Balanced Brackets

Problem Information:

problem's link

AC date : 2017-05-15

Category : data structure->stack


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.


  • $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





Sample Output







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)
            else if (!brackets.empty() && left[right.find(s[i])] ==
            else {
                flag = 0;
        if (!brackets.empty())
            flag = 0;

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

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