leetcode - 823. Binary Trees With Factors

KpLn_HJL2023-10-27 7:03

Description

Given an array of unique integers, arr, where each integer arr[i] 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 its children.

Return the number of binary trees we can make. The answer may be too large so return the answer modulo 10^9 + 7.

Example 1:

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

Example 2:

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Input: arr = [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].

Constraints:

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1 <= arr.length <= 1000
2 <= arr[i] <= 10^9
All the values of arr are unique.

Solution

Use dp, dp[i] = dp[j] * dp[k] + 1, for all j, k if j * k == i

Time complexity: o ( n 2 ) o(n^2) o(n2)

Space complexity: o ( n ) o(n) o(n)

Code

python3 复制代码

class Solution:
    def numFactoredBinaryTrees(self, arr: List[int]) -> int:
        dp = {each_num: 1 for each_num in arr}
        arr.sort()
        modulo = 1000000007
        for i in range(len(arr)):
            for j in range(i):
                if arr[i] % arr[j] == 0 and arr[i] // arr[j] in dp:
                    dp[arr[i]] += dp[arr[j]] * dp[arr[i] // arr[j]]
                    dp[arr[i]] %= modulo
        return sum(dp.values()) % modulo