Valid Parentheses in an Expression
Last Updated : 13 Jan, 2025
Given a string s representing an expression containing various types of brackets: {}, (), and [], the task is to determine whether the brackets in the expression are balanced or not. A balanced expression is one where every opening bracket has a corresponding closing bracket in the correct order.
Example:
Input: s = "[{()}]"
Output: true
Explanation: All the brackets are well-formed.
Input: s = "[()()]{}"
Output: true
Explanation: All the brackets are well-formed.
Input: s = "([]"
Output: false
Explanation: The expression is not balanced as there is a missing ')' at the end.
Input: s = "([{]})"
Output: false
Explanation: The expression is not balanced because there is a closing ']' before the closing '}'.
[Expected Approach 1] Using Stack - O(n) Time and O(n) Space
The idea is to put all the opening brackets in the stack. Whenever you hit a closing bracket, search if the top of the stack is the opening bracket of the same nature. If this holds then pop the stack and continue the iteration. In the end if the stack is empty, it means all brackets are balanced or well-formed. Otherwise, they are not balanced.
Step-by-step approach:
- Declare a character stack (say temp).
- Now traverse the string s.
- If the current character is an opening bracket ( '(' or '{' or '[' ) then push it to stack.
- If the current character is a closing bracket ( ')' or '}' or ']' ) and the closing bracket matches with the opening bracket at the top of stack, then pop the opening bracket. Else s is not balanced.
- After complete traversal, if some starting brackets are left in the stack then the expression is not balanced, else balanced.
Illustration:
C++ // C++ program to check if parentheses are balanced #include <bits/stdc++.h> using namespace std; bool isBalanced(const string& s) { // Declare a stack to store the opening brackets stack<char> st; for (int i = 0; i < s.length(); i++) { // Check if the character is an opening bracket if (s[i] == '(' || s[i] == '{' || s[i] == '[') { st.push(s[i]); } else { // If it's a closing bracket, check if the stack is non-empty // and if the top of the stack is a matching opening bracket if (!st.empty() && ((st.top() == '(' && s[i] == ')') || (st.top() == '{' && s[i] == '}') || (st.top() == '[' && s[i] == ']'))) { // Pop the matching opening bracket st.pop(); } else { // Unmatched closing bracket return false; } } } // If stack is empty, return true (balanced), otherwise false return st.empty(); } int main() { string s = "{([])}"; if (isBalanced(s)) cout << "true"; else cout << "false"; return 0; } // Java program to check if parentheses are balanced import java.util.Stack; class GfG { static boolean isBalanced(String s) { // Declare a stack to store the opening brackets Stack<Character> st = new Stack<>(); for (int i = 0; i < s.length(); i++) { // Check if the character is an opening bracket if (s.charAt(i) == '(' || s.charAt(i) == '{' || s.charAt(i) == '[') { st.push(s.charAt(i)); } else { // If it's a closing bracket, check if the stack is non-empty // and if the top of the stack is a matching opening bracket if (!st.empty() && ((st.peek() == '(' && s.charAt(i) == ')') || (st.peek() == '{' && s.charAt(i) == '}') || (st.peek() == '[' && s.charAt(i) == ']'))) { st.pop(); } else { // Unmatched closing bracket return false; } } } // If stack is empty, return true (balanced), // otherwise false return st.empty(); } public static void main(String[] args) { String s = "{([])}"; if (isBalanced(s)) System.out.println("true"); else System.out.println("false"); } } # Python program to check if parentheses are balanced def isBalanced(s): # Declare a stack to store the opening brackets st = [] for i in range(len(s)): # Check if the character is an opening bracket if s[i] == '(' or s[i] == '{' or s[i] == '[': st.append(s[i]) else: # If it's a closing bracket, check if the stack is non-empty # and if the top of the stack is a matching opening bracket if st and ((st[-1] == '(' and s[i] == ')') or (st[-1] == '{' and s[i] == '}') or (st[-1] == '[' and s[i] == ']')): # Pop the matching opening bracket st.pop() else: # Unmatched closing bracket return False # If stack is empty, return True (balanced), otherwise False return not st if __name__ == "__main__": s = "{([])}" if isBalanced(s): print("true") else: print("false") // C# program to