<p>There is a row of <code>m</code> houses in a small city, each house must be painted with one of the <code>n</code> colors (labeled from <code>1</code> to <code>n</code>), some houses that have been painted last summer should not be painted again.</p>

<p>A neighborhood is a maximal group of continuous houses that are painted with the same color.</p>

<ul>
	<li>For example: <code>houses = [1,2,2,3,3,2,1,1]</code> contains <code>5</code> neighborhoods <code>[{1}, {2,2}, {3,3}, {2}, {1,1}]</code>.</li>
</ul>

<p>Given an array <code>houses</code>, an <code>m x n</code> matrix <code>cost</code> and an integer <code>target</code> where:</p>

<ul>
	<li><code>houses[i]</code>: is the color of the house <code>i</code>, and <code>0</code> if the house is not painted yet.</li>
	<li><code>cost[i][j]</code>: is the cost of paint the house <code>i</code> with the color <code>j + 1</code>.</li>
</ul>

<p>Return <em>the minimum cost of painting all the remaining houses in such a way that there are exactly</em> <code>target</code> <em>neighborhoods</em>. If it is not possible, return <code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
<strong>Output:</strong> 9
<strong>Explanation:</strong> Paint houses of this way [1,2,2,1,1]
This array contains target = 3 neighborhoods, [{1}, {2,2}, {1,1}].
Cost of paint all houses (1 + 1 + 1 + 1 + 5) = 9.
</pre>

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

<pre>
<strong>Input:</strong> houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
<strong>Output:</strong> 11
<strong>Explanation:</strong> Some houses are already painted, Paint the houses of this way [2,2,1,2,2]
This array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}]. 
Cost of paint the first and last house (10 + 1) = 11.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3
<strong>Output:</strong> -1
<strong>Explanation:</strong> Houses are already painted with a total of 4 neighborhoods [{3},{1},{2},{3}] different of target = 3.
</pre>

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

<ul>
	<li><code>m == houses.length == cost.length</code></li>
	<li><code>n == cost[i].length</code></li>
	<li><code>1 &lt;= m &lt;= 100</code></li>
	<li><code>1 &lt;= n &lt;= 20</code></li>
	<li><code>1 &lt;= target &lt;= m</code></li>
	<li><code>0 &lt;= houses[i] &lt;= n</code></li>
	<li><code>1 &lt;= cost[i][j] &lt;= 10<sup>4</sup></code></li>
</ul>

const auto __ = []() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  return true;
}();
class Solution {
public:


    int minCost(vector<int>& houses, vector<vector<int>>& cost, int m, int n, int target) {
    const auto& house_cnt = m;
    const auto& color_cnt = n;
    vector<vector<int32_t>> v(target + 1, vector<int32_t>(color_cnt + 3, kOutOfBounds));
    fill(v[1].begin(), v[1].end(), 0);
    for (int32_t i = 0; i < house_cnt; ++i) {
      // Check groups from painting all different to painting all with the same color
      for (int32_t grp = min(target, i + 1); grp > 0; --grp) {
        v[grp][color_cnt + 1] = v[grp][color_cnt + 2] = kOutOfBounds;
        for (int32_t color = 0; color < color_cnt; ++color) {
          if (houses[i] != kUnpainted && houses[i] - 1 != color) {
            v[grp][color] = kOutOfBounds;
          } else {
            const int32_t costs =
                (houses[i] != kUnpainted ? 0 : cost[i][color]) +
                min(v[grp][color], v[grp - 1][color_cnt + 1 + (v[grp - 1][color_cnt] == color ? 1 : 0)]);
            if (costs <= v[grp][color_cnt + 1]) {
              v[grp][color_cnt + 2] = v[grp][color_cnt + 1];
              v[grp][color_cnt + 1] = costs;
              v[grp][color_cnt] = color;
            } else if (costs < v[grp][color_cnt + 2]) {
              v[grp][color_cnt + 2] = costs;
            }
            v[grp][color] = costs;
          }
        }
      }
    }
    return v[target][color_cnt + 1] < kOutOfBounds ? v[target][color_cnt + 1] : -1;
    }

    private:
      static constexpr int32_t kUnpainted = 0;
  static constexpr int32_t kOutOfBounds = 1e7;    
};
12965