Java Program To Check For Balanced Brackets In An Expression (Well-Formedness) Using Stack - GeeksforGeeks

14 Dec, 2021
Given an expression string exp, write a program to examine whether the pairs and the orders of “{“, “}”, “(“, “)”, “[“, “]” are correct in exp.
Example: 
Input: exp = “[()]{}{[()()]()}” Output: Balanced
Input: exp = “[(])” Output: Not Balanced 
Algorithm: 
Declare a character stack S.
Now traverse the expression string exp. If the current character is a starting bracket (‘(‘ or ‘{‘ or ‘[‘) then push it to stack.If the current character is a closing bracket (‘)’ or ‘}’ or ‘]’) then pop from stack and if the popped character is the matching starting bracket then fine else brackets are not balanced.
If the current character is a starting bracket (‘(‘ or ‘{‘ or ‘[‘) then push it to stack.If the current character is a closing bracket (‘)’ or ‘}’ or ‘]’) then pop from stack and if the popped character is the matching starting bracket then fine else brackets are not balanced.
If the current character is a starting bracket (‘(‘ or ‘{‘ or ‘[‘) then push it to stack.
If the current character is a closing bracket (‘)’ or ‘}’ or ‘]’) then pop from stack and if the popped character is the matching starting bracket then fine else brackets are not balanced.
After complete traversal, if there is some starting bracket left in stack then “not balanced”
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
Java
// Java program for checking// balanced bracketsimport java.util.*;  public class BalancedBrackets {      // function to check if brackets are balanced    static boolean areBracketsBalanced(String expr)    {        // Using ArrayDeque is faster than using Stack class        Deque<Character> stack            = new ArrayDeque<Character>();          // Traversing the Expression        for (int i = 0; i < expr.length(); i++)         {            char x = expr.charAt(i);              if (x == '(' || x == '[' || x == '{')             {                // Push the element in the stack                stack.push(x);                continue;            }              // If current character is not opening            // bracket, then it must be closing. So stack            // cannot be empty at this point.            if (stack.isEmpty())                return false;            char check;            switch (x) {            case ')':                check = stack.pop();                if (check == '{' || check == '[')                    return false;                break;              case '}':                check = stack.pop();                if (check == '(' || check == '[')                    return false;                break;              case ']':                check = stack.pop();                if (check == '(' || check == '{')                    return false;                break;            }        }          // Check Empty Stack        return (stack.isEmpty());    }      // Driver code    public static void main(String[] args)    {        String expr = "([{}])";          // Function call        if (areBracketsBalanced(expr))            System.out.println("Balanced ");        else            System.out.println("Not Balanced ");    }}
Balanced
Time Complexity: O(n) Auxiliary Space: O(n) for stack. 
Please refer complete article on Check for Balanced Brackets in an expression (well-formedness) using Stack for more details!
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