<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>

from collections import defaultdict
class Solution:
    def getMaxGridHappiness(self, m: int, n: int, introvertsCount: int, extrovertsCount: int) -> int:
        # Dynamic programming
        # Time complexity: O((m*n)^3*3^n), space complexity:
        # O((m*n)^2*3^n).
        s0 = tuple(0 for _ in range(n))
        dp = dict({(s0,0,0): 0})
        for i in range(m):
            for j in range(n):
            # k[0] is a length-n tuple in which k[0][c] denotes
            # the status of cell (i,c) for c < j or cell (i - 1,
            # c) for c >= j. k[0][c] = 0 if no one lives in the
            # cell, k[0][c] = 1 if an introvert lives in the cell,
            # and k[0][c] = 2 if an extrovert lives there.
                dp1 = defaultdict(int)
                for k,v in dp.items():
                    # Do not have anyone live in cell (i,j).
                    s0 = list(k[0])
                    s0[j] = 0
                    k0 = (tuple(s0), k[1], k[2])
                    dp1[k0] = max(dp1[k0], v)
                    
                    if k[1] < introvertsCount:
                    # Assign an introvert to cell (i,j).
                        v1 = v + 120
                        for c in [j - 1, j]:
                            if c < 0: continue
                            # Each introvert loses 30 happiness.
                            # Each extrovert gains 20 happiness.
                            if k[0][c] == 1: v1 -= 60
                            elif k[0][c] == 2: v1 -= 10
                        s1 = list(k[0])
                        s1[j] = 1
                        k1 = (tuple(s1), k[1] + 1, k[2])
                        dp1[k1] = max(dp1[k1], v1)
                    
                    if k[2] < extrovertsCount:
                    # Assign an extrovert to cell (i,j).
                        v2 = v + 40
                        for c in [j - 1, j]:
                            if c < 0: continue
                            if k[0][c] == 1: v2 -= 10
                            elif k[0][c] == 2: v2 += 40
                        s2 = list(k[0])
                        s2[j] = 2
                        k2 = (tuple(s2), k[1], k[2] + 1)
                        dp1[k2] = max(dp1[k2], v2)
                dp = dp1
        return max(dp.values())
19800