<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) {
        vector<vector<vector<long long>>>dynamic(
            m, vector<vector<long long>>(n + 1, vector<long long>(target + 1, -1)));
        auto dp = [&](auto && self, int currentindex, int lastcolor,
                      int target)->long long {
            if (currentindex == m && target == 1) {
                return 0;
            }
            if (target <= 0||currentindex==m) {
                return INT_MAX;
            }
            if (dynamic[currentindex][lastcolor][target] != -1) {
                return dynamic[currentindex][lastcolor][target];
            }
           long long count = INT_MAX;
            if (houses[currentindex] != 0) {
                if (houses[currentindex] == lastcolor) {
                    count = self(self, currentindex + 1, lastcolor, target);
                } else {
                    count = self(self, currentindex + 1, houses[currentindex],
                                 target - 1);
                }
                dynamic[currentindex][lastcolor][target]=count;
                return dynamic[currentindex][lastcolor][target];
            }

            for (int j = 0; j < n; j++) {

                if (lastcolor == j+1) {
                    count = min(count,(long long) cost[currentindex][j]+self(self, currentindex + 1, j+1, target));

                } else {
                    count =
                        min(count,(long long)cost[currentindex][j]+self(self, currentindex + 1, j+1, target - 1));
                }
            }
            dynamic[currentindex][lastcolor][target]=count;
            return dynamic[currentindex][lastcolor][target];
        };
        if(houses[0]!=0){
            long long ans=dp(dp,1,houses[0],target);
            if(ans>=INT_MAX){
                return -1;
            }
            else{
                return ans;
            }
        }
        else {
            long long mint =LLONG_MAX;
            for(int j=0;j<n;j++){
            mint=min(cost[0][j]+ dp(dp,1,j+1,target),mint);
            }
            if(mint>=INT_MAX){
                return -1;
            }
            else{
                return mint;
            }
           
        }
    
    }
};
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