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.

Not Sure What to Study? Take the 2-min Quiz to Find Your Missing Piece:

Given a sorted array of integers and an integer called target, find the element that equals to the target and return its index. Select the correct code that fills the ___ in the given code snippet.

1def binary_search(arr, target):
2    left, right = 0, len(arr) - 1
3    while left ___ right:
4        mid = (left + right) // 2
5        if arr[mid] == target:
6            return mid
7        if arr[mid] < target:
8            ___ = mid + 1
9        else:
10            ___ = mid - 1
11    return -1
12
1public static int binarySearch(int[] arr, int target) {
2    int left = 0;
3    int right = arr.length - 1;
4
5    while (left ___ right) {
6        int mid = left + (right - left) / 2;
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16
1function binarySearch(arr, target) {
2    let left = 0;
3    let right = arr.length - 1;
4
5    while (left ___ right) {
6        let mid = left + Math.trunc((right - left) / 2);
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16

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.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Which of the following is the prefix sum of array [1, 2, 3, 4, 5]?

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.

Not Sure What to Study? Take the 2-min Quiz:

What are the most two important steps in writing a depth first search function? (Select 2)

Python Solution

1class Solution:
2    def num_factored_binary_trees(self, arr: List[int]) -> int:
3        MODULO = 10**9 + 7  # The modulo value to keep the numbers in a range
4        n = len(arr)  # Length of input array
5        arr.sort()  # Sorting the array
6      
7        # Creating a dictionary to map array values to their indices for O(1) access
8        index_by_value = {value: index for index, value in enumerate(arr)}
9      
10        # Initialize a dp (dynamic programming) array to store the number of ways
11        # to make each element the root of a binary tree. Start with 1 for each element
12        # as each can be a tree by itself.
13        dp = [1] * n
14      
15        # Loop through each element 'a' in arr
16        for i, a in enumerate(arr):
17            # Try to form trees by checking all the elements 'b' that are factors of 'a' (a % b == 0)
18            # Here we are only checking for factors up to the current element 'a'
19            for j in range(i):
20                b = arr[j]
21                if a % b == 0:  # Check if 'b' is a factor of 'a'
22                    c = a // b  # Calculate the other factor
23                  
24                    # If the other factor 'c' is in the index_by_value, it means that 'b' and 'c'
25                    # can be children of a binary tree with 'a' as the root
26                    if c in index_by_value:
27                        # Update the number of ways to create a binary tree with 'a' as the root
28                        # by adding the product of the number of ways to create 'b' and 'c'
29                        # Take the modulo to manage large numbers
30                        dp[i] = (dp[i] + dp[j] * dp[index_by_value[c]]) % MODULO
31      
32        # Finally, sum up all the possible ways to form binary trees for each element as the root
33        # and take the modulo
34        return sum(dp) % MODULO
35

Java Solution

1import java.util.*;
2
3public class Solution {
4    // Function to calculate the number of possible binary trees formed by the elements of the array,
5    // where each node is equal to the product of two other nodes in the tree (if they exist in the array).
6    public int numFactoredBinaryTrees(int[] arr) {
7        final int MODULO = 1_000_000_007; // The value to mod the result with to prevent overflow
8        Arrays.sort(arr); // Sort the array in ascending order
9        Map<Integer, Integer> indexMap = new HashMap<>(); // Map each value to its index in the array
10        int arrayLength = arr.length;
11        // Populate the indexMap with array elements
12        for (int i = 0; i < arrayLength; i++) {
13            indexMap.put(arr[i], i);
14        }
15        long[] treeCounts = new long[arrayLength]; // Initialize tree count array for each node
16        Arrays.fill(treeCounts, 1); // Set initial tree count to 1 for each node
17      
18        // Iterate over the sorted array to count the number of trees
19        for (int i = 0; i < arrayLength; i++) {
20            int valueA = arr[i]; // Current value for which we are counting trees
21            for (int j = 0; j < i; j++) {
22                int valueB = arr[j]; // Possible factor of valueA
23                // Check if valueB is a factor of valueA
24                if (valueA % valueB == 0) {
25                    int valueC = valueA / valueB; // The other factor
26                    // Check if the other factor exists in the array using indexMap
27                    Integer indexC = indexMap.get(valueC);
28                    if (indexC != null) {
29                        // Update tree count for valueA by adding the product of tree counts for valueB and valueC
30                        treeCounts[i] = (treeCounts[i] + treeCounts[j] * treeCounts[indexC]) % MODULO;
31                    }
32                }
33            }
34        }
35        // Sum up the tree counts for all values and return the result modulo MODULO
36        long sum = 0;
37        for (long count : treeCounts) {
38            sum = (sum + count) % MODULO;
39        }
40        return (int)sum; // Cast sum to int type as final result is within int range due to modulo operation
41    }
42}
43

