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  1. Prepare
  2. Data Structures
  3. Stacks
  4. Balanced Brackets
  5. Discussions

Balanced Brackets

  • sowmi546
     + 4 comments

    My Java Solution 

    import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static boolean isBalanced(String s) {
   int len=s.length();
    if(len==0 || s==null) return true;
      Stack<Character> stack = new Stack<Character>();
    for(int i=0;i<s.length();i++)
    {
        if(s.charAt(i)=='(' || s.charAt(i)=='[' || s.charAt(i)=='{')  stack.push(s.charAt(i));
        else if(s.charAt(i)==')' && !stack.empty() && stack.peek()=='(') stack.pop();
        else if(s.charAt(i)==']' && !stack.empty() && stack.peek()=='[') stack.pop();

        else if(s.charAt(i)=='}' && !stack.empty() && stack.peek()=='{') stack.pop();
        else return false;


    }
    return stack.empty();
    }

    public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    for (int a0 = 0; a0 < t; a0++) {
        String expression = in.next();
        System.out.println( (isBalanced(expression)) ? "YES" : "NO" );
    }
}
}

    • saif01md
       + 1 comment
      else if(str.charAt(i)==']' && s.peek()=='[' && !s.empty())

      Please reply .. If we write the statement on either side of && as the order shown above ( peek before and empty after) , it is showing runtime error . Is the order of statement also important in writing these expressions ?

      • claytonsibanda60
         + 0 comments

        Yes it is. If !s.empty() is placed first and the stack is empty, s.peek()=='[' will not be executed, hence no runtime error.This is due to the short circuit evaluation nature of if statements. However the order you gave above gives runtime error because s.peek()=='[' is executed first and this may cause errors if the stack is empty.

    • saif01md
       + 2 comments

      Another questionisthat why !s.empty() is even required in the ?

      • TusharSh
         + 0 comments

        I think You can't pop is stack is empty

      • aleksei_maide
         + 0 comments

        If stack is not empty... then there is an extra unbalanced parenthesis on the stack...

    • b_vsbrk
       + 1 comment

      My code looks exactly like your code but still I'm not able to pass 3 test cases...

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

/**
 * Created by BK on 06-08-2017.
 */

public class BalancedParanthesis {
    public static void main(String... strings) {
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();
        for (int i = 0; i < tc; i++) {
            printAnswer(sc.next());
        }
    }

    private static void printAnswer(String input) {
        Stack<Character> stack = new Stack<>();
        boolean isValid=true;
        for (int i = 0; i < input.length(); i++) {
            char currentChar = input.charAt(i);
            if (currentChar == '{' || currentChar == '(' || currentChar == '[') stack.push(currentChar);
            else if (currentChar == '}') {
                if (stack.isEmpty())isValid=false;
                else {
                    if (stack.pop() != '{') {
                        isValid=false;
                    }
                }
            } else if (currentChar == ')') {
                if (stack.isEmpty())isValid=false;
                else {
                    if (stack.pop() != '(') {
                        isValid=false;
                    }
                }
            } else if (currentChar == ']') {
                if (stack.isEmpty())isValid=false;
                else {
                    if (stack.pop() != '[') {
                        isValid=false;
                    }
                }
            }
        }
        System.out.println(isValid?"YES":"NO");
    }
}

      • aishik_swarez
         + 1 comment

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

        public class Solution {

        static String isBalanced(String s) {
    Stack stack = new Stack();
    stack.push(s.charAt(0));
    char peek;
    char charAt;

    for (int i = 1; i < s.length(); i++) {
        try {
            peek = (char) stack.peek();
            charAt = s.charAt(i);
            if(!stack.empty()) {
            if ((charAt == '}' && peek == '{') || (charAt == ')' && peek == '(')
                    || (charAt == ']' && peek == '[')) {
                stack.pop();
            } else
                stack.push(charAt);
            }
            else
                stack.push(i);
        } catch (Exception e) {
            //i++;
            charAt = s.charAt(i);
            stack.push(charAt);
        }
    }
    if (stack.empty())
        return "YES";
    else
        return "NO";

}

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    in.nextLine();
    String result[] = new String[t];
    for (int a0 = 0; a0 < t; a0++) {
        String s = in.next();
        result[a0] = isBalanced(s);
    }
    in.close();
    for (int i = 0; i < t; i++)
        System.out.println(result[i]);
}

        }

        • amitwcsprof
           + 0 comments

          one query: If stack is empty then peek should return null. Is there any need for this check "!stack.empty() "

    • rangareddy_25
       + 0 comments

      if you put one more check i.e, **if (!(s.length()%2 ==0) ) ** before for loop it will improve it's performance and many cases straight forward it will be passed

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