<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:
    vector<vector<vector<int>>> dp;

int f(vector<int>& houses, vector<vector<int>>& cost, int target, int idx, int count, int prev){
    //base cases
    if(idx == houses.size()) return count==target ? 0 : 1e9;
    
    //memoization
    if(dp[idx][count][prev]!=-1) return dp[idx][count][prev];
    
    int minCost = 1e9;
    //If already painted
    if(houses[idx]){
        
        //if prev house is of different color, increase the count by 1
        if(houses[idx]!=prev) minCost = f(houses,cost,target,idx+1,count+1,houses[idx]);
        
        else minCost = f(houses,cost,target,idx+1,count,houses[idx]);
        
    }else{
        //we have to try all the colors
        for(int j=0;j<cost[0].size();j++){
            int tmp;  //store the current cost
            
            //if prev house is of different color, increase the count by 1
            if((j+1)!=prev) tmp = cost[idx][j] + f(houses,cost,target,idx+1,count+1,j+1);
            
            else tmp = cost[idx][j] + f(houses,cost,target,idx+1,count,j+1);
            
            minCost = min(minCost,tmp);
        }
    }
    //store before returning
    return dp[idx][count][prev] = minCost;
}

int minCost(vector<int>& houses, vector<vector<int>>& cost, int m, int n, int target) {
    dp.resize(m+1, vector<vector<int>>(m+1, vector<int>(n+1, -1)));
    int ans = f(houses,cost,target, 0, 0, 0);
    if (ans == 1e9) return -1;
    return ans;
}
};
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