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: BalancedInput: 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
- If we encounter an opening parentheses (, increase count by 1
- If we encounter a closing parentheses ), decrease count by 1
- 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; } |
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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; } |
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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 |
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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 |
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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 |
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Output:
Balanced Not Balanced
Time complexity: O(N)
Auxiliary Space: O(1)
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