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`

.

### Sample Input

3

{[()]}

{[(])}

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

### Sample Output

YES

NO

YES

## Answer

### 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; }