<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 {
int stringToInteger(string s) {
        int ans = 0;
        for (char nxt : s) {
            ans *= 10;
            ans += nxt - '0';
        }
        return ans;
    }

   public:
    string decodeString(string s) {
        string ans = "";
        int prev = 0;
        int repetitions = 0;
        int depth = 0;  // keeps track of # open bracket - # close bracket
        for (int i = 0; i < s.size(); i++) {
            if (depth == 0 && 'a' <= s[i] && s[i] <= 'z') {
            	  // case with lowercase letters
                ans.push_back(s[i]);
                prev = i + 1;
            }
            if (s[i] == '[') {
                depth++;
                if (depth == 1) {  // open bracket for the case "k[s]"
                    repetitions = stringToInteger(s.substr(prev, i - prev));
                    prev = i + 1;
                }
            } else if (s[i] == ']') {
                depth--;
                if (depth == 0) {              // close bracket for the case "k[s]"
                    while (repetitions > 0) {  // add k copies of s
                        ans += decodeString(s.substr(prev, i - prev));
                        repetitions--;
                    }
                    prev = i + 1;
                }
            }
        }
        return ans;
    }
};
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