<p>Given a string <code>s</code>, partition the string into one or more <strong>substrings</strong> such that the characters in each substring are <strong>unique</strong>. That is, no letter appears in a single substring more than <strong>once</strong>.</p>

<p>Return <em>the <strong>minimum</strong> number of substrings in such a partition.</em></p>

<p>Note that each character should belong to exactly one substring in a partition.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abacaba&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong>
Two possible partitions are (&quot;a&quot;,&quot;ba&quot;,&quot;cab&quot;,&quot;a&quot;) and (&quot;ab&quot;,&quot;a&quot;,&quot;ca&quot;,&quot;ba&quot;).
It can be shown that 4 is the minimum number of substrings needed.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ssssss&quot;
<strong>Output:</strong> 6
<strong>Explanation:
</strong>The only valid partition is (&quot;s&quot;,&quot;s&quot;,&quot;s&quot;,&quot;s&quot;,&quot;s&quot;,&quot;s&quot;).
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of only English lowercase letters.</li>
</ul>

class Solution {
public:
    int partitionString(string s) {
        int n = s.size();
        vector<int>vec(n,0);
        int head = -1;
        int tail =0 ;
        vector<int>cnt(26,0);

        while(tail < n){
            while(head+1 < n && cnt[s[head+1]-'a'] == 0){
              head++;
              cnt[s[head]-'a']++;
            }
            
            // cout<<tail<<" "<<head<<endl;

            vec[tail] = head ;

            if(head < tail){
                tail++;
                head = tail-1;
            }
            else{
                cnt[s[tail]-'a']--;
                tail++;
            }
        }

        int ans = 0 ;

        vector<int>dp(n+1,0);
        dp[n] = 0;
        
        // for(int i =0 ; i<n ;i++){
        //     cout<<vec[i]<<" ";
        // }
        // cout<<endl;

        for(int i = n-1; i>=0 ; i--){    
            dp[i] = 1 + dp[vec[i]+1];    
        }
        
        return dp[0];
    }
};
16803