Bracket Matching

 /*

Bracket Matching

Example 1:

Input:
{([])}
Output:
true
Explanation:
{ ( [ ] ) }. Same brackets can form
balanced pairs, with 0 number of
unbalanced bracket.

Example 2:

Input:
()
Output:
true
Explanation:
(). Same bracket can form balanced pairs,
and here only 1 type of bracket is
present and in balanced way.

Example 3:

Input:
([]
Output:
false
Explanation:
([]. Here square bracket is balanced but
the small bracket is not balanced and
Hence , the output will be unbalanced.

Your Task:
This is a function problem. You only need to complete the function ispar() that takes a string as a parameter and returns a boolean value true if brackets are balanced else returns false. The printing is done automatically by the driver code.

Expected Time Complexity: O(|x|)
Expected Auxiliary Space: O(|x|)

Constraints:
1 ≤ |x| ≤ 32000


*/


#include <bits/stdc++.h>
using namespace std;



class Solution
{
public:

    bool ispar(string x)
    {

        int open_parentheses = 0, open_curly = 0, open_square = 0;
        int close_parentheses = 0, close_curly = 0, close_square = 0;
        int y;
        for (int i = 0; i < x.size(); i++)
        {

            if (x[i] == '(')
                open_parentheses += 1;
            if (x[i] == '{')
                open_curly += 1;
            if (x[i] == '[')
                open_square += 1;
            if (x[i] == ']')
                close_square += 1;
            if (x[i] == '}')
                close_curly += 1;
            if (x[i] == ')')
                close_parenthesis += 1;
            if ((open_parenthesis < close_parentheses) || (open_square < close_square) || (open_curly < close_curly))
                return false;
        }
        if (((open_parentheses - close_parentheses) == 0) && ((open_square - close_square) == 0) && ((open_curly - close_curly) == 0))
            return true;
        return false;
    }
};


int main()
{
    int t;
    string a;
    cin >> t;
    while (t--)
    {
        cin >> a;
        Solution obj;
        if (obj.ispar(a))
            cout << "balanced" << endl;
        else
            cout << "not balanced" << endl;
    }
}