823. Binary Trees With Factors

Problem Description

In this problem, you are provided with an array of unique integers called arr, where each integer arr[i] is strictly greater than 1. You are tasked with creating binary trees using these integers, where integers can be used multiple times across different trees. The rule for constructing the trees is that each non-leaf node (i.e., internal node) must have a value equal to the product of the values of its left and right child nodes.

What’s interesting is that you're not being asked to construct the trees per se, but rather compute the count of different binary trees that can be formed following this product rule. Due to the potentially large number of trees, the final count should be returned modulo 10^9 + 7, which is a practice used to keep the answer within the bounds of typical integer storage limits in computational systems.

As an example, consider arr = [2, 4]. We can make trees with root values of 2 and 4 directly, each as a single node. Additionally, we can form a tree with 4 as the root and two children, both with the value 2, since 2 * 2 equals 4. Thus, several trees can be formed with different structures based on the given integers.

Intuition

The solution leverages dynamic programming to count the number of possible binary trees. The primary intuition is that for any number a within the array, if it can be factored into two numbers b and c such that a = b * c and both b and c exist within the array, it can serve as a root of a binary tree whose left child is b and right child is c.

Here is the step-by-step thinking process to arrive at this solution:

1. Sorting the Array: The algorithm begins by sorting the array, which allows us to consider pairs (b, c) that can be factors of any a in an orderly manner. Since b * c = a, once a is fixed, if b is also fixed, c is determined. Sorting ensures we don't miss any factor pairs due to their relative ordering.

2. Creating an Index Map: An index map (idx) is created to easily find the index of any number in arr after it's been sorted. This index is important because we will use it to look up the count of potential subtrees from the dynamic programming array f.

3. Dynamic Programming (DP) Array: A dynamic programming array f is initialized, where f[i] will eventually hold the count of binary trees that can be made with arr[i] as the root node.

4. Building the DP Array: The algorithm then iterates through each element a in the array. For each a, it checks every element b that comes before a in the sorted order. If a is perfectly divisible by b, and the quotient c = a / b also exists in the array, it means a can be the root of a binary tree with children b and c. The count of such trees is contributed by the product of counts of b and c's binary trees, which have been stored in f[j] and f[idx[c]] respectively.

5. Avoiding Large Numbers: As it computes the counts, the algorithm continually takes modulos of intermediate results by 10^9 + 7 to ensure the numbers do not exceed the storage limits of integer types.

6. Summarizing the Result: Finally, the total count of binary trees we can form is the sum of counts for all f[i], taken modulo 10^9 + 7.

This approach efficiently computes the total count of binary trees by considering all possible pairs of factors for each element in the array, multiplying the counts of trees that can be formed from those factor elements, and cumulatively adding them up while keeping the large number under control by modulus operation.

Learn more about Dynamic Programming and Sorting patterns.

Solution Approach

The following is a detailed walkthrough of the implementation of the solution which includes the algorithms, data structures, or patterns used:

1. Sort the Array:

   1arr.sort()

   The array arr is sorted to consider potential factor pairs in a systematic manner. Sorting is crucial because we want to ensure that we consider all potential child nodes for a given root value a in an ascending order.

2. Create an Index Map:

   1idx = {v: i for i, v in enumerate(arr)}

   An index map idx is created, mapping each value v in arr to its index i. This map is later used to quickly find the DP array index corresponding to a specific integer value from arr.

3. Initialize the Dynamic Programming Array:

   1f = [1] * n

   A dynamic programming array f is initialized with the length equal to the number of elements in arr, filled with 1's. This represents the fact that each individual number can form a binary tree with just one node.

4. Build the DP Array: The implementation uses a nested for-loop to build the DP array f based on the possible combinations of factors:

   1for i, a in enumerate(arr):
   2    for j in range(i):
   3        b = arr[j]
   4        if a % b == 0 and (c := (a // b)) in idx:
   5            f[i] = (f[i] + f[j] * f[idx[c]]) % mod

   For each number a in arr, we check each number b that comes before a in the array (due to the previous sorting). If a is divisible by b and the division result c is also in arr, then a can be constructed as a root node with b and c as children. The current count of trees with root a is updated by adding the product of counts of trees with roots b and c.

   Note the syntax (c := (a // b)) in idx, which is a walrus operator introduced in Python 3.8 that assigns the value to c and then immediately checks if c exists in the idx map.

