Given a string of balanced expressions, find if it contains a redundant parenthesis or not. A set of parenthesis is redundant if the 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.
Examples:
Input: str = “((a+b))”
Output: YES
Explanation: ((a+b)) can reduced to (a+b), this RedundantInput: str = “(a+(b)/c)”
Output: YES
Explanation: (a+(b)/c) can reduced to (a+b/c) because b is surrounded by () which is redundant.
Checking Redundant Bracket using Stack
The 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:
- We iterate through the given expression and for each character in the expression
- if the character is an 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 conditions will arise while popping.
- If immediate pop hits an open parenthesis ‘(‘, then we have found a duplicate parenthesis. For example, (((a+b))+c) has duplicate brackets around a+b. When we reach the second “)” after a+b, we have “((” in the stack. Since the top of the stack is an 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 contains unwanted () around a thus it is redundant.
Below is the implementation of the above approach:
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 expressionbool 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 (!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); // push open parenthesis '(', // operators and operands to stack } return false;}// Function to check redundant bracketsvoid findRedundant(string& str){ bool ans = checkRedundancy(str); if (ans == true) cout << "Yes\n"; else cout << "No\n";}// Driver codeint main(){ string str = "((a+b))"; 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); }} |
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) # 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); }}/* This code contributed by PrinciRaj1992 */ |
Javascript
<script>/* JavaScript Program to check whether valid expression is redundant or not*/// Function to check redundant brackets in a// balanced expressionfunction checkRedundancy(str){ // create a stack of characters var st = []; var ans = false; // Iterate through the given expression str.split('').forEach(ch => { // if current character is close parenthesis ')' if (ch == ')') { var top = st[st.length-1]; st.pop(); // If immediate pop have 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) ans = true; } else st.push(ch); // push open parenthesis '(', // operators and operands to stack }); return ans;}// Function to check redundant bracketsfunction findRedundant(str){ var ans = checkRedundancy(str); if (ans == true) document.write( "Yes<br>"); else document.write( "No<br>");}// Driver codevar str = "((a+b))";findRedundant(str);</script> |
Output
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)
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