening parenthesis
    private static final Set<Character> openingParenSet = new HashSet<>();

    static {
         matchingParenMap.put(')','(');
         matchingParenMap.put(']','['); 
         matchingParenMap.put('}','{'); 
         openingParenSet.addAll(matchingParenMap.values());  
    }

    //check if parenthesis match
    public static boolean hasMatchingParen(String input) {
      try {
         //stack to store opening parenthesis
         Stack<Character> parenStack = new Stack<>();

         for(int i=0; i< input.length(); i++) {
            char ch = input.charAt(i);

            //if an opening parenthesis then push to the stack
            if(openingParenSet.contains(ch)) {
                 parenStack.push(ch);
            } 

            //for closing parenthesis
            if(matchingParenMap.containsKey(ch)) {
                 Character lastParen = parenStack.pop();
                 if(lastParen != matchingParenMap.get(ch)) {
                    return false;
                 } 
            }
         }

         //returns true if the stack is empty else false
         return parenStack.isEmpty();
       }
         catch(StackOverflowException s) {}
         catch(StackUnderflowException s1) {}
         return false;
    }

I have explained the code snippet and the algorithm used on blog http://hetalrachh.home.blog/2019/12/25/stack-data-structure/

Answer 6

It's important to use a stack to push opening symbols onto it, then when you come across a closing brace you pop the element off the top of the stack and then you check it to see if it matches the type of closing brace. Here is a java implementation.

import java.util.Stack;

public class Balanced {
    public static void main (String [] args)
    {
        String test_good = "()(){}{}{()}";
        String test_bad = "((({}{}))()";

        System.out.println(checkBalanced(test_good));
        System.out.println(checkBalanced(test_bad));
    }

    public static boolean checkBalanced(String check)
    {
        Stack<Character> S = new Stack<Character>();
        for(int a = 0; a < check.length(); a++)
        {
            char let = check.charAt(a);
            if(let == '[' || let == '{' || let == '(')
                S.push(let);
            else if(let == ']' || let == '}' || let == ')')
            {
                if(S.empty())
                    return false;
                switch(let)
                {
                    // Opening square brace
                    case ']':
                        if (S.pop() != '[')
                            return false;
                        break;
                    // Opening curly brace
                    case '}':
                        if (S.pop() != '{')
                            return false;
                        break;
                    // Opening paren brace
                    case ')':
                        if (S.pop() != '(')
                            return false;
                        break;
                    default:
                        break;
                }
            }
        }
        if(S.empty())
            return true;
        return false;
    }
}