Maximize Grid Happiness
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1600-1699/1659.Maximize%20Grid%20Happiness/images/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>
Bit Manipulation; Memoization; Dynamic Programming; Bitmask
class Solution {
    private int m;
    private int mx;
    private int[] f;
    private int[][] g;
    private int[][] bits;
    private int[] ix;
    private int[] ex;
    private Integer[][][][] memo;
    private final int[][] h = {{0, 0, 0}, {0, -60, -10}, {0, -10, 40}};

    public int getMaxGridHappiness(int m, int n, int introvertsCount, int extrovertsCount) {
        this.m = m;
        mx = (int) Math.pow(3, n);
        f = new int[mx];
        g = new int[mx][mx];
        bits = new int[mx][n];
        ix = new int[mx];
        ex = new int[mx];
        memo = new Integer[m][mx][introvertsCount + 1][extrovertsCount + 1];
        for (int i = 0; i < mx; ++i) {
            int mask = i;
            for (int j = 0; j < n; ++j) {
                int x = mask % 3;
                mask /= 3;
                bits[i][j] = x;
                if (x == 1) {
                    ix[i]++;
                    f[i] += 120;
                } else if (x == 2) {
                    ex[i]++;
                    f[i] += 40;
                }
                if (j > 0) {
                    f[i] += h[x][bits[i][j - 1]];
                }
            }
        }
        for (int i = 0; i < mx; ++i) {
            for (int j = 0; j < mx; ++j) {
                for (int k = 0; k < n; ++k) {
                    g[i][j] += h[bits[i][k]][bits[j][k]];
                }
            }
        }
        return dfs(0, 0, introvertsCount, extrovertsCount);
    }

    private int dfs(int i, int pre, int ic, int ec) {
        if (i == m || (ic == 0 && ec == 0)) {
            return 0;
        }
        if (memo[i][pre][ic][ec] != null) {
            return memo[i][pre][ic][ec];
        }
        int ans = 0;
        for (int cur = 0; cur < mx; ++cur) {
            if (ix[cur] <= ic && ex[cur] <= ec) {
                ans = Math.max(
                    ans, f[cur] + g[pre][cur] + dfs(i + 1, cur, ic - ix[cur], ec - ex[cur]));
            }
        }
        return memo[i][pre][ic][ec] = ans;
    }
}