<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>

#include <vector>
using namespace std;

class Solution {
public:
    int lengthOfLongestSubstring(string s) 
    {
        if(s.length() == 0)
            return 0;
        

        int finalMax = 0;

        vector<int> usedLetters;

        for(int i = 0; i < 127; i++)
            usedLetters.push_back(-1);

        vector<int> repeatMax[s.length()];

        for(int i = 0; i < s.length(); i++)
            repeatMax[i] = usedLetters;

        int startValue = 0;
        int max;

        for(int i = 0; i < s.length(); i++)
        {
            cout << "This is starting value: " << startValue << endl;
            for(int j = startValue; j < i+1; j++)
            {
                if(repeatMax[j][s[i]] == 1)
                {
                    max = i - j;
                    if(max > finalMax)
                        finalMax = max;

                    startValue++;
                }   
                else
                { 
                    repeatMax[j][s[i]] = 1;
                    
                    if(repeatMax[j][s[i]] == 1)
                    {
                        max = (i+1) - j;
                        if(max > finalMax)
                            finalMax = max;

                    }   


                }
            }
        }
        
        


        return finalMax;

    }
};
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