C++ Solution

1#include <vector>
2#include <unordered_map>
3#include <algorithm>
4
5class Solution {
6public:
7    int numFactoredBinaryTrees(std::vector<int>& arr) {
8        const int MOD = 1e9 + 7; // Modulo to avoid integer overflow
9        std::sort(arr.begin(), arr.end()); // Sort the array for orderly processing
10        std::unordered_map<int, int> indexMap; // Map to store the index of each element
11        int n = arr.size(); // Number of elements in the array
12        // Populate the index map with the elements from the array
13        for (int i = 0; i < n; ++i) {
14            indexMap[arr[i]] = i;
15        }
16        std::vector<long> dp(n, 1); // Dynamic programming array to store number of trees
17      
18        // Iterate over each element to compute the number of trees with the element as root
19        for (int i = 0; i < n; ++i) {
20            int rootValue = arr[i];
21            // Iterate over potential left children
22            for (int j = 0; j < i; ++j) {
23                int leftChildValue = arr[j];
24                // Check if the current element can be factored by the left child
25                if (rootValue % leftChildValue == 0) {
26                    int rightValue = rootValue / leftChildValue;
27                    // Check if right child is present in the array
28                    if (indexMap.count(rightValue)) {
29                        int rightChildIndex = indexMap[rightValue];
30                        // Multiply the number of trees from left and right child and add to dp array for root
31                        dp[i] = (dp[i] + dp[j] * dp[rightChildIndex]) % MOD;
32                    }
33                }
34            }
35        }
36      
37        long answer = 0; // Initialize the answer variable to store the final result
38        // Sum up all possible trees for each element to get the final result
39        for (long value : dp) {
40            answer = (answer + value) % MOD;
41        }
42        return answer; // Return the final answer
43    }
44};
45

Typescript Solution

1// Function to calculate the number of possible binary trees formed by the elements of the array
2// where each node is equal to the product of two other nodes in the tree (if they exist in the array).
3function numFactoredBinaryTrees(arr: number[]): number {
4    const MODULO = 10 ** 9 + 7; // The value to mod the result with to prevent overflow
5    arr.sort((a, b) => a - b); // Sort the array in ascending order
6    const indexMap: Map<number, number> = new Map(); // Map each value to its index in the array
7    const arrayLength = arr.length;
8    // Populate the indexMap with array elements
9    for (let i = 0; i < arrayLength; ++i) {
10        indexMap.set(arr[i], i);
11    }
12    const treeCounts: number[] = new Array(arrayLength).fill(1); // Initialize tree count to 1 for each node
13  
14    // Iterate over the sorted array to count the number of trees
15    for (let i = 0; i < arrayLength; ++i) {
16        const valueA = arr[i]; // Current value for which we are counting trees
17        for (let j = 0; j < i; ++j) {
18            const valueB = arr[j]; // Possible factor of valueA
19            // Check if valueB is a factor of valueA
20            if (valueA % valueB === 0) {
21                const valueC = valueA / valueB; // The other factor
22                // Check if the other factor exists in the array using indexMap
23                if (indexMap.has(valueC)) {
24                    const indexC = indexMap.get(valueC)!; // Index of the other factor
25                    // Update tree count for valueA by adding the product of tree counts for valueB and valueC
26                    treeCounts[i] = (treeCounts[i] + treeCounts[j] * treeCounts[indexC]) % MODULO;
27                }
28            }
29        }
30    }
31    // Sum up the tree counts for all values and return the result modulo MODULO
32    return treeCounts.reduce((sum, count) => (sum + count) % MODULO, 0);
33}
34
Fast Track Your Learning with Our Quick Skills Quiz:

Which of the following is the prefix sum of array [1, 2, 3, 4, 5]?

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.


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