<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 {
    int dp[101][101][21];
    vector<vector<int>>cost;
    int target;
    vector<int>houses;
    int m,n;
    int solve(int i,int t,int prev){
        // cout<<i<<' '<<t<<' '<<prev<<endl;
        if(i==m && target==t)return 0;
        if(t>target)return (1e6+1);
        if(i==m)return (int)(1e6+1);
        if(dp[i][t][prev+1]!=-1)return dp[i][t][prev+1];
        int ans=1e6+1;
        if(houses[i]!=0){
            if(houses[i]==prev+1)return dp[i][t][prev+1]=solve(i+1,t,prev);
            else return dp[i][t][prev+1]=solve(i+1,t+1,houses[i]-1);
        }
        for(int c=0;c<n;c++){
            
            if(c==prev){
                ans=min(ans,cost[i][c]+solve(i+1,t,c));
            }
            else ans=min(ans,cost[i][c]+solve(i+1,t+1,c));
        }
        return dp[i][t][prev+1]=ans;

    }
public:
    int minCost(vector<int>& houses, vector<vector<int>>& cost, int m, int n, int target) {
        this->cost=cost;
        this->m=m;
        this->n=n;
        this->target=target;
        this->houses=houses;
        memset(dp,-1,sizeof(dp));
        int ans=solve(0,0,-1);
        return ans==1e6+1?-1:ans;
    }
};
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