How do you solve Most Profitable Path in a Tree on LeetCode?

Most Profitable Path in a Tree (LeetCode #2467) 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: Alice's path is flexible, while Bob's path is fixed.

What is the brute force solution for Most Profitable Path in a Tree?

The brute force approach for Most Profitable Path in a Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we explore all possible paths Alice can take to each leaf node and calculate the net income for each path. We then compare these values to find the maximum net income Alice can achieve. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Most Profitable Path in a Tree?

Clarify the problem statement and constraints. Discuss edge cases, such as when Alice or Bob are at the same node. Use clear variable names and maintain clean code.

What are common mistakes when solving Most Profitable Path in a Tree?

Not considering simultaneous arrivals at nodes. Failing to account for leaf nodes correctly.

#2467

Most Profitable Path in a Tree

Medium

ArrayTreeDepth-First SearchBreadth-First SearchGraph TheoryDepth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

In the optimal approach, we use a single DFS traversal to calculate the maximum net income for Alice while considering Bob's path. This allows us to efficiently compute the result in linear time.

⚙️

Algorithm

3 steps

1Step 1: Build the tree structure from the edges.
2Step 2: Use DFS to calculate the maximum income for Alice while keeping track of Bob's position.
3Step 3: Adjust the income based on whether Alice and Bob reach the same node.

solution.py20 lines

1def maxProfit(edges, amount, bob):
2    from collections import defaultdict
3    tree = defaultdict(list)
4    for a, b in edges:
5        tree[a].append(b)
6        tree[b].append(a)
7
8    def dfs(node, parent, bob_pos):
9        if node == bob_pos:
10            return amount[node] // 2
11        if len(tree[node]) == 1 and node != 0:
12            return amount[node]
13        max_income = float('-inf')
14        for neighbor in tree[node]:
15            if neighbor != parent:
16                income = dfs(neighbor, node, bob_pos)
17                max_income = max(max_income, income)
18        return max_income + (amount[node] // 2)
19
20    return dfs(0, -1, bob)

ℹ

Complexity note: The time complexity is O(n) because we perform a single DFS traversal of the tree, visiting each node once. The space complexity is O(n) due to the recursion stack and the tree structure.

1Alice's path is flexible, while Bob's path is fixed.
2Simultaneous arrival at nodes affects income calculations.

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