<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 f[100][100];

    int strangePrinter(string s) {
        int n = s.size();
        for (int l = 1; l <= n; l++) {
            // substring with length of l
            for (int i = 0; i + l <= n; i++) {
                // starting at index i
                if (l == 1) f[i][i] = 1;
                else if (l == 2) f[i][i + 1] = s[i] == s[i + 1]? 1 : 2;
                else {
                    int res = l;
                    int curr = 1;
                    vector<pair<int, int>> gaps;
                    int start = -1;
                    for (int j = 1; j <= l; j++) {
                        if (s[i + j - 1] != s[i]) {
                            if (start == -1) {
                                start = i + j - 1;
                            }
                        } else {
                            if (start != -1) {
                                gaps.emplace_back(start, i + j - 2);
                            }
                            start = -1;
                        }
                    }
                    if (start != -1) {
                        gaps.emplace_back(start, i + l - 1);
                    }
                    f[i][i + l - 1] = search(gaps, 0) + 1;
                }
            }
        }
        return f[0][n - 1];
    }

    int search(vector<pair<int, int>>& gaps, int start) {
        if (start == gaps.size()) return 0;
        int ans = INT_MAX;
        for (int i = start; i < gaps.size(); i++) {
            ans = min(ans, f[gaps[start].first][gaps[i].second] + search(gaps, i + 1));
        }
        return ans;
    }
};
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