<p>You are given four integers, <code>m</code>, <code>n</code>, <code>introvertsCount</code>, and <code>extrovertsCount</code>. You have an <code>m x n</code> grid, and there are two types of people: introverts and extroverts. There are <code>introvertsCount</code> introverts and <code>extrovertsCount</code> extroverts.</p>

<p>You should decide how many people you want to live in the grid and assign each of them one grid cell. Note that you <strong>do not</strong> have to have all the people living in the grid.</p>

<p>The <strong>happiness</strong> of each person is calculated as follows:</p>

<ul>
	<li>Introverts <strong>start</strong> with <code>120</code> happiness and <strong>lose</strong> <code>30</code> happiness for each neighbor (introvert or extrovert).</li>
	<li>Extroverts <strong>start</strong> with <code>40</code> happiness and <strong>gain</strong> <code>20</code> happiness for each neighbor (introvert or extrovert).</li>
</ul>

<p>Neighbors live in the directly adjacent cells north, east, south, and west of a person&#39;s cell.</p>

<p>The <strong>grid happiness</strong> is the <strong>sum</strong> of each person&#39;s happiness. Return<em> the <strong>maximum possible grid happiness</strong>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/05/grid_happiness.png" style="width: 261px; height: 121px;" />
<pre>
<strong>Input:</strong> m = 2, n = 3, introvertsCount = 1, extrovertsCount = 2
<strong>Output:</strong> 240
<strong>Explanation:</strong> Assume the grid is 1-indexed with coordinates (row, column).
We can put the introvert in cell (1,1) and put the extroverts in cells (1,3) and (2,3).
- Introvert at (1,1) happiness: 120 (starting happiness) - (0 * 30) (0 neighbors) = 120
- Extrovert at (1,3) happiness: 40 (starting happiness) + (1 * 20) (1 neighbor) = 60
- Extrovert at (2,3) happiness: 40 (starting happiness) + (1 * 20) (1 neighbor) = 60
The grid happiness is 120 + 60 + 60 = 240.
The above figure shows the grid in this example with each person&#39;s happiness. The introvert stays in the light green cell while the extroverts live on the light purple cells.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> m = 3, n = 1, introvertsCount = 2, extrovertsCount = 1
<strong>Output:</strong> 260
<strong>Explanation:</strong> Place the two introverts in (1,1) and (3,1) and the extrovert at (2,1).
- Introvert at (1,1) happiness: 120 (starting happiness) - (1 * 30) (1 neighbor) = 90
- Extrovert at (2,1) happiness: 40 (starting happiness) + (2 * 20) (2 neighbors) = 80
- Introvert at (3,1) happiness: 120 (starting happiness) - (1 * 30) (1 neighbor) = 90
The grid happiness is 90 + 80 + 90 = 260.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> m = 2, n = 2, introvertsCount = 4, extrovertsCount = 0
<strong>Output:</strong> 240
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m, n &lt;= 5</code></li>
	<li><code>0 &lt;= introvertsCount, extrovertsCount &lt;= min(m * n, 6)</code></li>
</ul>

import math

class Solution:
    def getMaxGridHappiness(self, m: int, n: int, introvertsCount: int, extrovertsCount: int) -> int:
        n3 = int(math.pow(3, n))
        lineStatusList = [[0]*n for _ in range(n3)]
        introvertsCountInner = [0]*n3
        extrovertsCountInner = [0]*n3
        scoreInner = [0]*n3
        scoreOuter = [[0]*n3 for _ in range(n3)]
        dp = [[[[-1]*7 for _ in range(7)] for _ in range(6)] for _ in range(n3)]

        for lineStatus in range(n3):
            tmp = lineStatus
            for i in range(n):
                lineStatusList[lineStatus][i] = tmp % 3
                tmp //= 3
            introvertsCountInner[lineStatus], extrovertsCountInner[lineStatus], scoreInner[lineStatus] = 0, 0, 0
            for i in range(n):
                if lineStatusList[lineStatus][i] != 0:
                    if lineStatusList[lineStatus][i] == 1:
                        introvertsCountInner[lineStatus] += 1
                        scoreInner[lineStatus] += 120
                    else:
                        extrovertsCountInner[lineStatus] += 1
                        scoreInner[lineStatus] += 40
                    if i-1 >= 0:
                        scoreInner[lineStatus] += self.closeScore(lineStatusList[lineStatus][i], lineStatusList[lineStatus][i-1])

        for lineStatus0 in range(n3):
            for lineStatus1 in range(n3):
                scoreOuter[lineStatus0][lineStatus1] = 0
                for i in range(n):
                    scoreOuter[lineStatus0][lineStatus1] += self.closeScore(lineStatusList[lineStatus0][i], lineStatusList[lineStatus1][i])

        return self.dfs(0, 0, introvertsCount, extrovertsCount, m, n3, dp, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter)

    def closeScore(self, x, y):
        if x == 0 or y == 0:
            return 0
        if x == 1 and y == 1:
            return -60
        if x == 2 and y == 2:
            return 40
        return -10

    def dfs(self, lineStatusLast, row, introvertsCount, extrovertsCount, m, n3, dp, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter):
        if row == m or introvertsCount+extrovertsCount == 0:
            return 0
        if dp[lineStatusLast][row][introvertsCount][extrovertsCount] != -1:
            return dp[lineStatusLast][row][introvertsCount][extrovertsCount]
        best = 0
        for lineStatus in range(n3):
            if introvertsCountInner[lineStatus] > introvertsCount or extrovertsCountInner[lineStatus] > extrovertsCount:
                continue
            score = scoreInner[lineStatus] + scoreOuter[lineStatus][lineStatusLast]
            best = max(best, score + self.dfs(lineStatus, row+1, introvertsCount-introvertsCountInner[lineStatus], extrovertsCount-extrovertsCountInner[lineStatus], m, n3, dp, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter))
        dp[lineStatusLast][row][introvertsCount][extrovertsCount] = best
        return best

    def max(self, a, b):
        return a if a > b else b
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