<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>

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class Solution:
    def getMaxGridHappiness(self, m: int, n: int, introvertsCount: int, extrovertsCount: int) -> int:
        
        # return max happiness starting at grid[i][j] with previous state and intro extro left
        @cache
        def dfs(i, j, state, intro, extro):
            if i == m:
                return 0
            if j == n:
                return dfs(i + 1, 0, state, intro, extro)
            
            res = dfs(i, j + 1, (state * 3) % (3 ** n), intro, extro)
            
            tri_up = state // (3 ** (n - 1))
            tri_left = state % 3
            up = tri_up if i != 0 else 0
            left = tri_left if j != 0 else 0
            if intro:
                diff = 120
                if up == 1:
                    diff -= 60
                elif up == 2:
                    diff -= 10
                
                if left == 1:
                    diff -= 60
                elif left == 2:
                    diff -= 10
                    
                res = max(res, dfs(i, j + 1, (state * 3 + 1) % (3 ** n), intro - 1, extro) + diff)
                    
            if extro:
                diff = 40
                if up == 1:
                    diff -= 10
                elif up == 2:
                    diff += 40
                
                if left == 1:
                    diff -= 10
                elif left == 2:
                    diff += 40
                    
                res = max(res, dfs(i, j + 1, (state * 3 + 2) % (3 ** n), intro, extro - 1) + diff)
            return res
        
        return dfs(0, 0, 0, introvertsCount, extrovertsCount)
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