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Problem solution · Python

Sum of Perfect Square Ancestors

Sum of Perfect Square Ancestors: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
120 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Sum of Perfect Square Ancestors, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 120 lines of Python from the credited upstream file sum-of-perfect-square-ancestors.py.
  • The implementation visibly relies on sequence storage, hash lookup.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSum of Perfect Square Ancestors · PythonPython
Use this to learn the idea, then write your own version.
# Time:  precompute: O(r)#        runtime:    O(nlogx)# Space: O(r + n) import collections  def linear_sieve_of_eratosthenes(n):  # Time: O(n), Space: O(n)    primes = []    spf = [-1]*(n+1)  # the smallest prime factor    for i in xrange(2, n+1):        if spf[i] == -1:            spf[i] = i            primes.append(i)        for p in primes:            if i*p > n or p > spf[i]:                break            spf[i*p] = p    return spf  MAX_NUMS = 10**5SPF = linear_sieve_of_eratosthenes(MAX_NUMS) # number theory, iterative dfs, freq tableclass Solution(object):    def sumOfAncestors(self, n, edges, nums):        """        :type n: int        :type edges: List[List[int]]        :type nums: List[int]        :rtype: int        """        def prime_factors(x):            result = 1            while x != 1:                if result%SPF[x] == 0:                    result //= SPF[x]                else:                    result *= SPF[x]                x //= SPF[x]            return result         def iter_dfs():            result = 0            stk = [(1, (0, -1))]            while stk:                step, args = stk.pop()                if step == 1:                    u, p = args                    x = prime_factors(nums[u])                    result += cnt[x]                    cnt[x] += 1                    stk.append((3, (x,)))                    stk.append((2, (u, p, 0)))                elif step == 2:                    u, p, i = args                    if i == len(adj[u]):                        continue                    stk.append((2, (u, p, i+1)))                    v = adj[u][i]                    if v == p:                        continue                    stk.append((1, (v, u)))                elif step == 3:                    x = args[0]                    cnt[x] -= 1            return result         adj = [[] for _ in xrange(n)]        for u, v in edges:            adj[u].append(v)            adj[v].append(u)        cnt = collections.defaultdict(int)        return iter_dfs()  # Time:  precompute: O(r)#        runtime:    O(nlogx)# Space: O(r + n)import collections  # number theory, dfs, freq tableclass Solution2(object):    def sumOfAncestors(self, n, edges, nums):        """        :type n: int        :type edges: List[List[int]]        :type nums: List[int]        :rtype: int        """        def prime_factors(x):            result = 1            while x != 1:                if result%SPF[x] == 0:                    result //= SPF[x]                else:                    result *= SPF[x]                x //= SPF[x]            return result         def dfs(u, p):            x = prime_factors(nums[u])            result = cnt[x]            cnt[x] += 1            for v in adj[u]:                if v == p:                    continue                result += dfs(v, u)            cnt[x] -= 1            return result         adj = [[] for _ in xrange(n)]        for u, v in edges:            adj[u].append(v)            adj[v].append(u)        cnt = collections.defaultdict(int)        return dfs(0, -1)

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