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

// 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; 
} 

Output:

Balanced 
Not Balanced

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

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My Personal Notes

C

C++

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