<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> stack;
        for (int i = 0; i < s.length(); i++) {
            if (s[i] == ']') {
                string decodedString = "";
                // get the encoded string
                while (stack.top() != '[') {
                    decodedString += stack.top();
                    stack.pop();
                }
                // pop [ from stack
                stack.pop();
                int base = 1;
                int k = 0;
                // get the number k
                while (!stack.empty() && isdigit(stack.top())) {
                    k = k + (stack.top() - '0') * base;
                    stack.pop();
                    base *= 10;
                }
                int currentLen = decodedString.size();
                // decode k[decodedString], by pushing decodedString k times into stack
                while (k != 0) {
                    for (int j = decodedString.size() - 1; j >= 0; j--) {
                        stack.push(decodedString[j]);
                    }
                    k--;
                }
            }
            // push the current character to stack
            else {
                stack.push(s[i]);
            }
        }
        // get the result from stack
        string result;
        for (int i = stack.size() - 1; i >= 0; i--) {
            result = stack.top() + result;
            stack.pop();
        }
        return result;
    }
};


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