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

    import java.io.IOException;
    import java.util.ArrayList;
    import java.util.List;
    import java.util.Scanner;
    import java.util.Stack;
    public class BalancedParenthesisWithStack {

    /*This is purely Java Stack based solutions without using additonal 
      data structure like array/Map */

    public static void main(String[] args) throws IOException {

        Scanner sc = new Scanner(System.in);

        /*Take list of String inputs (parenthesis expressions both valid and 
         invalid from console*/

        List<String> inputs=new ArrayList<>();
        while (sc.hasNext()) {

            String input=sc.next();
            inputs.add(input);

        }

        //For every input in above list display whether it is valid or 
         //invalid parenthesis expression

        for(String input:inputs){



        System.out.println("\nisBalancedParenthesis:"+isBalancedParenthesis
        (input));
        }
    }

    //This method identifies whether expression is valid parenthesis or not

    public static boolean isBalancedParenthesis(String expression){

        //sequence of opening parenthesis according to its precedence
         //i.e. '[' has higher precedence than '{' or '('
        String openingParenthesis="[{(";

        //sequence of closing parenthesis according to its precedence
        String closingParenthesis=")}]";

        //Stack will be pushed on opening parenthesis and popped on closing.
        Stack<Character> parenthesisStack=new Stack<>();


          /*For expression to be valid :
          CHECK :
          1. it must start with opening parenthesis [()...
          2. precedence of parenthesis  should be proper (eg. "{[" invalid  
                                                              "[{(" valid  ) 


          3. matching pair if(  '(' => ')')  i.e. [{()}(())] ->valid [{)]not 
          */
         if(closingParenthesis.contains
         (((Character)expression.charAt(0)).toString())){
            return false;
        }else{
        for(int i=0;i<expression.length();i++){

        char ch= (Character)expression.charAt(i);

        //if parenthesis is opening(ie any of '[','{','(') push on stack
        if(openingParenthesis.contains(ch.toString())){
                parenthesisStack.push(ch);
            }else if(closingParenthesis.contains(ch.toString())){
        //if parenthesis is closing (ie any of ']','}',')') pop stack
        //depending upon check-3 
                if(parenthesisStack.peek()=='(' && (ch==')') || 
                    parenthesisStack.peek()=='{' && (ch=='}') ||    
                    parenthesisStack.peek()=='[' && (ch==']')
                        ){
                parenthesisStack.pop();
                }
            }
        }

        return (parenthesisStack.isEmpty())? true : false;
        }
    }

Answer 2

The improved method, from @Smartoop.

public boolean balancedParenthensies(String str) {
    List<Character> leftKeys = Arrays.asList('{', '(', '<', '[');
    List<Character> rightKeys = Arrays.asList('}', ')', '>', ']');

    Stack<Character> stack = new Stack<>();
    for (int i = 0; i < str.length(); i++) {
        char c = str.charAt(i);
        if (leftKeys.contains(c)) {
            stack.push(c);
        } else if (rightKeys.contains(c)) {
            int index = rightKeys.indexOf(c);
            if (stack.isEmpty() || stack.pop() != leftKeys.get(index)) {
                return false;
            }
        }
    }
    return stack.isEmpty();
}

Answer 3

This is my own implementation. I tried to make it the shortest an clearest way possible:

public static boolean isBraceBalanced(String braces) {
    Stack<Character> stack = new Stack<Character>();

    for(char c : braces.toCharArray()) {
        if(c == '(' || c == '[' || c == '{') {
            stack.push(c);
        } else if((c == ')' && (stack.isEmpty() || stack.pop() != '(')) ||
                  (c == ']' && (stack.isEmpty() || stack.pop() != '[')) ||
                  (c == '}' && (stack.isEmpty() || stack.pop() != '{'))) {
            return false;
        }
    }

    return stack.isEmpty();
}

Answer 4

This is my implementation for this question. This program allows numbers, alphabets and special characters with input string but simply ignore them while processing the string.

CODE:

import java.util.Scanner;
import java.util.Stack;

public class StringCheck {

    public static void main(String[] args) {
        boolean flag =false;
        Stack<Character> input = new Stack<Character>();
        System.out.println("Enter your String to check:");
        Scanner scanner = new Scanner(System.in);
        String sinput = scanner.nextLine();
        char[] c = new char[15];
        c = sinput.toCharArray();
        for (int i = 0; i < c.length; i++) {
            if (c[i] == '{' || c[i] == '(' || c[i] == '[')
                input.push(c[i]);
            else if (c[i] == ']') {
                if (input.pop() == '[') {
                    flag = true;
                    continue;
                } else {
                    flag = false;
                    break;
                }
            } else if (c[i] == ')') {
                if (input.pop() == '(') {
                    flag = true;
                    continue;
                } else {
                    flag = false;
                    break;
                }
            } else if (c[i] == '}') {
                if (input.pop() == '{') {
                    flag = true;
                    continue;
                } else {
                    flag = false;
                    break;
                }
            }
        }
        if (flag == true)
            System.out.println("Valid String");
        else
            System.out.println("Invalid String");
        scanner.close();

    }

}

Answer 5

public static boolean isBalanced(String expression) {
  if ((expression.length() % 2) == 1) return false;
  else {
    Stack<Character> s = new Stack<>();
    for (char bracket : expression.toCharArray())
      switch (bracket) {
        case '{': s.push('}'); break;
        case '(': s.push(')'); break;
        case '[': s.push(']'); break;
        default :
          if (s.isEmpty() || bracket != s.peek()) { return false;}
          s.pop();
      }
    return s.isEmpty();
  }
}

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    String expression = in.nextLine();
    boolean answer = isBalanced(expression);
    if (answer) { System.out.println("YES");}
    else { System.out.println("NO");}

}

Answer 6

Here is the Code. I have tested all the possible test case on Hacker Rank.

static String isBalanced(String input) {

    Stack<Character> stack = new Stack<Character>();
    for (int i = 0; i < input.length(); i++) {
        Character ch = input.charAt(i);
        if (input.charAt(i) == '{' || input.charAt(i) == '['
                || input.charAt(i) == '(') {
            stack.push(input.charAt(i));
        } else {
            if (stack.isEmpty() 
                    || (stack.peek() == '[' && ch != ']')
                    || (stack.peek() == '{' && ch != '}')
                    || (stack.peek() == '(' && ch != ')')) {
                return "NO";
            } else {
                stack.pop();
            }
        }
    }
    if (stack.empty())
        return "YES";
    return "NO";

}