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Balance a string after removing extra brackets in C++
A string is an array of characters. In this problem, we are given a string which has opening and closing brackets. And we will balance this string by removing extra brackets from the string.
Let’s take an example,
Input : “)Tutor)ials(p(oin)t(...)” Output : “Tutorials(p(oin)t(...))”
To solve this problem, we will traverse through the string and check for matching brackets. For unmatched brackets eliminate the closing brackets.
Algorithm
Step 1 : Traverse the string from left to right. Step 2 : For opening bracket ‘(’ , print it and increase the count. Step 3 : For occurence of closing bracket ‘)’ , print it only if count is greater than 0 and decrease the count. Step 4 : Print all characters other than brackets are to be printed in the array. Step 5 : In the last add closing brackets ‘)’ , to make the count 0 by decrementing count with every bracket.
Example
#include<iostream>
#include<string.h>
using namespace std;
void balancedbrackets(string str){
int count = 0, i;
int n = str.length();
for (i = 0; i < n; i++) {
if (str[i] == '(') {
cout << str[i];
count++;
}
else if (str[i] == ')' && count != 0) {
cout << str[i];
count--;
}
else if (str[i] != ')')
cout << str[i];
}
if (count != 0)
for (i = 0; i < count; i++)
cout << ")";
}
int main() {
string str = ")Tutor)ials(p(oin)t(...)";
cout<<"Original string : "<<str;
cout<<"\nBalanced string : ";
balancedbrackets(str);
return 0;
}Output
Original string : )Tutor)ials(p(oin)t(...) Balanced string : Tutorials(p(oin)t(...))
Updated on: 24-Oct-2019
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