Check for Balanced Brackets in an expression (well-formedness) using Stack

Difficulty Level : Easy
Last Updated : 22 Mar, 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

check-for-balanced-parentheses-in-an-expression

Algorithm:

Declare a character stack S.
Now traverse the expression string exp.
1. If the current character is a starting bracket (‘(‘ or ‘{‘ or ‘[‘) then push it to stack.
2. 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:

C++

// CPP program to check for balanced brackets.
#include <bits/stdc++.h>
using namespace std;
 
// function to check if brackets are balanced
bool areBracketsBalanced(string expr)
{ 
    stack<char> s;
    char x;
 
    // Traversing the Expression
    for (int i = 0; i < expr.length(); i++)
    {
        if (expr[i] == '(' || expr[i] == '['
            || expr[i] == '{')
        {
            // Push the element in the stack
            s.push(expr[i]);
            continue;
        }
 
        // IF current current character is not opening
        // bracket, then it must be closing. So stack
        // cannot be empty at this point.
        if (s.empty())
            return false;
 
        switch (expr[i]) {
        case ')':
             
            // Store the top element in a
            x = s.top();
            s.pop();
            if (x == '{' || x == '[')
                return false;
            break;
 
        case '}':
 
            // Store the top element in b
            x = s.top();
            s.pop();
            if (x == '(' || x == '[')
                return false;
            break;
 
        case ']':
 
            // Store the top element in c
            x = s.top();
            s.pop();
            if (x == '(' || x == '{')
                return false;
            break;
        }
    }
 
    // Check Empty Stack
    return (s.empty());
}
 
// Driver code
int main()
{
    string expr = "{()}[]";
 
    // Function call
    if (areBracketsBalanced(expr))
        cout << "Balanced";
    else
        cout << "Not Balanced";
    return 0;
}

C

#include <stdio.h>
#include <stdlib.h>
#define bool int
 
// structure of a stack node
struct sNode {
    char data;
    struct sNode* next;
};
 
// Function to push an item to stack
void push(struct sNode** top_ref, int new_data);
 
// Function to pop an item from stack
int pop(struct sNode** top_ref);
 
// Returns 1 if character1 and character2 are matching left
// and right Brackets
bool isMatchingPair(char character1, char character2)
{
    if (character1 == '(' && character2 == ')')
        return 1;
    else if (character1 == '{' && character2 == '}')
        return 1;
    else if (character1 == '[' && character2 == ']')
        return 1;
    else
        return 0;
}
 
// Return 1 if expression has balanced Brackets
bool areBracketsBalanced(char exp[])
{
    int i = 0;
 
    // Declare an empty character stack
    struct sNode* stack = NULL;
 
    // Traverse the given expression to check matching
    // brackets
    while (exp[i])
    {
        // If the exp[i] is a starting bracket then push
        // it
        if (exp[i] == '{' || exp[i] == '(' || exp[i] == '[')
            push(&stack;, exp[i]);
 
        // If exp[i] is an ending bracket then pop from
        // stack and check if the popped bracket is a
        // matching pair*/
        if (exp[i] == '}' || exp[i] == ')'
            || exp[i] == ']') {
 
            // If we see an ending bracket without a pair
            // then return false
            if (stack == NULL)
                return 0;
 
            // Pop the top element from stack, if it is not
            // a pair bracket of character then there is a
            // mismatch.
            // his happens for expressions like {(})
            else if (!isMatchingPair(pop(&stack;), exp[i]))
                return 0;
        }
        i++;
    }
 
    // If there is something left in expression then there
    // is a starting bracket without a closing
    // bracket
    if (stack == NULL)
        return 1; // balanced
    else
        return 0; // not balanced
}
 
// Driver code
int main()
{
    char exp[100] = "{()}[]";
 
    // Function call
    if (areBracketsBalanced(exp))
        printf("Balanced \n");
    else
        printf("Not Balanced \n");
    return 0;
}
 
// Function to push an item to stack
void push(struct sNode** top_ref, int new_data)
{
    // allocate node
    struct sNode* new_node
        = (struct sNode*)malloc(sizeof(struct sNode));
 
    if (new_node == NULL) {
        printf("Stack overflow n");
        getchar();
        exit(0);
    }
 
    // put in the data
    new_node->data = new_data;
 
    // link the old list off the new node
    new_node->next = (*top_ref);
 
