<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>

class Solution {
public:
    string decodeString(string s) 
    {
        if(s.empty()) return "";
    
        std::string res = "";
        int i = 0, j = 0;
        
        while( j < s.size() )
        {
            while( j < s.size() && isalpha(s[j]) ) 
                j++;

            res += s.substr(i, j - i);
            i = j;

            if(j == s.size())
                break;

            while( isdigit(s[j]) ) j++;

            int k = stoi(s.substr(i, j - i));
            int cnt = 1;
            i = j + 1;
            
            while(cnt != 0)
            {
                if(s[++j] == ']')
                {
                    cnt--;
                }
                else if(s[j] == '[') 
                {
                    cnt++;
                }
            }
            while(k--)
            {
                res += decodeString(s.substr(i, j - i));
            }
            i = ++j;
        }
        return res;
    }
};
10000