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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