Remove brackets from an algebraic string containing + and - operators

Simplify a given algebraic string of characters, ‘+’, ‘-‘ operators and parentheses. Output the simplified string without parentheses.

Examples:

Input : "a-(b+c)"
Output : "a-b-c"

Input : "a-(b-c-(d+e))-f"
Output : "a-b+c+d+e-f"

The idea is to check operators just before starting of bracket, i.e., before character ‘(‘. If operator is -, we need to toggle all operators inside the bracket. A stack is used which stores only two integers 0 and 1 to indicate whether to toggle or not.
We iterate for every character of input string. Initially push 0 to stack. Whenever the character is an operator (‘+’ or ‘-‘), check top of stack. If top of stack is 0, append the same operator in the resultant string. If top of stack is 1, append the other operator (if ‘+’ append ‘-‘) in the resultant string.

C++

// C++ program to simplify algebraic string 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to simplify the string 
char* simplify(string str) 
{ 
    int len = str.length(); 
  
    // resultant string of max length equal 
    // to length of input string 
    char* res = new char(len); 
    int index = 0, i = 0; 
  
    // create empty stack 
    stack<int> s; 
    s.push(0); 
  
    while (i < len) { 
        if (str[i] == '+') { 
  
            // If top is 1, flip the operator 
            if (s.top() == 1) 
                res[index++] = '-'; 
  
            // If top is 0, append the same operator 
            if (s.top() == 0) 
                res[index++] = '+'; 
  
        } else if (str[i] == '-') { 
            if (s.top() == 1) 
                res[index++] = '+'; 
            else if (s.top() == 0) 
                res[index++] = '-'; 
        } else if (str[i] == '(' && i > 0) { 
            if (str[i - 1] == '-') { 
  
                // x is opposite to the top of stack 
                int x = (s.top() == 1) ? 0 : 1; 
                s.push(x); 
            } 
  
            // push value equal to top of the stack 
            else if (str[i - 1] == '+') 
                s.push(s.top()); 
        } 
  
        // If closing parentheses pop the stack once 
        else if (str[i] == ')') 
            s.pop(); 
  
        // copy the character to the result 
        else
            res[index++] = str[i]; 
        i++; 
    } 
    return res; 
} 
  
// Driver program 
int main() 
{ 
    string s1 = "a-(b+c)"; 
    string s2 = "a-(b-c-(d+e))-f"; 
    cout << simplify(s1) << endl; 
    cout << simplify(s2) << endl; 
    return 0; 
} 

Java

// Java program to simplify algebraic string 
import java.util.*; 
class GfG {  
  
// Function to simplify the string  
static String simplify(String str)  
{  
    int len = str.length();  
  
    // resultant string of max length equal  
    // to length of input string  
    char res[] = new char[len];  
    int index = 0, i = 0;  
  
    // create empty stack  
    Stack<Integer> s = new Stack<Integer> ();  
    s.push(0);  
  
    while (i < len) {  
        if (str.charAt(i) == '+') {  
  
            // If top is 1, flip the operator  
            if (s.peek() == 1)  
                res[index++] = '-';  
  
            // If top is 0, append the same operator  
            if (s.peek() == 0)  
                res[index++] = '+';  
  
        } else if (str.charAt(i) == '-') {  
            if (s.peek() == 1)  
                res[index++] = '+';  
            else if (s.peek() == 0)  
                res[index++] = '-';  
        } else if (str.charAt(i) == '(' && i > 0) {  
            if (str.charAt(i - 1) == '-') {  
  
                // x is opposite to the top of stack  
                int x = (s.peek() == 1) ? 0 : 1;  
                s.push(x);  
            }  
  
            // push value equal to top of the stack  
            else if (str.charAt(i - 1) == '+')  
                s.push(s.peek());  
        }  
  
        // If closing parentheses pop the stack once  
        else if (str.charAt(i) == ')')  
            s.pop();  
  
        // copy the character to the result  
        else
            res[index++] = str.charAt(i);  
        i++;  
    }  
    return new String(res);  
}  
  
