Check if given Parentheses expression is balanced or not

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

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// 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;
}

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C++

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// 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;
}

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Output:

Balanced 
Not Balanced

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

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