algoadvance

Clarifying Questions

1. Input Format:
   • Are all numbers in the input array unique?
   • What is the range of values for the numbers in the input array and the length of the array?
2. Output Format:
   • Should the result be returned as a single integer representing the number of possible binary trees?

Strategy

1. Understanding the Problem:
   • We need to find the number of binary trees we can form using values from the input array such that each node value is a product of any two node values from the tree.
   • For example, given an array [2, 4], we can form the trees 2 and 4(2,2).
2. Steps:
   • Sort the array to make the processing simpler.
   • Use dynamic programming to count the number of binary trees that can be formed with each element as the root.
   • Use a dictionary to store the number of ways to form a tree with a particular root.
   • Iterate through the sorted array, and for each element, iteratively check every pair (x, y) such that x * y = element.
   • Sum up all possible trees considering all valid pairs for each element.
3. Dynamic Programming Approach:
   • Use a dictionary dp where dp[x] represents the number of binary trees with root x.
   • Initialize each dp[x] to 1 to account for the single node tree.
   • Iterate through each element in the sorted array and check for pairs (a, b) such that a * b = element.
     • For each valid pair, multiply the number of trees that can be formed with roots a and b (dp[a] * dp[b]).
     • Keep a running total and add it to dp[element].
   • The final result is the sum of all values in the dp dictionary modulo 10^9 + 7.

Code

def numFactoredBinaryTrees(arr):
    MOD = 10**9 + 7
    arr.sort()
    dp = {}
    
    for num in arr:
        dp[num] = 1  # Each number can at least form a single node tree by itself.
    
    for i, num in enumerate(arr):
        for j in range(i):
            if num % arr[j] == 0:  # arr[j] is the left child
                right = num // arr[j]
                if right in dp:
                    dp[num] = (dp[num] + dp[arr[j]] * dp[right]) % MOD

    return sum(dp.values()) % MOD

# Example usage
arr = [2, 4, 5, 10]
print(numFactoredBinaryTrees(arr))  # Output: 7

Time Complexity

• Sorting: O(n log n), where n is the length of the array.
• Main loop: Each number is processed and for each number, we’re checking pairs up to n, resulting in a nested loop which is O(n^2).
• Overall Complexity: O(n^2) due to the nested loop after sorting.

Conclusion

This approach efficiently calculates the number of binary trees with factors using dynamic programming while adhering to the constraints and providing an optimal solution.

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