How do you solve Sum of Perfect Square Ancestors on LeetCode?

Sum of Perfect Square Ancestors (LeetCode #3715) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Perfect squares relate to prime factorization and square-free kernels.

What is the time complexity of Sum of Perfect Square Ancestors?

The optimal solution for Sum of Perfect Square Ancestors runs in O(n) time and uses O(n) space. Each node is processed once, and we use a map to track kernels, leading to linear complexity.

What is the brute force solution for Sum of Perfect Square Ancestors?

The brute force approach for Sum of Perfect Square Ancestors is Brute Force, with O(n²) time complexity and O(1) space complexity. For each node, check all its ancestors and compute the product with their values. Count how many products are perfect squares. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Sum of Perfect Square Ancestors?

Explain your thought process clearly. Use diagrams to illustrate tree structures. Practice prime factorization problems.

What are common mistakes when solving Sum of Perfect Square Ancestors?

Not considering the square-free representation correctly. Failing to manage state during DFS traversal.

#3715

Sum of Perfect Square Ancestors

Hard

ArrayHash TableMathTreeDepth-First SearchCountingNumber TheoryHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Use prime factorization to determine the square-free kernel of each node's value. Maintain a count of these kernels while traversing the tree.

⚙️

Algorithm

3 steps

1Step 1: Precompute the square-free representation of each node's value using prime factorization.
2Step 2: Perform a DFS traversal, maintaining a count of the square-free kernels encountered.
3Step 3: For each node, check if its kernel matches any ancestor's kernel and update the count accordingly.

solution.py30 lines

1from collections import defaultdict
2import math
3
4def prime_factors(n):
5    factors = defaultdict(int)
6    for i in range(2, int(math.sqrt(n)) + 1):
7        while n % i == 0:
8            factors[i] += 1
9            n //= i
10    if n > 1:
11        factors[n] += 1
12    return frozenset(p for p, exp in factors.items() if exp % 2 == 1)
13
14def sum_of_perfect_square_ancestors(n, edges, nums):
15    graph = [[] for _ in range(n)]
16    for u, v in edges:
17        graph[u].append(v)
18        graph[v].append(u)
19    total = 0
20    def dfs(node, parent, kernels):
21        nonlocal total
22        current_kernel = prime_factors(nums[node])
23        total += kernels[current_kernel]
24        kernels[current_kernel] += 1
25        for neighbor in graph[node]:
26            if neighbor != parent:
27                dfs(neighbor, node, kernels)
28        kernels[current_kernel] -= 1
29    dfs(0, -1, defaultdict(int))
30    return total

ℹ

Complexity note: Each node is processed once, and we use a map to track kernels, leading to linear complexity.

1Perfect squares relate to prime factorization and square-free kernels.
2Efficient ancestor tracking can reduce redundant calculations.

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