<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) {
        stack<char> st;
        for(int i = 0; i < s.size(); ++i){
            if(s[i] == ']'){
                stack<char> aux;
                while(st.top() != '['){
                    aux.push(st.top());
                    st.pop();
                }
                st.pop();
                int x = 1;
                int check = st.top() - '0';
                int build = 0;
                while(!st.empty() && st.top() - '0' >= 0 && st.top() - '0' < 10){
                    check = st.top() - '0';
                    build += (check) * x;
                    st.pop();
                    x = x * 10;
                }
                cout << build << '\n';
                for(int i = 0; i < build; ++i){
                    stack<char> x = aux;
                    while(!x.empty()){
                        st.push(x.top());
                        x.pop();
                    }
                }
            }
            else{
                st.push(s[i]);
            }
        }
        string ans = "";
        while(!st.empty()){
            ans.push_back(st.top());
            st.pop();
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};;
10400