<p>Given a string <code>s</code>, find the length of the <strong>longest</strong> <span data-keyword="substring-nonempty"><strong>substring</strong></span> without repeating characters.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcabcbb&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The answer is &quot;abc&quot;, with the length of 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bbbbb&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> The answer is &quot;b&quot;, with the length of 1.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;pwwkew&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The answer is &quot;wke&quot;, with the length of 3.
Notice that the answer must be a substring, &quot;pwke&quot; is a subsequence and not a substring.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= s.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>s</code> consists of English letters, digits, symbols and spaces.</li>
</ul>


class Solution
{
public:
    int lengthOfLongestSubstring(string s)
    {
        string bestSubstring = "";
        int startIndex = 0;
        unordered_map<char, int> existingCharacters = {};

        while (startIndex < s.size())
        {
            int nextIndex = startIndex + 1;
            existingCharacters.clear();
            existingCharacters[s.at(startIndex)] = startIndex;
            string currentSubstring = "";
            currentSubstring.push_back(s.at(startIndex));
            while (nextIndex < s.size())
            {
                if (existingCharacters.count(s.at(nextIndex)))
                {
                    break;
                }

                currentSubstring.push_back(s.at(nextIndex));
                existingCharacters[s.at(nextIndex)] = nextIndex;
                nextIndex++;
            }

            if (currentSubstring.size() > bestSubstring.size())
            {
                bestSubstring = currentSubstring;
            }

            startIndex++;
        }

        return bestSubstring.size();
    }
};
162035