<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) {
        if(s.length()==0)
        {
            return 0;
        }
        unordered_map<int, int> mp;
        string temp="";
        int maxi=0;
        for (int i=0; i<s.length(); i++)
        {
            if(mp.find(s[i])==mp.end())
            {
               mp[s[i]]++;
               if (mp.size()>maxi)
               {
                    maxi=mp.size();
               }
               temp+=s[i];
            }
            else
            {
                int key=s[i];
                while (temp.length()!=0&&temp[0]!=key)
                {
                    mp.erase(temp[0]);
                    temp=temp.substr(1, temp.length());
                }
                if (temp.length()!=0&&temp[0]==key)
                {
                    mp.erase(temp[0]);
                    temp=temp.substr(1, temp.length());
                    temp+=key;
                    mp[key]++;
                }
            }
        }
        if (mp.size()>maxi)
        {
             maxi=mp.size();
        }
        return maxi;
    }
};
30090