<p>Given an encoded string, return its decoded string.</p>

<p>The encoding rule is: <code>k[encoded_string]</code>, where the <code>encoded_string</code> inside the square brackets is being repeated exactly <code>k</code> times. Note that <code>k</code> is guaranteed to be a positive integer.</p>

<p>You may assume that the input string is always valid; there are no extra white spaces, square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, <code>k</code>. For example, there will not be input like <code>3a</code> or <code>2[4]</code>.</p>

<p>The test cases are generated so that the length of the output will never exceed <code>10<sup>5</sup></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;3[a]2[bc]&quot;
<strong>Output:</strong> &quot;aaabcbc&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;3[a2[c]]&quot;
<strong>Output:</strong> &quot;accaccacc&quot;
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;2[abc]3[cd]ef&quot;
<strong>Output:</strong> &quot;abcabccdcdcdef&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 30</code></li>
	<li><code>s</code> consists of lowercase English letters, digits, and square brackets <code>&#39;[]&#39;</code>.</li>
	<li><code>s</code> is guaranteed to be <strong>a valid</strong> input.</li>
	<li>All the integers in <code>s</code> are in the range <code>[1, 300]</code>.</li>
</ul>

#define PB push_back
class Solution {
public:
    string decodeString(string s) {
        vector<int> ns;
        vector<char> cs;
        string num;
        for(auto ch : s) {
            if(ch=='[') {
                cs.PB(ch);
                ns.PB(stoi(num));
                num=""s;
            }
            else if(ch==']') {
                vector<char> sub;
                while(cs.back()!='[') {
                    sub.PB(cs.back());
                    cs.pop_back();
                }
                cs.pop_back();
                for(int i=0;i<ns.back();i++)
                    cs.insert(cs.end(), sub.rbegin(), sub.rend());
                ns.pop_back();
            }
            else if(ch<='z' && ch>='a')
                cs.PB(ch);
            else
                num.PB(ch);
        }
        return string(cs.begin(), cs.end());
    }
};
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