<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 solve(string s, int left, int right, vector<vector<int>> &dp) {
        if (left == right) {
            return 1;
        }

        if (left > right) {
            return 0;
        }
        
        if (dp[left][right] != -1) {
            return dp[left][right];
        }

        int i = left + 1;

        if (i > right) {
            return dp[left][right] = 1;
        }

        int stringBreak = 1 + solve(s, i, right, dp);

        int stringNotBreak = INT_MAX;
        for (int j = i; j <= right; j++) {
            if (s[j] == s[left]) {
                stringNotBreak = min(stringNotBreak, solve(s, i, j - 1, dp) + solve(s, j, right, dp));
            }
        }

        return dp[left][right] = min(stringNotBreak, stringBreak);
    }

    int strangePrinter(string s) {
        string u = "";
        u += s[0];
        for (int i = 1; i < s.size(); i++) {
            if (s[i] != u[u.size() - 1]) {
                u += s[i];
            }
        }
        int n = u.size();
        vector<vector<int>> dp(n, vector<int>(n, -1));
        return solve(u, 0, n - 1, dp);
    }
};
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