# Balanced Brackets

Problem Information:

AC date : 2017-05-15

Category : data structure->stack

## Question

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ].

Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and ().

A matching pair of brackets is not balanced if the set of brackets it encloses are not matched. For example, {[(])} is not balanced because the contents in between { and } are not balanced. The pair of square brackets encloses a single, unbalanced opening bracket, (, and the pair of parentheses encloses a single, unbalanced closing square bracket, ].

By this logic, we say a sequence of brackets is considered to be balanced if the following conditions are met:

• It contains no unmatched brackets.
• The subset of brackets enclosed within the confines of a matched pair of brackets is also a matched pair of brackets.

Given strings of brackets, determine whether each sequence of brackets is balanced. If a string is balanced, print YES on a new line; otherwise, print NO on a new line.

### Input Format

The first line contains a single integer, $$n$$, denoting the number of strings. Each line of the subsequent lines consists of a single string, , denoting a sequence of brackets.

### Constraints

• $1 \le n \le 10^3$
• $$1 \le len_s \le 10^3$$, where $$len_s$$ is the length of the sequence.
• Each character in the sequence will be a bracket (i.e., {, }, (, ), [, and ]).

### Output Format

For each string, print whether or not the string of brackets is balanced on a new line. If the brackets are balanced, print YES; otherwise, print NO.

3

{[()]}

{[(])}

{{[[(())]]}}

YES

NO

YES

### C++

//head.h

using namespace std;

int main() {
int t;
cin >> t;
string left = "([{";
string right = ")]}";

for (int a0 = 0; a0 < t; a0++) {
string s;
cin >> s;
int i = 0;
bool flag = 1;
stack<char> brackets;

while (i < s.size()) {
if (left.find(s[i]) != string::npos)
brackets.push(s[i]);
else if (!brackets.empty() && left[right.find(s[i])] == brackets.top())
brackets.pop();
else {
flag = 0;
break;
}
i++;
}
if (!brackets.empty())
flag = 0;

if (flag)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}


2017-05-15 17:10