Introduction

In this blog post, we will explore how to solve the LeetCode problem "823" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 109 + 7. Example 1: Input: arr = [2,4] Output: 3 Explanation: We can make these trees: [2], [4], [4, 2, 2] Example 2: 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: 1 <= arr.length <= 1000 2 <= arr[i] <= 109 All the values of arr are unique.

Explanation

Here's a breakdown of the approach and the Python code:

• Dynamic Programming: We use dynamic programming to store the number of trees that can be formed with each number as the root.
• Bottom-Up Approach: We iterate through the sorted array. For each number, we check if it can be formed by multiplying two other numbers in the array.

• Modulo Arithmetic: We apply the modulo operator to prevent integer overflow.

• Time Complexity: O(n²), where n is the length of the array. Space Complexity: O(n).

Code

    def numFactoredBinaryTrees(arr):
    """
    Calculates the number of factored binary trees that can be formed from the given array.

    Args:
        arr: A list of unique integers, each greater than 1.

    Returns:
        The number of factored binary trees modulo 10^9 + 7.
    """
    MOD = 10**9 + 7
    n = len(arr)
    arr.sort()
    dp = {}  # Store the number of trees with each number as the root
    for x in arr:
        dp[x] = 1  # Each number can form a tree by itself

    for i in range(n):
        for j in range(i):
            if arr[i] % arr[j] == 0:
                other = arr[i] // arr[j]
                if other in dp:
                    dp[arr[i]] = (dp[arr[i]] + dp[arr[j]] * dp[other]) % MOD

    total_trees = sum(dp.values()) % MOD
    return total_trees