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Identify and mark unmatched parenthesis in an expression
  • Difficulty Level : Easy
  • Last Updated : 18 Sep, 2020

Given an expression, find and mark matched and unmatched parenthesis in it. We need to replace all balanced opening parenthesis with 0, balanced closing parenthesis with 1, and all unbalanced with -1.
Examples:  

Input : ((a) 
Output : -10a1

Input : (a))
Output : 0a1-1

Input  : (((abc))((d)))))
Output : 000abc1100d111-1-1


The idea is based on a stack. We run a loop from the start of the string Up to end and for every ‘(‘, we push it into a stack. If the stack is empty, and we encounter a closing bracket ‘)’ we replace -1 at that index of the string. Else we replace all opening brackets ‘(‘ with 0 and closing brackets with 1. Then pop from the stack. 
 

C++

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// CPP program to mark balanced and unbalanced
// parenthesis.
#include <bits/stdc++.h>
using namespace std;
 
void identifyParenthesis(string a)
{
    stack<int> st;
 
    // run the loop upto end of the string
    for (int i = 0; i < a.length(); i++) {
 
        // if a[i] is opening bracket then push
        // into stack
        if (a[i] == '(')
            st.push(i);
         
        // if a[i] is closing bracket ')'
        else if (a[i] == ')') {
 
            // If this closing bracket is unmatched
            if (st.empty())
                a.replace(i, 1, "-1");
             
            else {
 
                // replace all opening brackets with 0
                // and closing brackets with 1
                a.replace(i, 1, "1");
                a.replace(st.top(), 1, "0");
                st.pop();
            }
        }
    }
 
    // if stack is not empty then pop out all
    // elements from it and replace -1 at that
    // index of the string
    while (!st.empty()) {
        a.replace(st.top(), 1, "-1");
        st.pop();
    }
 
    // print final string
    cout << a << endl;
}
 
// Driver code
int main()
{
    string str = "(a))";
    identifyParenthesis(str);
    return 0;
}

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Java

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// Java program to mark balanced and
// unbalanced parenthesis.
import java.util.*;
 
class GFG
{
static void identifyParenthesis(StringBuffer a)
{
    Stack<Integer> st = new Stack<Integer>();
 
    // run the loop upto end of the string
    for (int i = 0; i < a.length(); i++)
    {
 
        // if a[i] is opening bracket then push
        // into stack
        if (a.charAt(i) == '(')
            st.push(i);
         
        // if a[i] is closing bracket ')'
        else if (a.charAt(i) == ')')
        {
 
            // If this closing bracket is unmatched
            if (st.empty())
                a.replace(i, i + 1, "-1");
             
            else
            {
 
                // replace all opening brackets with 0
                // and closing brackets with 1
                a.replace(i, i + 1, "1");
                a.replace(st.peek(), st.peek() + 1, "0");
                st.pop();
            }
        }
    }
 
    // if stack is not empty then pop out all
    // elements from it and replace -1 at that
    // index of the string
    while (!st.empty())
    {
        a.replace(st.peek(), 1, "-1");
        st.pop();
    }
 
    // print final string
    System.out.println(a);
}
 
// Driver code
public static void main(String[] args)
{
    StringBuffer str = new StringBuffer("(a))");
    identifyParenthesis(str);
}
}
 
// This code is contributed by Princi Singh

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Python3

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# Python3 program to
# mark balanced and
# unbalanced parenthesis.
def identifyParenthesis(a):
    st = []
 
    # run the loop upto
    # end of the string
    for i in range (len(a)):
 
        # if a[i] is opening
        # bracket then push
        # into stack
        if (a[i] == '('):
            st.append(a[i])
         
        # if a[i] is closing bracket ')'
        elif (a[i] == ')'):
 
            # If this closing bracket
            # is unmatched
            if (len(st) == 0):
                a = a.replace(a[i], "-1", 1)
             
            else:
 
                # replace all opening brackets with 0
                # and closing brackets with 1
                a = a.replace(a[i], "1", 1)
                a = a.replace(st[-1], "0", 1)
                st.pop()
           
    # if stack is not empty
    # then pop out all
    # elements from it and
    # replace -1 at that
    # index of the string
    while (len(st) != 0):
        a = a.replace(st[-1], 1, "-1");
        st.pop()
 
    # print final string
    print(a)
 
# Driver code
if __name__ == "__main__":
 
    st = "(a))"
    identifyParenthesis(st)
     
# This code is contributed by Chitranayal

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Output: 
 

0a1-1

 

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