check if parentheses are balanced using System; using System.Collections.Generic; class GfG { static bool isBalanced(string s) { // Declare a stack to store the opening brackets Stack<char> st = new Stack<char>(); for (int i = 0; i < s.Length; i++) { // Check if the character is an opening bracket if (s[i] == '(' || s[i] == '{' || s[i] == '[') { st.Push(s[i]); } else { // If it's a closing bracket, check if the stack is non-empty // and if the top of the stack is a matching opening bracket if (st.Count > 0 && ((st.Peek() == '(' && s[i] == ')') || (st.Peek() == '{' && s[i] == '}') || (st.Peek() == '[' && s[i] == ']'))) { st.Pop(); } else { // Unmatched closing bracket return false; } } } // If stack is empty, return true (balanced), // otherwise false return st.Count == 0; } public static void Main(string[] args) { string s = "{([])}"; if (isBalanced(s)) Console.WriteLine("true"); else Console.WriteLine("false"); } } // Javascript program to check if parentheses are balanced function isBalanced(s) { // Declare a stack to store the opening brackets let st = []; for (let i = 0; i < s.length; i++) { // Check if the character is an opening bracket if (s[i] === '(' || s[i] === '{' || s[i] === '[') { st.push(s[i]); } else { // If it's a closing bracket, check if the stack is non-empty // and if the top of the stack is a matching opening bracket if (st.length > 0 && ((st[st.length - 1] === '(' && s[i] === ')') || (st[st.length - 1] === '{' && s[i] === '}') || (st[st.length - 1] === '[' && s[i] === ']'))) { // Pop the matching opening bracket st.pop(); } else { // Unmatched closing bracket return false; } } } // If stack is empty, return true (balanced), otherwise false return st.length === 0; } // Driver Code let s = "{([])}"; console.log(isBalanced(s) ? "true" : "false"); [Expected Approach 2] Without using Stack - O(n) Time and O(1) Space
Instead of using actual Stack, we can uses the input string s itself to simulate stack behavior. We can use a top variable to keep track of the "top" of this virtual stack. This approach makes use of the existing string to avoid the need for additional memory to store stack elements.
Note: Strings are immutable in Java, Python, C#, and JavaScript. Therefore, we cannot modify them in place, making this approach unsuitable for these languages.
Step-by-Step approach:
- Initialize top = -1 to represent an empty stack.
- Traverse over the given string and for each character:
- If top is -1 or the current character doesn’t match the top, increment top and store the character at s[top].
- If the current character matches s[top], decrement top to remove the last unmatched opening parenthesis.
- After processing, if top is -1, the string is balanced. Otherwise, it is unbalanced.
C++ // C++ program to check if parentheses are balanced #include <bits/stdc++.h> using namespace std; // Check if characters match bool checkMatch(char c1, char c2){ if (c1 == '(' && c2 == ')') return true; if (c1 == '[' && c2 == ']') return true; if (c1 == '{' && c2 == '}') return true; return false; } // Check if parentheses are balanced bool isBalanced(string& s){ // Initialize top to -1 int top = -1; for (int i = 0; i < s.length(); ++i){ // Push char if stack is empty or no match if (top < 0 || !checkMatch(s[top], s[i])){ ++top; s[top] = s[i]; } else{ // Pop from stack if match found --top; } } // Return true if stack is empty (balanced) return top == -1; } int main(){ string s = "{([])}"; cout << (isBalanced(s) ? "true" : "false") << endl; return 0; } // C program to check if parentheses are balanced #include <stdio.h> #include <stdbool.h> #include <string.h> // Check if characters match bool checkMatch(char c1, char c2){ if (c1 == '(' && c2 == ')') return true; if (c1 == '[' && c2 == ']') return true; if (c1 == '{' && c2 == '}') return true; return false; } // Check if parentheses are balanced bool isBalanced(char s[]){ // Initialize top as -1 (empty stack simulation) int top = -1; for (int i = 0; i < strlen(s); ++i){ // Push char if stack is empty or no match if (top < 0 || !checkMatch(s[top], s[i])){ ++top; s[top] = s[i]; } else{ // Pop from stack if match found --top; } } // Return true if stack is empty (balanced) return top == -1; } int main(){ char s[] = "{([])}"; printf("%s\n", isBalanced(s) ? "true" : "false"); return 0; } Similar Reads
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