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.
// 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;
} |
// 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 |
# 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 |
// C# program to mark balanced and // unbalanced parenthesis. using System;
using System.Collections.Generic;
class GFG {
static void identifyParenthesis( string a)
{
Stack< int > st = new Stack< int >();
// 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.Count == 0)
{
a = a.Substring(0, i) + "-1" + a.Substring(i + 1);
}
else
{
// replace all opening brackets with 0
// and closing brackets with 1
a = a.Substring(0, i) + "1" + a.Substring(i + 1);
a = a.Substring(0, st.Peek()) + "0" + a.Substring(st.Peek() + 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 (st.Count > 0)
{
a = a.Substring(0, st.Peek()) + "-1" + a.Substring(st.Peek() + 1);
st.Pop();
}
// print final string
Console.Write( new string (a));
}
static void Main() {
string str = "(a))" ;
identifyParenthesis(str);
}
} // This code is contributed by divyesh072019. |
<script> // Javascript program to mark balanced and
// unbalanced parenthesis.
function identifyParenthesis(a)
{
let st = [];
// run the loop upto end of the string
for (let 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.length == 0)
{
a[i] = "-1" ;
}
else
{
// replace all opening brackets with 0
// and closing brackets with 1
a[i] = "1" ;
a[st[st.length - 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.length > 0)
{
a[st[st.length - 1]] = "-1" ;
st.pop();
}
// print final string
document.write(a.join( "" ));
}
let str = "(a))" ;
identifyParenthesis(str.split( '' ));
// This code is contributed by suresh07. </script> |
Output:
0a1-1
Time Complexity: O(n)
Auxiliary Space : O(n)