How do you solve Binary Trees With Factors on LeetCode?

Binary Trees With Factors (LeetCode #823) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Each non-leaf node's value must be the product of its children.

What is the brute force solution for Binary Trees With Factors?

The brute force approach for Binary Trees With Factors is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible binary trees using the integers in the array. For each integer, we check if it can be formed by multiplying two smaller integers from the array, recursively building trees until we exhaust all possibilities. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Binary Trees With Factors?

Binary Trees With Factors uses the following concepts: Array, Hash Table, Dynamic Programming, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Binary Trees With Factors?

Think about how to break down the problem into smaller subproblems. Consider edge cases, such as the smallest input size. Practice explaining your thought process clearly while coding.

What are common mistakes when solving Binary Trees With Factors?

Forgetting to check if the right factor exists in the map. Not handling large numbers correctly due to overflow.

#823

Binary Trees With Factors

Medium

ArrayHash TableDynamic ProgrammingSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

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Intuition

Time O(n²)Space O(n)

The optimal solution leverages dynamic programming and a hash map to efficiently count the number of trees. By sorting the array and using a hash map to store the number of trees for each integer, we can quickly find factors and update counts without redundant calculations.

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Algorithm

4 steps

1Step 1: Sort the array to ensure we can find factors in order.
2Step 2: Create a hash map to store the number of trees for each integer.
3Step 3: For each integer, check all previous integers to see if they can form the current integer by multiplication.
4Step 4: Update the count for the current integer based on the counts of its factors.

solution.py12 lines

1def numFactoredBinaryTrees(arr):
2    MOD = 10**9 + 7
3    arr.sort()
4    count = {x: 1 for x in arr}
5    for i in range(len(arr)):
6        for j in range(i):
7            if arr[i] % arr[j] == 0:
8                right = arr[i] // arr[j]
9                if right in count:
10                    count[arr[i]] += count[arr[j]] * count[right]
11                    count[arr[i]] %= MOD
12    return sum(count.values()) % MOD

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we use a hash map for O(1) access to counts, improving efficiency in updates.

1Each non-leaf node's value must be the product of its children.
2Using a hash map allows for efficient counting of possible trees.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.