<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 dp[101][101][101];
    int sol(int i,int j,int n,vector<int>& houses,vector<vector<int>>& cost, int target){
        if(i>=houses.size()){
            if(target==0){
                return 0;
            }
            return 1e9;
        }
        if(target<0){
            return 1e9;
        }
        if(dp[i][j][target]!=-1){
            return dp[i][j][target];
        }
        int ans=1e9;
        if(houses[i]==0){
            //backtrack
            for(int color=1;color<=n;color++){
                int c=0;
                houses[i]=color;
                if(color!=j){
                    c=sol(i+1,color,n,houses,cost,target-1);
                }
                else{
                    c=sol(i+1,color,n,houses,cost,target);
                }
                houses[i]=0;
                ans=min(ans,c + cost[i][color-1]);
            }
        }
        else{
            int c=0;
            if(houses[i]!=j){
                c=sol(i+1,houses[i],n,houses,cost,target-1);
            }
            else{
                c=sol(i+1,j,n,houses,cost,target);
            }
            ans=min(ans,c);
        }

        return dp[i][j][target]=ans;

    }
    int minCost(vector<int>& houses, vector<vector<int>>& cost, int m, int n, int target) {
        //base case
        int c=0;
        int prev=-1;
        int fl=0;
        for(int i=0;i<m;i++){
            if(houses[i]==0){
                fl=1;
            }
            if(prev==-1){
                c++;
                prev=houses[i];
            }
            else if(prev==houses[i]){

            }
            else{
                c++;
                prev=houses[i];
            }
        }
        if(!fl && c!=target){
            return -1;
        }
        memset(dp,-1,sizeof(dp));
        return sol(0,0,n,houses,cost,target)==1e9?-1:sol(0,0,n,houses,cost,target);        
        // return 0;
    }
};
15759