Expression contains redundant bracket or not

16 Jun, 2022
Given a string of balanced expression, find if it contains a redundant parenthesis or not. A set of parenthesis are 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.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 are 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 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 condition 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 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 contain unwanted () around a thus it is redundant.  
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 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 contain unwanted () around a thus it is redundant.  
 
C++
Java
Python3
C#
Javascript
/* 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);     str = "(a+(b)/c)";    findRedundant(str);     str = "(a+b*(c-d))";    findRedundant(str);     return 0;}
/* Java Program to check whether validexpression 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 Program to check whether valid# expression is redundant or not # Function to check redundant brackets# in a balanced expressiondef 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 bracketsdef findRedundant(Str):    ans = checkRedundancy(Str)    if (ans == True):        print("Yes")    else:        print("No") # Driver codeif __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# Program to check whether validexpression 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 */
<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 code var str = "((a+b))";findRedundant(str); str = "(a+(b)/c)";findRedundant(str); str = "(a+b*(c-d))";findRedundant(str);  </script>
Yes
Yes
No
Time complexity : O(n) Auxiliary Space : O(n)
Rajput-Ji
PranchalKatiyar
princiraj1992
randomuserhere
rutvik_56
rishabhdevba
2002sukhdevsuthar
pushpendrayadav1057
mahendrabagul569
expression-evaluation
Stack
Strings
Strings
Stack
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