Given a balanced expression, find if it contains duplicate parenthesis or not. A set of parenthesis are duplicate if the same subexpression is surrounded by multiple parenthesis.
Examples:
Below expressions have duplicate parenthesis - ((a+b)+((c+d))) The subexpression "c+d" is surrounded by two pairs of brackets. (((a+(b)))+(c+d)) The subexpression "a+(b)" is surrounded by two pairs of brackets. (((a+(b))+c+d)) The whole expression is surrounded by two pairs of brackets. ((a+(b))+(c+d)) (b) and ((a+(b)) is surrounded by two pairs of brackets but, it will not be counted as duplicate. Below expressions don't have any duplicate parenthesis - ((a+b)+(c+d)) No subexpression is surrounded by duplicate brackets.
It may be assumed that the given expression is valid and there are not any white spaces present.
The idea is to use stack. 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, push it to the top of the stack. If the character is close parenthesis ‘)’, then pop characters from the stack till matching open parenthesis ‘(‘ is found and a counter is used, whose value is incremented for every character encountered till the opening parenthesis ‘(‘ is found. If the number of characters encountered between the opening and closing parenthesis pair, which is equal to the value of the counter, is less than 1, then a pair of duplicate parenthesis is found else there is no occurrence of redundant parenthesis pairs. For example, (((a+b))+c) has duplicate brackets around “a+b”. When the second “)” after a+b is encountered, the stack contains “((“. Since the top of stack is a opening bracket, it can be concluded that there are duplicate brackets.
Below is the implementation of above idea :
// C++ program to find duplicate parenthesis in a // balanced expression #include <bits/stdc++.h> using namespace std;
// Function to find duplicate parenthesis in a // balanced expression bool findDuplicateparenthesis(string str)
{ // create a stack of characters
stack< char > Stack;
// Iterate through the given expression
for ( char ch : str)
{
// if current character is close parenthesis ')'
if (ch == ')' )
{
// pop character from the stack
char top = Stack.top();
Stack.pop();
// stores the number of characters between a
// closing and opening parenthesis
// if this count is less than or equal to 1
// then the brackets are redundant else not
int elementsInside = 0;
while (top != '(' )
{
elementsInside++;
top = Stack.top();
Stack.pop();
}
if (elementsInside < 1) {
return 1;
}
}
// push open parenthesis '(', operators and
// operands to stack
else
Stack.push(ch);
}
// No duplicates found
return false ;
} // Driver code int main()
{ // input balanced expression
string str = "(((a+(b))+(c+d)))" ;
if (findDuplicateparenthesis(str))
cout << "Duplicate Found " ;
else
cout << "No Duplicates Found " ;
return 0;
} |
import java.util.Stack;
// Java program to find duplicate parenthesis in a // balanced expression public class GFG {
// Function to find duplicate parenthesis in a // balanced expression static boolean findDuplicateparenthesis(String s) {
// create a stack of characters
Stack<Character> Stack = new Stack<>();
// Iterate through the given expression
char [] str = s.toCharArray();
for ( char ch : str) {
// if current character is close parenthesis ')'
if (ch == ')' ) {
// pop character from the stack
char top = Stack.peek();
Stack.pop();
// stores the number of characters between a
// closing and opening parenthesis
// if this count is less than or equal to 1
// then the brackets are redundant else not
int elementsInside = 0 ;
while (top != '(' ) {
elementsInside++;
top = Stack.peek();
Stack.pop();
}
if (elementsInside < 1 ) {
return true ;
}
} // push open parenthesis '(', operators and
// operands to stack
else {
Stack.push(ch);
}
}
// No duplicates found
return false ;
}
// Driver code public static void main(String[] args) {
// input balanced expression
String str = "(((a+(b))+(c+d)))" ;
if (findDuplicateparenthesis(str)) {
System.out.println( "Duplicate Found " );
} else {
System.out.println( "No Duplicates Found " );
}
}
} |
# Python3 program to find duplicate # parenthesis in a balanced expression # Function to find duplicate parenthesis # in a balanced expression def findDuplicateparenthesis(string):
# create a stack of characters
Stack = []
# Iterate through the given expression
for ch in string:
# if current character is
# close parenthesis ')'
if ch = = ')' :
# pop character from the stack
top = Stack.pop()
# stores the number of characters between
# a closing and opening parenthesis
# if this count is less than or equal to 1
# then the brackets are redundant else not
elementsInside = 0
while top ! = '(' :
elementsInside + = 1
top = Stack.pop()
if elementsInside < 1 :
return True
# push open parenthesis '(', operators
# and operands to stack
else :
Stack.append(ch)
# No duplicates found
return False
# Driver Code if __name__ = = "__main__" :
# input balanced expression
string = "(((a+(b))+(c+d)))"
if findDuplicateparenthesis(string) = = True :
print ( "Duplicate Found" )
else :
print ( "No Duplicates Found" )
# This code is contributed by Rituraj Jain |
// C# program to find duplicate parenthesis // in a balanced expression using System;
using System.Collections.Generic;
class GFG
{ // Function to find duplicate parenthesis // in a balanced expression static Boolean findDuplicateparenthesis(String s)
{ // create a stack of characters
Stack< char > Stack = new Stack< char >();
// Iterate through the given expression
char [] str = s.ToCharArray();
foreach ( char ch in str)
{
// if current character is
// close parenthesis ')'
if (ch == ')' )
{
// pop character from the stack
char top = Stack.Peek();
Stack.Pop();
// stores the number of characters between
// a closing and opening parenthesis
// if this count is less than or equal to 1
// then the brackets are redundant else not
int elementsInside = 0;
while (top != '(' )
{
elementsInside++;
top = Stack.Peek();
Stack.Pop();
}
if (elementsInside < 1)
{
return true ;
}
}
// push open parenthesis '(',
// operators and operands to stack
else
{
Stack.Push(ch);
}
}
// No duplicates found
return false ;
} // Driver code public static void Main(String[] args)
{ // input balanced expression
String str = "(((a+(b))+(c+d)))" ;
if (findDuplicateparenthesis(str))
{
Console.WriteLine( "Duplicate Found " );
}
else
{
Console.WriteLine( "No Duplicates Found " );
}
} } // This code is contributed by 29AjayKumar |
<script> // Javascript program to find duplicate // parenthesis in a balanced expression // Function to find duplicate parenthesis // in a balanced expression function findDuplicateparenthesis(s)
{ // Create a stack of characters
let Stack = [];
// Iterate through the given expression
let str = s.split( "" );
for (let ch = 0; ch < str.length;ch++)
{
// If current character is close
// parenthesis ')'
if (str[ch] == ')' )
{
// pop character from the stack
let top = Stack.pop();
// Stores the number of characters between a
// closing and opening parenthesis
// if this count is less than or equal to 1
// then the brackets are redundant else not
let elementsInside = 0;
while (top != '(' )
{
elementsInside++;
top = Stack.pop();
}
if (elementsInside < 1)
{
return true ;
}
}
// push open parenthesis '(', operators
// and operands to stack
else
{
Stack.push(str[ch]);
}
}
// No duplicates found
return false ;
} // Driver code let str = "(((a+(b))+(c+d)))" ;
// Input balanced expression if (findDuplicateparenthesis(str))
{ document.write( "Duplicate Found " );
} else { document.write( "No Duplicates Found " );
} // This code is contributed by rag2127 </script> |
Output:
Duplicate Found
Time complexity of above solution is O(n).
Auxiliary space used by the program is O(n).