<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 {
private:
    int solve(string &s , int idx , vector<int> &dp){

        unordered_map<char,int> hash;
        if(dp[idx] != -1) return dp[idx]; 

        int ct = 0;
        int ans = 0;

        for(int i=idx;i<s.size();i++){
            hash[s[i]]++;
             
            if(hash[s[i]] < 2){
                ct++;
                ans = max({ct , solve(s,i+1 , dp) , ans});
            }
            else {
                break;
            }

        }
        hash.clear();
        return dp[idx] = ans;
    }

    bool subtract(vector<int> &f , vector<int> &s){

        for(int i=0;i<256;i++){
            int x = f[i]-s[i];
            if(x>=2) return false;
        }

        return true;

    }
public:
    int lengthOfLongestSubstring(string s) {
        int n = s.size();

        vector<int> hash(256,0);

        vector<vector<int>> dp(n+1);
        dp[0] = hash;

        // storing prefix count of characters
        for(int i=0;i<n;i++){
            hash[s[i]-0]++;
            dp[i+1] = hash;
        }


        // finding our ans
        int i=0;
        int j=1;
        int len = 0;
        while(j<=n && i<=j){
            if(subtract(dp[j] , dp[i])){
                len = max(len , j-i);
                j++;
            } else i++;
        }
        return len;
    }
};
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