<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[103][103][103];

    int help(vector<int> &houses, vector<vector<int>> &cost, int m, int n, int target, int i, int prev) {

        if (i == m) {
            if (!target) return 0;
            else return 1e9;
        }

        if (target < 0) return  1e9;

        if (dp[i][prev][target] != -1) return dp[i][prev][target];


        int mn = 1e9;

        // case 1:
        // no need to colorize
        if (houses[i]) {
            if (houses[i] != prev) // nbrs increase, hence target decreases
                return dp[i][prev][target] = help(houses, cost, m, n, target - 1, i + 1, houses[i]);

            else return dp[i][prev][target] = help(houses, cost, m, n, target, i + 1, houses[i]);
        }

        else {

            // case 2: color them

            // I need to paint with colors [1,n] such that the current color isn't the same as the prev
            // if they are the same, nbrs remain the same (no change in target)
            // else, nbrs increase, so target-1

            // colors: 1 to n
            // cost[i][j]: cost to paint ith building with color j+1


            for (int c = 1; c <= n; c++) {
                int cur = cost[i][c - 1];

                if (prev == c) cur += help(houses, cost, m , n, target, i + 1, c);
                else {

                    // explore the rest (next rows)
                    cur += help(houses, cost, m , n, target - 1, i + 1, c);
                }
                mn = min(mn, cur);
            }

            return dp[i][prev][target] = mn;
        }


    }

    int minCost(vector<int>& houses, vector<vector<int>>& cost, int m, int n, int target) {

        memset(dp, -1, sizeof(dp));
        int mm =  help(houses, cost, m, n, target, 0, 0);
        return (mm >= 1e9) ? -1 : mm;
    }
};
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