823. Binary Trees With Factors

Given an array of unique integers, each integer is strictly greater than 1.

We make a binary tree using these integers and each number may be used for any number of times.

Each non-leaf node’s value should be equal to the product of the values of it’s children.

How many binary trees can we make? Return the answer modulo 10 ** 9 + 7.

Example 1:

Input: A = [2, 4]
Output: 3
Explanation: We can make these trees: [2], [4], [4, 2, 2]

Example 2:

Input: A = [2, 4, 5, 10]
Output: 7
Explanation: We can make these trees: [2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2].

Note:

1 <= A.length <= 1000.
2 <= A[i] <= 10 ^ 9.

class Solution {
    public int numFactoredBinaryTrees(int[] A) {
        int MOD = 1_000_000_007;
        int N = A.length;
        Arrays.sort(A);
        long[] dp = new long[N];
        Arrays.fill(dp, 1);

        Map index = new HashMap();
        for (int i = 0; i < N; ++i)
            index.put(A[i], i);

        for (int i = 0; i < N; ++i)
            for (int j = 0; j < i; ++j) {
                if (A[i] % A[j] == 0) { // A[j] is left child
                    int right = A[i] / A[j];
                    if (index.containsKey(right)) {
                        dp[i] = (dp[i] + dp[j] * dp[index.get(right)]) % MOD;
                    }
                }
            }

        long ans = 0;
        for (long x: dp) ans += x;
        return (int) (ans % MOD);
    }
}