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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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-

1. 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.
2. 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 #include using namespace std;    // Function to check redundant brackets in a // balanced expression bool checkRedundancy(string& str) {     // create a stack of characters     stack 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 ";     else         cout << "No "; }    // 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 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);     } }

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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 (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

Output

Yes
Yes
No

Time complexity: O(n)
Auxiliary space: O(n)