Expression contains redundant bracket or not Last Updated : 08 Feb, 2025 Summarize Comments Improve Suggest changes Share Like Article Like Report Try it on GfG Practice Given a string of balanced expressions, find if it contains a redundant parenthesis or not. Print 'Yes' if redundant, else 'No'.What are Redundant Parentheses?A pair of parentheses is redundant if removing them does not change the expression. This happens in two cases:Extra Parentheses Around an Expression: ((a+b)) → The outermost brackets are unnecessary.Unnecessary Parentheses Around a Single Element: (a)+b → The parentheses around a are not needed.Note: Expression may contain '+', '*', '-' and '/' operators. Given expression is valid and there are no white spaces present.Examples: Input: s = "((a+b))"Output: TrueExplanation: ((a+b)) can reduced to (a+b), this RedundantInput: s = "(a+(b)/c)"Output: FalseExplanation: (a+(b)/c) can reduced to (a+b/c) because b is surrounded by () which is redundant.Approach: Using StackThe idea is to use the stack, For any sub-expression of expression, if we are able to pick any sub-expression of expression surrounded by (), then we are again left with ( ) as part of the string, we have redundant braces. Follow the steps mentioned below to implement the approach:Iterate through each character in the given expression.If the character is an open parenthesis (, an operand (a-z), or an operator (+, -, *, /), push it onto the stack.If the character is a closing parenthesis ), we need to check if the parentheses are redundant.When a closing parenthesis ) is found, pop elements from the stack until an opening parenthesis ( is encountered.While popping, check if there is at least one operator (+, -, *, /).If no operator is found before reaching (, it means the parentheses are redundant.If the top of the stack is an opening ( immediately when popping starts, it also means redundancy.If redundancy is found, return "True". Otherwise, after checking the entire expression, return "False". C++ /* C++ Program to check whether valid expression is redundant or not*/ #include <iostream> #include <vector> #include <stack> using namespace std; // Function to check redundant brackets in a // balanced expression bool checkRedundancy(string& s) { // create a stack of characters stack<char> st; // Iterate through the given expression for (auto& ch : s) { // if current character is close parenthesis ')' if (ch == ')') { char top = st.top(); st.pop(); // If immediate pop have open parenthesis '(' // duplicate brackets found bool flag = true; while (!st.empty() and top != '(') { // Check for operators in expression if (top == '+' || top == '-' || top == '*' || top == '/') flag = false; // Fetch top element of stack top = st.top(); st.pop(); } // If operators not found if (flag == true) return true; } else st.push(ch); } return false; } // Function to check redundant brackets void findRedundant(string& s) { bool res = checkRedundancy(s); if (res == true) cout << "True\n"; else cout << "False\n"; } int main() { string s = "((a+b))"; findRedundant(s); return 0; } Java /* Java Program to check whether valid expression is redundant or not*/ import java.util.Stack; public class GFG { // Function to check redundant brackets in a // balanced expression static boolean checkRedundancy(String s) { // create a stack of characters Stack<Character> st = new Stack<>(); char[] str = s.toCharArray(); // Iterate through the given expression for (char ch : str) { // if current character is close parenthesis ')' if (ch == ')') { char top = st.peek(); st.pop(); // If immediate pop has open parenthesis '(' // duplicate brackets found boolean flag = true; while (top != '(') { // Check for operators in expression if (top == '+' || top == '-' || top == '*' || top == '/') { flag = false; } // Fetch top element of stack top = st.peek(); st.pop(); } // If operators not found if (flag) { return true; } } else { st.push(ch); } } return false; } // Function to check redundant brackets static void findRedundant(String s) { boolean res = checkRedundancy(s); if (res) { System.out.println("True"); } else { System.out.println("False"); } } public static void main(String[] args) { String s = "((a+b))"; findRedundant(s); } } Python # Python3 Program to check whether valid # expression is redundant or not # Function to check redundant brackets # in a balanced expression def checkRedundancy(s): # create a stack of characters st = [] # Iterate through the given expression for ch in s: # if current character is close # parenthesis ')' if ch == ')': top = st[-1] st.pop() # If immediate pop has open parenthesis # '(' duplicate brackets found flag = True while top != '(': # Check for operators in expression if top in {'+', '-', '*', '/'}: flag = False # Fetch top element of stack top = st[-1] st.pop() # If operators not found if flag: return True else: st.append(ch) return False # Function to check redundant brackets def findRedundant(s): res = checkRedundancy(s) if res: print("True") else: print("False") if __name__ == '__main__': s = "((a+b))" findRedundant(s) C# /* C# Program to check whether valid expression is redundant or not*/ using System; using System.Collections.Generic; class GFG { // Function to check redundant brackets in a // balanced expression static bool checkRedundancy(String s) { // create a stack of characters Stack<char> st = new Stack<char>(); char[] str = s.ToCharArray(); // Iterate through the given expression foreach (char ch in str) { // if current character is close parenthesis ')' if (ch == ')') { char top = st.Peek(); st.Pop(); // If immediate pop has open parenthesis '(' // duplicate brackets found bool flag = true; while (top != '(') { // Check for operators in expression if (top == '+' || top == '-' || top == '*' || top == '/') { flag = false; } // Fetch top element of stack top = st.Peek(); st.Pop(); } // If operators not found if (flag == true) { return true; } } else { st.Push(ch); } } return false; } // Function to check redundant brackets static void findRedundant(String s) { bool res = checkRedundancy(s); if (res == true) { Console.WriteLine("True"); } else { Console.WriteLine("False"); } } public static void Main(String[] args) { String s = "((a+b))"; findRedundant(s); } } JavaScript /* JavaScript Program to check whether valid expression is redundant or not*/ // Function to check redundant brackets in a // balanced expression function checkRedundancy(s) { // create a stack of characters var st = []; var res = false; // Iterate through the given expression s.split('').forEach(ch => { // if current character is close parenthesis ')' if (ch == ')') { var top = st[st.length-1]; st.pop(); // If immediate pop has open parenthesis '(' // duplicate brackets found var flag = true; while (st.length!=0 && top != '(') { // Check for operators in expression if (top == '+' || top == '-' || top == '*' || top == '/') flag = false; // Fetch top element of stack top = st[st.length-1]; st.pop(); } // If operators not found if (flag == true) res = true; } else st.push(ch); }); return res; } // Function to check redundant brackets function findRedundant(s) { var res = checkRedundancy(s); if (res == true) console.log( "True"); else console.log( "False"); } // Driver code var s = "((a+b))"; findRedundant(s); OutputTrue Time Complexity: O(n)Auxiliary Space: O(n) Comment More infoAdvertise with us Next Article What is Stack Data Structure? 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