5. Modulo Operation:

   1% mod

   Modulo operation is used to ensure the count values in the DP array f do not exceed the integer storage limits.

6. Sum and Modulo for the Final Result:

   1return sum(f) % mod

   To get the final result, the counts from the DP array f are summed up, and the sum is returned with one final modulo operation to stay within the storage limits. This is the total count of binary trees that can be created from the array arr.

The algorithm efficiently calculates the count by exploiting the mathematical properties of factors and products, using dynamic programming for optimization, and ensuring the numbers are managed within a reasonable range using the modulo operation.

Example Walkthrough

Let's take a small example and walk through the solution approach described above. Suppose we are given the following array of unique integers: arr = [2, 4, 8].

1. Sort the Array: We start by sorting the array, resulting in arr remaining the same since it is already sorted: [2, 4, 8]. Sorting is not impactful in this specific case but is essential in the algorithm to consider all pairs in order.

2. Create an Index Map: We create an index map for easy lookup:

   1idx = {2: 0, 4: 1, 8: 2}

   This allows us to find the index of any integer present in arr.

3. Initialize the Dynamic Programming Array: We initialize f to [1, 1, 1]. Each number can at least form a binary tree as a single node.

4. Build the DP Array:

   • For a = 2 (f[0]): There are no factors of 2 in the array before it, so f[0] remains 1.
   • For a = 4 (f[1]): We have one factor 2 which comes before 4. Since 4 % 2 == 0 and c = 4 // 2 = 2 is in arr, we can have a binary tree with root 4 and children 2 and 2. So, f[1] becomes 1 (initial) + 1 (count for 2) * 1 (count for 2) = 2.
   • For a = 8 (f[2]): First, we consider the factor 2, 8 % 2 == 0 and c = 8 // 2 = 4 is in arr, so we can form a tree with root 8 and children 2 and 4. The count f[2] becomes 1 (initial) + 1 (count for 2) * 2 (count for 4) = 3. Then, for the factor 4, 8 % 4 == 0 and c = 8 // 4 = 2 is also in arr, adding another tree with root 8 and children 4 and 2: f[2] becomes 3 (previous) + 2 (count for 4) * 1 (count for 2) = 5.

5. Modulo Operation: Since all the numbers obtained are smaller than 10^9 + 7, there is no need for a modulo operation in this example. However, in the algorithm, we apply % mod at each update for safety.

6. Sum and Modulo for the Final Result: We sum the counts in the dynamic programming array:

   1sum(f) % mod

   Which results in (1 + 2 + 5) % mod = 8. Hence, there are a total of 8 unique binary trees that can be formed using the integers from arr.

This demonstrates how the algorithm would count the unique binary trees that can be formed by considering all possible factor pairs for each number in the array, updating the count in the dynamic programming array, and continually using modulo operations to manage large numbers.

Solution Implementation

Time and Space Complexity

The provided code defines a function numFactoredBinaryTrees to calculate the number of binary trees where every node is a factor of its children.

Time Complexity:

The algorithm's time complexity is mainly determined by the double nested loop. Here are the steps, broken down:

1. Sorting: The arr.sort() operation has a time complexity of O(n log n), where n is the length of the input list arr.

2. Building a dictionary (idx): This operation is linear with respect to the number of elements in arr, leading to a time complexity of O(n).

3. Nested Loops:

   • The outer loop runs n times, once for each element in arr.
   • For each element in the outer loop, the inner loop runs at most i times, where i ranges from 0 to n-1.
   • As a result, in the worst case, the total iterations of the nested loop can be calculated by summing up from 1 to n-1, leading to a complexity of O(n^2/2) which simplifies to O(n^2).

Combining these, the dominant term is the nested loops with O(n^2), leading to an overall time complexity of O(n log n + n + n^2), which simplifies to O(n^2) since n^2 is the most significant term for large n.

Space Complexity:

The space complexity is determined by the additional space allocated for the computations, not including the input:

1. Dictionary (idx): Space complexity of O(n) for storing indices of elements in arr.

2. Array (f): Space complexity of O(n) for storing the number of possible binary trees for each element in arr.

Therefore, the total space complexity of the algorithm is O(n + n), which simplifies to O(n).

In conclusion, the algorithm has a time complexity of O(n^2) and a space complexity of O(n).

Learn more about how to find time and space complexity quickly using problem constraints.