<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:
    bool hasRepeatingCharacters(string s)
    {
        unordered_map<char, int> count;

        for(auto c: s)
        {
            if(count.find(c) == count.end()) // if not found
                count[c] = 1;
            else
                return true;
        }

        return false;
    }

    int lengthOfLongestSubstring(string s) {
        int n = s.length();
        int x = min(95, n);
        int left = 1, right=x;
        int potentialAns = (left + right) >> 1;

        while(left <= right)
        {
            bool found = false;
            for(int i = 0; i <= n - potentialAns; i++)
            {
                if(!(hasRepeatingCharacters(s.substr(i, potentialAns))))
                {
                    found = true;
                    break;
                }
            }

            if(found)
            {
                left = potentialAns + 1;
                potentialAns = (left + right) >> 1;
            }
            else
            {
                right = potentialAns - 1;
                potentialAns = (left + right) >> 1;
            }
        }
        return potentialAns;
    }
};
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