Count Number of Ways to Place Houses
<p>There is a street with <code>n * 2</code> <strong>plots</strong>, where there are <code>n</code> plots on each side of the street. The plots on each side are numbered from <code>1</code> to <code>n</code>. On each plot, a house can be placed.</p>

<p>Return <em>the number of ways houses can be placed such that no two houses are adjacent to each other on the same side of the street</em>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Note that if a house is placed on the <code>i<sup>th</sup></code> plot on one side of the street, a house can also be placed on the <code>i<sup>th</sup></code> plot on the other side of the street.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 4
<strong>Explanation:</strong> 
Possible arrangements:
1. All plots are empty.
2. A house is placed on one side of the street.
3. A house is placed on the other side of the street.
4. Two houses are placed, one on each side of the street.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2300-2399/2320.Count%20Number%20of%20Ways%20to%20Place%20Houses/images/arrangements.png" style="width: 500px; height: 500px;" />
<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 9
<strong>Explanation:</strong> The 9 possible arrangements are shown in the diagram above.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
</ul>
Dynamic Programming
class Solution {
    public int countHousePlacements(int n) {
        final int mod = (int) 1e9 + 7;
        int[] f = new int[n];
        int[] g = new int[n];
        f[0] = 1;
        g[0] = 1;
        for (int i = 1; i < n; ++i) {
            f[i] = g[i - 1];
            g[i] = (f[i - 1] + g[i - 1]) % mod;
        }
        long v = (f[n - 1] + g[n - 1]) % mod;
        return (int) (v * v % mod);
    }
}