<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>

class Solution {
public:
    int minCost(vector<int>& houses, vector<vector<int>>& cost, int m, int n, int target) {
        // dp[i][j][k] - i-th house painted j-th color with k neighbourhoods
        // if houses[i] == 0 || houses[i] == j then 
        //      dp[i][j][k] =  min(x=1..n, x != j min(dp[i - 1][x][k - 1]),
        //                         min[i - 1][j][k]) + cost[i][j], 
        vector<vector<vector<int>>> dp(m + 1, vector<vector<int>>(n + 1, vector<int>(target + 1, 1e9)));
        
        for(int j = 0; j <= n; ++j) {
           dp[0][j][0] = 0;
        }
        for(int h = 1; h <= target; ++h) {
            for(int i = 1; i <= m; ++i) {
                
                int min1 = n, min2 = n;
                if(i == 1 || houses[i - 2] == 0) {
                    // find two best costs of painting up to i-1 th house
                    priority_queue<pair<int, int>> pq;
                    for(int j = 0; j < n; ++j) {
                        if(pq.size() < 2) {
                            pq.push({dp[i - 1][j][h - 1], j});
                        } else if(pq.top().first > dp[i - 1][j][h - 1]) {
                            pq.pop();
                            pq.push({dp[i - 1][j][h - 1], j});
                        }
                    }
                    min1 = pq.top().second;
                    pq.pop();
                    if(!pq.empty())
                        min2 = pq.top().second;
                    else
                        min2 = n;
                        
                } else {
                    min1 = min2 = houses[i - 2] - 1;
                }
                // if already painted
                if(houses[i - 1] > 0) {
                    int color = houses[i - 1] - 1;
                    dp[i][color][h] = min({dp[i - 1][min1][h - (min1 != color)],
                                          dp[i - 1][min2][h - (min2 != color)],
                                          dp[i - 1][color][h]});
                } else {
                    for(int j = 0; j < n; ++j) {
                        dp[i][j][h] = min({dp[i - 1][min1][h - (j != min1)],
                                          dp[i - 1][min2][h - (j != min2)],
                                          dp[i - 1][j][h]}) + cost[i - 1][j];
                    }
                }
            }
        }
        int ans = 1e9;
        for(int j = 0; j < n; ++j) {
            ans = min(ans, dp[m][j][target]);
        }
        
        return ans == 1e9 ? -1 : ans;
    }
};
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