<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:
    bool isDig(char c) {
        return c>='0' && c<='9';
    }
    bool isAlpha(char c) {
        return c>='a' && c<='z';
    }
    string f(string &s, int &i) {
        int op = 0;
        string ans = "";
        while(i<s.size() && isAlpha(s[i])) ans+=s[i++];
        string x = "";
        int y = 0;
        while(i<s.size() && op>=0) {
            if (s[i] == '[') {
                y = stoi(x);
                op++;i++;
                x = f(s, i);
                while(y--)ans+=x;
                x="";
            } else if (s[i] == ']') {
                op--;
                if (op>=0)i++;
            } else {
                if (isDig(s[i])) x+=s[i];
                else ans+=s[i];
                i++;
            }
        }
        return ans;
    }
    string decodeString(string s) {
        stack<char> st;
        for (int i = 0; i < s.size(); i++) {
            if (s[i]!=']') {
                st.push(s[i]);
            } else {
                string cur = "";
                while(st.top()!='[') {
                    cur+=st.top();
                    st.pop();
                }
                reverse(cur.begin(), cur.end());
                st.pop();
                int y = 0, x = 1;
                while(!st.empty() && isDig(st.top())) {
                    // cout<<st.top()<<" "<<x<<endl;
                    y+=(st.top()-'0')*x;
                    x*=10;
                    st.pop();
                }
                // cout<<endl;
                // cout<<y<<endl;
                while(y--) {
                    for(auto c: cur)st.push(c);
                }
            }
        }
        string ans = "";
        while(!st.empty()){
            ans+=st.top();
            st.pop();
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};
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