What is the time complexity of Maximum Points After Collecting Coins From All Nodes?

The optimal solution for Maximum Points After Collecting Coins From All Nodes runs in O(n) time and uses O(n) space. The complexity is O(n) because we traverse each node once, and memoization allows us to avoid redundant calculations.

What is the brute force solution for Maximum Points After Collecting Coins From All Nodes?

The brute force approach for Maximum Points After Collecting Coins From All Nodes is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we explore all possible ways to collect coins from each node in the tree. This involves recursively calculating the points for each node based on both collection methods and their effects on the subtree. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Maximum Points After Collecting Coins From All Nodes?

Always clarify the problem constraints and edge cases before jumping into coding. Think about how you can optimize your solution from the start, especially with recursive problems. Practice explaining your thought process as you code, as this is often as important as the solution itself.

What are common mistakes when solving Maximum Points After Collecting Coins From All Nodes?

Failing to consider the impact of the halving operation on child nodes. Not using memoization effectively, leading to repeated calculations.

#2920

Maximum Points After Collecting Coins From All Nodes

Hard

ArrayDynamic ProgrammingBit ManipulationTreeDepth-First SearchMemoizationDynamic ProgrammingDepth-First SearchMemoization

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)

The optimal approach uses dynamic programming with memoization to efficiently calculate the maximum points from each node while avoiding redundant calculations. We keep track of how many times we've used the halving operation to ensure we maximize our points.

⚙️

Algorithm

3 steps

1Step 1: Build the tree using adjacency lists from the edges.
2Step 2: Use a depth-first search (DFS) to traverse the tree and calculate points for each node.
3Step 3: Maintain a memoization table to store results for each node and the number of times the halving operation has been used.

solution.py25 lines

1# Full working Python code
2class Solution:
3    def collectTheCoins(self, edges, coins, k):
4        from collections import defaultdict
5        tree = defaultdict(list)
6        for a, b in edges:
7            tree[a].append(b)
8            tree[b].append(a)
9
10        memo = {}  # Memoization dictionary
11
12        def dfs(node, parent, t):
13            if (node, t) in memo:
14                return memo[(node, t)]
15            total_points = 0
16            for child in tree[node]:
17                if child != parent:
18                    total_points += dfs(child, node, t)
19            collect_direct = coins[node] - k if coins[node] - k > 0 else -abs(coins[node] - k)
20            collect_half = coins[node] // 2
21            result = max(total_points + collect_direct, total_points + collect_half)
22            memo[(node, t)] = result
23            return result
24
25        return dfs(0, -1, 0)

ℹ

Complexity note: The complexity is O(n) because we traverse each node once, and memoization allows us to avoid redundant calculations.

1Understanding the effects of each collection method is crucial for maximizing points.
2Using memoization can significantly reduce the time complexity by avoiding redundant calculations.

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