Expression contains redundant bracket or not
Given a string of balanced expression, find if it contains a redundant parenthesis or not. A set of parenthesis are redundant if same sub-expression is surrounded by unnecessary or multiple brackets. Print ‘Yes’ if redundant else ‘No’.
Note: Expression may contain ‘+‘, ‘*‘, ‘–‘ and ‘/‘ operators. Given expression is valid and there are no white spaces present.
Example:
Input: ((a+b)) (a+(b)/c) (a+b*(c-d)) Output: Yes Yes No Explanation: 1. ((a+b)) can reduced to (a+b), this Redundant 2. (a+(b)/c) can reduced to (a+b/c) because b is surrounded by () which is redundant. 3. (a+b*(c-d)) doesn't have any redundant or multiple brackets.
The idea is to use stack which is discussed in this article. For any sub-expression of expression, if we able to pick any sub-expression of expression surrounded by (), then we again left with () as part of string, we have redundant braces.
We iterate through the given expression and for each character in the expression, if the character is a open parenthesis ‘(‘ or any of the operators or operands, we push it to the stack. If the character is close parenthesis ‘)’, then pop characters from the stack till matching open parenthesis ‘(‘ is found.
Now for redundancy two condition will arise while popping-
- If immediate pop hits a open parenthesis ‘(‘, then we have found a duplicate parenthesis. For example, (((a+b))+c) has duplicate brackets around a+b. When we reach second “)” after a+b, we have “((” in the stack. Since the top of stack is a opening bracket, we conclude that there are duplicate brackets.
- If immediate pop doesn’t hit any operand(‘*’, ‘+’, ‘/’, ‘-‘) then it indicates the presence of unwanted brackets surrounded by expression. For instance, (a)+b contain unwanted () around a thus it is redundant.
C++
/* C++ Program to check whether valid expression is redundant or not*/#include <bits/stdc++.h> using namespace std; // Function to check redundant brackets in a // balanced expression bool checkRedundancy(string& str) { // create a stack of characters stack<char> st; // Iterate through the given expression for (auto& ch : str) { // 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 (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); // push open parenthesis '(', // operators and operands to stack } return false; } // Function to check redundant brackets void findRedundant(string& str) { bool ans = checkRedundancy(str); if (ans == true) cout << "Yes\n"; else cout << "No\n"; } // Driver code int main() { string str = "((a+b))"; findRedundant(str); str = "(a+(b)/c)"; findRedundant(str); str = "(a+b*(c-d))"; findRedundant(str); 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 have 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 == true) { return true; } } else { st.push(ch); // push open parenthesis '(', } // operators and operands to stack } return false; } // Function to check redundant brackets static void findRedundant(String str) { boolean ans = checkRedundancy(str); if (ans == true) { System.out.println("Yes"); } else { System.out.println("No"); } } // Driver code public static void main(String[] args) { String str = "((a+b))"; findRedundant(str); str = "(a+(b)/c)"; findRedundant(str); str = "(a+b*(c-d))"; findRedundant(str); } } |
Python3
# Python3 Program to check whether valid # expression is redundant or not # Function to check redundant brackets # in a balanced expression def checkRedundancy(Str): # create a stack of characters st = [] # Iterate through the given expression for ch in Str: # if current character is close # parenthesis ')' if (ch == ')'): top = st[-1] st.pop() # If immediate pop have open parenthesis # '(' duplicate brackets found flag = True while (top != '('): # Check for operators in expression if (top == '+' or top == '-' or top == '*' or top == '/'): flag = False # Fetch top element of stack top = st[-1] st.pop() # If operators not found if (flag == True): return True else: st.append(ch) # append open parenthesis '(', # operators and operands to stack return False # Function to check redundant brackets def findRedundant(Str): ans = checkRedundancy(Str) if (ans == True): print("Yes") else: print("No") # Driver code if __name__ == '__main__': Str = "((a+b))" findRedundant(Str) Str = "(a+(b)/c)" findRedundant(Str) Str = "(a+b*(c-d))" findRedundant(Str) # This code is contributed by PranchalK |
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 have 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); // push open parenthesis '(', } // operators and operands to stack } return false; } // Function to check redundant brackets static void findRedundant(String str) { bool ans = checkRedundancy(str); if (ans == true) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } } // Driver code public static void Main(String[] args) { String str = "((a+b))"; findRedundant(str); str = "(a+(b)/c)"; findRedundant(str); str = "(a+b*(c-d))"; findRedundant(str); } } /* This code contributed by PrinciRaj1992 */ |
Output
Yes Yes No
Time complexity: O(n)
Auxiliary space: O(n)
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