823

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

Solution 1. DP

// OJ: https://leetcode.com/problems/binary-trees-with-factors/
// Author: github.com/lzl124631x
// 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;
    }
};