<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) {
      unsigned short result = 0;
    unsigned short counter = 0;

    unordered_map<char, unsigned short> existChar;
    for (unsigned short i = 0; i < s.length(); ++i){

        if (existChar.find(s[i]) == existChar.end()){
            existChar.insert(pair<char, unsigned short>(s[i], i));
            ++counter;
            if (i + 1 == s.length())
                result = result < counter ? counter : result;
            continue;
        }

        result = result < counter ? counter : result;
        counter = 0;
        i -= i - existChar[s[i]];
        existChar = unordered_map<char, unsigned short>();
    }
    return result;
    }
};
222010