<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;
        }
        vector<int> dp(s.length());
        dp[0] = 1;
        unordered_set<char> chars;
        chars.insert(s[0]);
        int max = 1;
        int newStart = 0;
        for(int i = 1; i < s.length(); i++) {
            if(chars.find(s[i]) == chars.end()) {
                chars.insert(s[i]);
                dp[i] = dp[i-1]+1;
                if(max < dp[i]) {
                    max = dp[i];
                }
            } else {
                chars.clear();
                chars.insert(s[i]);
                for(int j = i-1; j >= 0; j--) {
                    if(s[i] == s[j]) {
                        dp[i] = i-j;
                        if(max < dp[i]) {
                            max = dp[i];
                        }
                        break;
                    }
                    chars.insert(s[j]);
                }
            }
        }
        return max;
    }
};
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