<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>

#define vvi vector<vector<int>>
#define vvvi vector<vector<vector<int>>>

#pragma GCC optimize("O3, unroll-loops")
#pragma GCC target("avx2, bmi, bmi2, lzcnt, popcnt")

static auto _ = []() { 
    cin.tie(nullptr) -> ios_base::sync_with_stdio(false); 
    return nullptr; 
} ();

class Solution {
 public:
  int minCost(vector<int>& houses, vvi& cost, int m, int n, int T) {
    vvvi mem(T + 1, vvi(m, vector<int>(n + 1)));
    const int c = mincost(houses, cost, m, n, T, 0, 0, mem);
    return c == INF ? -1 : c;
  }

 private:
  static constexpr int INF = 1000001;
  // min cost to paint houses[i..m) into k neighborhoods, s.t. 
  // houses[i - 1] colors = prev_clr
  int mincost(const vector<int>& houses, const vvi& cost,
              const int& m, const int& n, int k, int i, int prev_clr,
              vvvi& mem) {
    if (i == m || k < 0) return k == 0 ? 0 : INF;
    if (mem[k][i][prev_clr] > 0) return mem[k][i][prev_clr];
    if (houses[i] > 0) return mincost(houses, cost, m, n, k - (prev_clr != houses[i]), i + 1, houses[i], mem);
    int res = INF;
    // paint houses[i] with color in [n]
    for (int clr = 1; clr <= n; ++clr) {
        res = min(res, cost[i][clr - 1] + mincost(houses, cost, m, n, k - (prev_clr != clr), i + 1, clr, mem));
    }
    return mem[k][i][prev_clr] = res;
  }
};
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