<p>There is a strange printer with the following two special properties:</p>

<ul>
	<li>The printer can only print a sequence of <strong>the same character</strong> each time.</li>
	<li>At each turn, the printer can print new characters starting from and ending at any place and will cover the original existing characters.</li>
</ul>

<p>Given a string <code>s</code>, return <em>the minimum number of turns the printer needed to print it</em>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;aaabbb&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Print &quot;aaa&quot; first and then print &quot;bbb&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Print &quot;aaa&quot; first and then print &quot;b&quot; from the second place of the string, which will cover the existing character &#39;a&#39;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    int strangePrinter(string s) {
        unordered_map<string, int> dp;
        int n = s.size();
        string result = "";

        for(int i=0;i<n;i++){
            if(i == 0){
                result += s[i];
            } else {
                if(result[result.size() - 1] != s[i]){
                    result += s[i];
                }
            }
        }

        return solve(result, dp, 0, result.size() - 1);
    }

    int solve(string& s, unordered_map<string, int>& dp, int l, int r){
        if(l > r) return 0;
        int n = s.size();
        string t = to_string(l) + "#" + to_string(r);
        if(dp.find(t) != dp.end()){
            return dp[t];
        }
        int ans = 1 + solve(s, dp, l+1, r);
        for(int i=l+1;i<=r;i++){
            if(s[l] == s[i]){
                ans = min(ans, solve(s, dp, l, i-1) + solve(s, dp, i+1, r));
            }
        }    
        dp[t] = ans;
        return dp[t];
    }
};
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