Formatted question description: https://leetcode.ca/all/823.html

# 823. Binary Trees With Factors (Medium)

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: , , [4, 2, 2]

Example 2:

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

Note:

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

## Solution 1. DP

// OJ: https://leetcode.com/problems/binary-trees-with-factors/

// Time: O(N^2)
// Space: O(N)
class Solution {
public:
int numFactoredBinaryTrees(vector<int>& A) {
int mod = 1e9 + 7, N = A.size();
sort(A.begin(), A.end());
vector<long long> dp(N, 1);
unordered_map<int, long long> m;
for (int i = 0; i < N; ++i) m[A[i]] = i;
for (int i = 0; i < N; ++i) {
for (int j = 0; j < i; ++j) {
if (A[i] % A[j] || m.find(A[i] / A[j]) == m.end()) continue;
dp[i] = (dp[i] + dp[j] * dp[m[A[i] / A[j]]]) % mod;
}
}
int ans = 0;
for (auto n : dp) ans = (ans + n) % mod;
return ans;
}
};


Java

class Solution {
public int numFactoredBinaryTrees(int[] A) {
final int MODULO = 1000000007;
Arrays.sort(A);
int length = A.length;
long[] counts = new long[length];
Arrays.fill(counts, 1);
Map<Integer, Integer> indexMap = new HashMap<Integer, Integer>();
for (int i = 0; i < length; i++)
indexMap.put(A[i], i);
for (int i = 1; i < length; i++) {
for (int j = 0; j < i; j++) {
if (A[i] % A[j] == 0) {
int leftChild = A[j];
int rightChild = A[i] / A[j];
if (indexMap.containsKey(rightChild)) {
int rightIndex = indexMap.get(rightChild);
counts[i] = (counts[i] + counts[j] * counts[rightIndex]) % MODULO;
}
}
}
}
long trees = 0;
for (int i = 0; i < length; i++)
trees = (trees + counts[i]) % MODULO;
return (int) trees;
}
}