Binary Trees With Factors
<p>Given an array of unique integers, <code>arr</code>, where each integer <code>arr[i]</code> is strictly greater than <code>1</code>.</p>

<p>We make a binary tree using these integers, and each number may be used for any number of times. Each non-leaf node&#39;s value should be equal to the product of the values of its children.</p>

<p>Return <em>the number of binary trees we can make</em>. The answer may be too large so return the answer <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [2,4]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can make these trees: <code>[2], [4], [4, 2, 2]</code></pre>

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

<pre>
<strong>Input:</strong> arr = [2,4,5,10]
<strong>Output:</strong> 7
<strong>Explanation:</strong> We can make these trees: <code>[2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2]</code>.</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 1000</code></li>
	<li><code>2 &lt;= arr[i] &lt;= 10<sup>9</sup></code></li>
	<li>All the values of <code>arr</code> are <strong>unique</strong>.</li>
</ul>
Array; Hash Table; Dynamic Programming; Sorting
class Solution:
    def numFactoredBinaryTrees(self, arr: List[int]) -> int:
        mod = 10**9 + 7
        n = len(arr)
        arr.sort()
        idx = {v: i for i, v in enumerate(arr)}
        f = [1] * n
        for i, a in enumerate(arr):
            for j in range(i):
                b = arr[j]
                if a % b == 0 and (c := (a // b)) in idx:
                    f[i] = (f[i] + f[j] * f[idx[c]]) % mod
        return sum(f) % mod