    // move the head to point to the new node
    (*top_ref) = new_node;
}
 
// Function to pop an item from stack
int pop(struct sNode** top_ref)
{
    char res;
    struct sNode* top;
 
    // If stack is empty then error
    if (*top_ref == NULL) {
        printf("Stack overflow n");
        getchar();
        exit(0);
    }
    else {
        top = *top_ref;
        res = top->data;
        *top_ref = top->next;
        free(top);
        return res;
    }
}

Java

// Java program for checking
// balanced brackets
import 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 ");
    }
}

Python3

# Python3 program to check for
# balanced brackets.
 
# function to check if
# brackets are balanced
 
 
def areBracketsBalanced(expr):
    stack = []
 
    # Traversing the Expression
    for char in expr:
        if char in ["(", "{", "["]:
 
            # Push the element in the stack
            stack.append(char)
        else:
 
            # IF current character is not opening
            # bracket, then it must be closing.
            # So stack cannot be empty at this point.
            if not stack:
                return False
            current_char = stack.pop()
            if current_char == '(':
                if char != ")":
                    return False
            if current_char == '{':
                if char != "}":
                    return False
            if current_char == '[':
                if char != "]":
                    return False
 
    # Check Empty Stack
    if stack:
        return False
    return True
 
 
# Driver Code
if __name__ == "__main__":
    expr = "{()}[]"
 
    # Function call
    if areBracketsBalanced(expr):
        print("Balanced")
    else:
        print("Not Balanced")
 
# This code is contributed by AnkitRai01 and improved
# by Raju Pitta

C#

// C# program for checking
// balanced Brackets
using System;
using System.Collections.Generic;
 
public class BalancedBrackets {
    public class stack {
        public int top = -1;
        public char[] items = new char[100];
 
        public void push(char x)
        {
            if (top == 99)
            {
                Console.WriteLine("Stack full");
            }
            else {
                items[++top] = x;
            }
        }
 
        char pop()
        {
            if (top == -1)
            {
                Console.WriteLine("Underflow error");
                return '\0';
            }
            else
            {
                char element = items[top];
                top--;
                return element;
            }
        }
 
        Boolean isEmpty()
        {
            return (top == -1) ? true : false;
        }
    }
 
    // Returns true if character1 and character2
    // are matching left and right brackets */
    static Boolean isMatchingPair(char character1,
                                  char character2)
    {
        if (character1 == '(' && character2 == ')')
            return true;
        else if (character1 == '{' && character2 == '}')
            return true;
        else if (character1 == '[' && character2 == ']')
            return true;
        else
            return false;
    }
 
    // Return true if expression has balanced
    // Brackets
    static Boolean areBracketsBalanced(char[] exp)
    {
        // Declare an empty character stack */
        Stack<char> st = new Stack<char>();
 
        // Traverse the given expression to
        //   check matching brackets
        for (int i = 0; i < exp.Length; i++)
        {
            // If the exp[i] is a starting
            // bracket then push it
            if (exp[i] == '{' || exp[i] == '('
                || exp[i] == '[')
                st.Push(exp[i]);
 
            //  If exp[i] is an ending bracket
            //  then pop from stack and check if the
            //   popped bracket is a matching pair
            if (exp[i] == '}' || exp[i] == ')'
                || exp[i] == ']') {
 
                // If we see an ending bracket without
                //   a pair then return false
                if (st.Count == 0)
                {
                    return false;
                }
 
                // Pop the top element from stack, if
                // it is not a pair brackets of
                // character then there is a mismatch. This
                // happens for expressions like {(})
                else if (!isMatchingPair(st.Pop(),
                                         exp[i])) {
                    return false;
                }
            }
        }
 
        // If there is something left in expression
        // then there is a starting bracket without
        // a closing bracket
 
        if (st.Count == 0)
            return true; // balanced
        else
        {
            // not balanced
            return false;
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        char[] exp = { '{', '(', ')', '}', '[', ']' };
 
        // Function call
        if (areBracketsBalanced(exp))
            Console.WriteLine("Balanced ");
        else
            Console.WriteLine("Not Balanced ");
    }
}
 
// This code is contributed by 29AjayKumar

Output

Balanced

Time Complexity: O(n)
Auxiliary Space: O(n) for stack.

Please write comments if you find any bug in above codes/algorithms, or find other ways to solve the same problem

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