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Check if given Parentheses expression is balanced or not

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  • Difficulty Level : Easy
  • Last Updated : 08 Apr, 2022

Given a string str of length N, consisting of ‘(‘ and ‘)‘ only, the task is to check whether it is balanced or not.
Examples:
 

Input: str = “((()))()()” 
Output: Balanced
Input: str = “())((())” 
Output: Not Balanced 
 

Approach: 
 

  • Declare a Flag variable which denotes expression is balanced or not.
  • Initialise Flag variable with true and Count variable with 0.
  • Traverse through the given expression
    1. If we encounter an opening parentheses (, increase count by 1
    2. If we encounter a closing parentheses ), decrease count by 1
    3. If Count becomes negative at any point, then expression is said to be not balanced, 
      so mark Flag as false and break from loop.
  • After traversing the expression, if Count is not equal to 0, 
    it means the expression is not balanced so mark Flag as false.
  • Finally, if Flag is true, expression is balanced else not balanced.

Below is the implementation of the above approach:
 

C




// C program of the above approach
#include <stdbool.h>
#include <stdio.h>
 
// Function to check if
// parentheses are balanced
bool isBalanced(char exp[])
{
    // Initialising Variables
    bool flag = true;
    int count = 0;
 
    // Traversing the Expression
    for (int i = 0; exp[i] != '\0'; i++) {
 
        if (exp[i] == '(') {
            count++;
        }
        else {
            // It is a closing parenthesis
            count--;
        }
        if (count < 0) {
            // This means there are
            // more Closing parenthesis
            // than opening ones
            flag = false;
            break;
        }
    }
 
    // If count is not zero,
    // It means there are more
    // opening parenthesis
    if (count != 0) {
        flag = false;
    }
 
    return flag;
}
 
// Driver code
int main()
{
    char exp1[] = "((()))()()";
 
    if (isBalanced(exp1))
        printf("Balanced \n");
    else
        printf("Not Balanced \n");
 
    char exp2[] = "())((())";
 
    if (isBalanced(exp2))
        printf("Balanced \n");
    else
        printf("Not Balanced \n");
 
    return 0;
}

C++




// C++ program for the above approach.
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check
// if parentheses are balanced
bool isBalanced(string exp)
{
 
    // Initialising Variables
    bool flag = true;
    int count = 0;
 
    // Traversing the Expression
    for (int i = 0; i < exp.length(); i++) {
 
        if (exp[i] == '(') {
            count++;
        }
        else {
 
            // It is a closing parenthesis
            count--;
        }
        if (count < 0) {
 
            // This means there are
            // more Closing parenthesis
            // than opening ones
            flag = false;
            break;
        }
    }
 
    // If count is not zero,
    // It means there are
    // more opening parenthesis
    if (count != 0) {
        flag = false;
    }
 
    return flag;
}
 
// Driver code
int main()
{
    string exp1 = "((()))()()";
 
    if (isBalanced(exp1))
        cout << "Balanced \n";
    else
        cout << "Not Balanced \n";
 
    string exp2 = "())((())";
 
    if (isBalanced(exp2))
        cout << "Balanced \n";
    else
        cout << "Not Balanced \n";
 
    return 0;
}

Java




// Java program for the above approach.
class GFG{
 
// Function to check
// if parentheses are balanced
public static boolean isBalanced(String exp)
{
     
    // Initialising variables
    boolean flag = true;
    int count = 0;
     
    // Traversing the expression
    for(int i = 0; i < exp.length(); i++)
    {
        if (exp.charAt(i) == '(')
        {
            count++;
        }
        else
        {
             
            // It is a closing parenthesis
            count--;
        }
        if (count < 0)
        {
             
            // This means there are
            // more Closing parenthesis
            // than opening ones
            flag = false;
            break;
        }
    }
     
    // If count is not zero,
    // It means there are
    // more opening parenthesis
    if (count != 0)
    {
        flag = false;
    }
    return flag;
}
 
// Driver code
public static void main(String[] args)
{
    String exp1 = "((()))()()";
     
    if (isBalanced(exp1))
        System.out.println("Balanced");
    else
        System.out.println("Not Balanced");
     
    String exp2 = "())((())";
     
    if (isBalanced(exp2))
        System.out.println("Balanced");
    else
        System.out.println("Not Balanced");
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python3 program for the above approach
 
# Function to check if
# parenthesis are balanced
def isBalanced(exp):
 
    # Initialising Variables
    flag = True
    count = 0
 
    # Traversing the Expression
    for i in range(len(exp)):
        if (exp[i] == '('):
            count += 1
        else:
             
            # It is a closing parenthesis
            count -= 1
 
        if (count < 0):
 
            # This means there are
            # more closing parenthesis
            # than opening
            flag = False
            break
 
    # If count is not zero ,
    # it means there are more
    # opening parenthesis
    if (count != 0):
        flag = False
 
    return flag
 
# Driver code
if __name__ == '__main__':
     
 
    exp1 = "((()))()()"
 
    if (isBalanced(exp1)):
        print("Balanced")
    else:
        print("Not Balanced")
 
    exp2 = "())((())"
 
    if (isBalanced(exp2)):
        print("Balanced")
    else:
        print("Not Balanced")
 
# This code is contributed by himanshu77

C#




// C# program for the above approach.
using System;
 
class GFG{
 
// Function to check
// if parentheses are balanced
public static bool isBalanced(String exp)
{
     
    // Initialising variables
    bool flag = true;
    int count = 0;
     
    // Traversing the expression
    for(int i = 0; i < exp.Length; i++)
    {
        if (exp[i] == '(')
        {
            count++;
        }
        else
        {
             
            // It is a closing parenthesis
            count--;
        }
        if (count < 0)
        {
             
            // This means there are
            // more Closing parenthesis
            // than opening ones
            flag = false;
            break;
        }
    }
     
    // If count is not zero,
    // It means there are
    // more opening parenthesis
    if (count != 0)
    {
        flag = false;
    }
    return flag;
}
 
// Driver code
public static void Main(String[] args)
{
    String exp1 = "((()))()()";
     
    if (isBalanced(exp1))
        Console.WriteLine("Balanced");
    else
        Console.WriteLine("Not Balanced");
     
    String exp2 = "())((())";
     
    if (isBalanced(exp2))
        Console.WriteLine("Balanced");
    else
        Console.WriteLine("Not Balanced");
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to check if
// parenthesis are balanced
function isBalanced(exp){
 
    // Initialising Variables
    let flag = true
    let count = 0
 
    // Traversing the Expression
    for(let i=0;i<exp.length;i++){
        if (exp[i] == '(')
            count += 1
        else
             
            // It is a closing parenthesis
            count -= 1
 
        if (count < 0){
 
            // This means there are
            // more closing parenthesis
            // than opening
            flag = false
            break
        }
    }
    // If count is not zero ,
    // it means there are more
    // opening parenthesis
    if (count != 0)
        flag = false
 
    return flag
}
 
// Driver code
     
let exp1 = "((()))()()"
 
if (isBalanced(exp1))
    document.write("Balanced","</br>")
else
    document.write("Not Balanced","</br>")
 
let exp2 = "())((())"
 
if (isBalanced(exp2))
    document.write("Balanced","</br>")
else
    document.write("Not Balanced","</br>")
 
// This code is contributed by shinjanpatra
 
</script>

Output: 

Balanced 
Not Balanced

 

Time complexity: O(N) 
Auxiliary Space: O(1)
 


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