<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 isValid(string s)
    {
        unordered_map<char, int> mp;
        for(auto ch : s)
        {
            mp[ch]++;
            if(mp[ch] > 1)
            {
                return false;
            }
        }
        return true;
    }

    int solve(int idx, string s)
    {
        if(idx == s.size())
        {
            return 0;
        }
        
        int ans = INT_MAX;
        for(int j = idx; j < s.size(); j++)
        {
            string sub = s.substr(idx, j - idx + 1);
            if(isValid(sub))
            {
                int mincut = 1 + solve(j + 1, s);
                ans = min(ans, mincut);
            }
        }
        return ans;
    }

    int partitionString(string s) {
        /*   Time Limit Exceeded
        
        return solve(0, s); 
        */

        unordered_map<char,int> mp;
        int i = 0, j = 0, mincuts = 0;

        while(j < s.size())
        {
            mp[s[j]]++;
            while(mp[s[j]] > 1)
            {
                mp.clear();
                mincuts++;
                j--;
            }
            j++;
        }
        return mincuts + 1;
    }
};
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