<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) {
        unordered_map<char, long> unique;
        int max_unique = 0;
        int current_count = 0;
        long left = 0;
        long right = s.size() - 1;
        bool left_to_right = true;
        
        while (left <= right) {
            if (left_to_right) {
                auto dupe = unique.find(s[left]);
                if (dupe == unique.end()) {
                    unique[s[left]] = left;
                    max_unique = max(max_unique, ++current_count);
                    left++;
                } else {
                    left = dupe->second + 1;
                    unique.clear();
                    current_count = 0;
                    left_to_right = false;
                }
            } else {
                auto dupe = unique.find(s[right]);
                if (dupe == unique.end()) {
                    unique[s[right]] = right;
                    max_unique = max(max_unique, ++current_count);
                    right--;
                } else {
                    right = dupe->second - 1;
                    unique.clear();
                    current_count = 0;
                    left_to_right = true;
                }
            }
        }
        
        return max_unique;
    }
};
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