<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) {
        int n = s.size();
        int l = 0,r = n+1;
        while(l < r-1)
        {
            int m = (l+r)/2;
            auto ok = [&](int m)->bool
            {
                map<char,int>mp;
                for(int i = 0;i < m;i++)
                    mp[s[i]]++;
                int x = 0,y = m;
                while(y < n and mp.size() != m)
                {
                    mp[s[x]]--;
                    if(!mp[s[x]])
                        mp.erase(s[x]);
                    mp[s[y]]++;
                    x++;
                    y++;
                }
                return mp.size() == m;
            };
            if(ok(m))
                l = m;
            else
                r = m;
        }
        return l;
    }
};
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