// Driver program  
public static void main(String[] args)  
{  
    String s1 = "a-(b+c)";  
    String s2 = "a-(b-c-(d+e))-f";  
    System.out.println(simplify(s1));  
    System.out.println(simplify(s2)); 
} 
}  

Python3

# Python3 program to simplify algebraic String  
  
# Function to simplify the String  
def simplify(Str): 
    Len = len(Str) 
  
    # resultant String of max Length 
    # equal to Length of input String  
    res = [None] * Len
    index = 0
    i = 0
  
    # create empty stack  
    s = [] 
    s.append(0)  
  
    while (i < Len):  
        if (Str[i] == '+'):  
  
            # If top is 1, flip the operator  
            if (s[-1] == 1):  
                res[index] = '-'
                index += 1
  
            # If top is 0, append the  
            # same operator  
            if (s[-1] == 0): 
                res[index] = '+'
                index += 1
  
        elif (Str[i] == '-'): 
            if (s[-1] == 1):  
                res[index] = '+'
                index += 1
            elif (s[-1] == 0):  
                res[index] = '-'
                index += 1
        elif (Str[i] == '(' and i > 0): 
            if (Str[i - 1] == '-'): 
  
                # x is opposite to the top of stack  
                x = 0 if (s[-1] == 1) else 1
                s.append(x) 
  
            # append value equal to top of the stack  
            elif (Str[i - 1] == '+'):  
                s.append(s[-1]) 
  
        # If closing parentheses pop 
        # the stack once  
        elif (Str[i] == ')'):  
            s.pop()  
  
        # copy the character to the result  
        else: 
            res[index] = Str[i] 
            index += 1
        i += 1
    return res 
  
# Driver Code 
if __name__ == '__main__': 
  
    s1 = "a-(b+c)"
    s2 = "a-(b-c-(d+e))-f"
    r1 = simplify(s1) 
    for i in r1: 
        if i != None: 
            print(i, end = " ") 
        else: 
            break
    print() 
    r2 = simplify(s2) 
    for i in r2: 
        if i != None: 
            print(i, end = " ") 
        else: 
            break
  
# This code is contributed by PranchalK 

C#

// C# program to simplify algebraic string 
using System; 
using System.Collections.Generic; 
  
class GfG  
{  
  
// Function to simplify the string  
static String simplify(String str)  
{  
    int len = str.Length;  
  
    // resultant string of max length equal  
    // to length of input string  
    char []res = new char[len];  
    int index = 0, i = 0;  
  
    // create empty stack  
    Stack<int> s = new Stack<int> ();  
    s.Push(0);  
  
    while (i < len) 
    {  
        if (str[i] == '+') 
        {  
  
            // If top is 1, flip the operator  
            if (s.Peek() == 1)  
                res[index++] = '-';  
  
            // If top is 0, append the same operator  
            if (s.Peek() == 0)  
                res[index++] = '+';  
  
        }  
        else if (str[i] == '-') 
        {  
            if (s.Peek() == 1)  
                res[index++] = '+';  
            else if (s.Peek() == 0)  
                res[index++] = '-';  
        } 
        else if (str[i] == '(' && i > 0)  
        {  
            if (str[i - 1] == '-')  
            {  
  
                // x is opposite to the top of stack  
                int x = (s.Peek() == 1) ? 0 : 1;  
                s.Push(x);  
            }  
  
            // push value equal to top of the stack  
            else if (str[i - 1] == '+')  
                s.Push(s.Peek());  
        }  
  
        // If closing parentheses pop the stack once  
        else if (str[i]== ')')  
            s.Pop();  
  
        // copy the character to the result  
        else
            res[index++] = str[i];  
        i++;  
    }  
    return new String(res);  
}  
  
// Driver code 
public static void Main(String[] args)  
{  
    String s1 = "a-(b+c)";  
    String s2 = "a-(b-c-(d+e))-f";  
    Console.WriteLine(simplify(s1));  
    Console.WriteLine(simplify(s2)); 
} 
}  
  
// This code has been contributed by 29AjayKumar 

Output:

a-b-c
a-b+c+d+e-f

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Remove brackets from an algebraic string containing + and – operators

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C#

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