<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, pair<int, int>> map;
        string longest = "";
        string strs = "";

        for(int i = 0; i<s.length(); i++) {
            if(map.find(s[i]) != map.end()) {
                if(strs.length() > longest.length())
                    longest = strs;
                
                if(s[i-1] != s[i]) {
                    i = map[s[i]].second + 1;
                }
                strs = "";
                map.clear();
            }
            
            strs+=s[i];
            map[s[i]] = make_pair(1, i);
        }

        if(strs.length() > longest.length())
            longest = strs;

        return longest.length();
    }
};
198020