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Problem solution · Python

Find Number of Coins to Place in Tree Nodes

Find Number of Coins to Place in Tree Nodes: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
50 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 Find Number of Coins to Place in Tree Nodes, 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

  • 50 lines of Python from the credited upstream file 2973.py.
  • The implementation visibly relies on sequence storage.
  • 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 walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind Number of Coins to Place in Tree Nodes · PythonPython
Use this to learn the idea, then write your own version.
class ChildCost:  def __init__(self, cost: int):    self.numNodes = 1    self.maxPosCosts = [cost] if cost > 0 else []    self.minNegCosts = [cost] if cost < 0 else []   def update(self, childCost: 'ChildCost') -> None:    self.numNodes += childCost.numNodes    self.maxPosCosts.extend(childCost.maxPosCosts)    self.minNegCosts.extend(childCost.minNegCosts)    self.maxPosCosts.sort(reverse=True)    self.minNegCosts.sort()    self.maxPosCosts = self.maxPosCosts[:3]    self.minNegCosts = self.minNegCosts[:2]   def maxProduct(self) -> int:    if self.numNodes < 3:      return 1    if not self.maxPosCosts:      return 0    res = 0    if len(self.maxPosCosts) == 3:      res = self.maxPosCosts[0] * self.maxPosCosts[1] * self.maxPosCosts[2]    if len(self.minNegCosts) == 2:      res = max(res,                self.minNegCosts[0] * self.minNegCosts[1] * self.maxPosCosts[0])    return res  class Solution:  def placedCoins(self, edges: list[list[int]], cost: list[int]) -> list[int]:    n = len(cost)    ans = [0] * n    tree = [[] for _ in range(n)]     for u, v in edges:      tree[u].append(v)      tree[v].append(u)     def dfs(u: int, prev: int) -> None:      res = ChildCost(cost[u])      for v in tree[u]:        if v != prev:          res.update(dfs(v, u))      ans[u] = res.maxProduct()      return res     dfs(0, -1)    return ans

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