#3562

Hard

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Maximum Profit from Trading Stocks with Discounts

Solve the Maximum Profit from Trading Stocks with Discounts problem using binary-tree traversal and dynamic state tracking for optimal gains.

array dynamic programming tree depth first search

Problem Statement

You are given n employees in a company, each with a unique ID from 1 to n, where employee 1 is the CEO. Each employee i has a stock present value present[i] and a future projected value future[i]. The company's management hierarchy is provided as a list hierarchy of pairs [ui, vi], indicating ui is the direct supervisor of vi. You also have a total budget for investments.

The goal is to determine the maximum profit achievable by investing in employees under the hierarchical constraints and budget, factoring in discounts where applicable. You must calculate the best investment strategy that maximizes total profit while respecting the tree structure and budget limitations.

Examples

Example 1

Input: n = 2, present = [1,2], future = [4,3], hierarchy = [[1,2]], budget = 3

Output: 5

Example 2

Input: n = 2, present = [3,4], future = [5,8], hierarchy = [[1,2]], budget = 4

Output: 4

Example 3

Input: n = 3, present = [4,6,8], future = [7,9,11], hierarchy = [[1,2],[1,3]], budget = 10

Output: 10

Constraints

1 <= n <= 160
present.length, future.length == n
1 <= present[i], future[i] <= 50
hierarchy.length == n - 1
hierarchy[i] == [ui, vi]
1 <= ui, vi <= n
ui != vi
1 <= budget <= 160
There are no duplicate edges.
Employee 1 is the direct or indirect boss of every employee.
The input graph hierarchy is guaranteed to have no cycles.

Solution Approach

Tree Construction and DFS Traversal

Build the employee hierarchy as an adjacency list. Perform depth-first search (DFS) from the CEO to process each employee, ensuring all child subtrees are evaluated before computing profits for the current node.

Dynamic State Tracking

For each node u, maintain two states: max_profit[u] without using discount and max_profit1[u] with discount applied. Combine child states carefully during DFS to respect budget constraints and hierarchical order, ensuring optimal profit aggregation.

Budget Allocation Optimization

Iterate over possible investment allocations across children using a DP table that tracks maximum profit for each budget amount. This allows merging child node profits while considering the total budget and discount effect to determine the global maximum.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n * budget^2) due to DFS traversal and DP merging of children states for each budget allocation. Space complexity is O(n * budget) for storing DP states per node.

What Interviewers Usually Probe

They may ask you to track two DP states per node to handle discounts effectively.
Expect a prompt on merging child node profits under budget constraints.
Clarification on handling hierarchical dependencies versus independent investments may be requested.

Common Pitfalls or Variants

Common pitfalls

Ignoring the distinction between max_profit and max_profit1 leads to incorrect discount calculations.
Merging child DP states incorrectly can underestimate total achievable profit.
Not respecting the tree structure can cause invalid profit combinations violating hierarchy rules.

Follow-up variants

Allowing multiple discount types per employee requiring additional DP states.
Changing the hierarchy to a forest of trees instead of a single CEO-rooted tree.
Introducing a cost for applying discounts, modifying profit calculations per node.

FAQ

What is the main strategy for Maximum Profit from Trading Stocks with Discounts?

Use a DFS traversal of the employee hierarchy and maintain two DP states per node to track profits with and without discounts.

How do I handle the budget constraint efficiently?

Maintain a DP table for each node that tracks maximum achievable profit for all possible budget allocations, merging child states carefully.

Can the hierarchy have cycles or multiple roots?

No, the hierarchy is guaranteed to be a single rooted tree with no cycles, so DFS traversal is safe.

Why do I need two profit states per employee?

One state tracks profit without discount, and the other tracks profit with a discount applied, ensuring optimal decision-making.

Is dynamic programming necessary for this problem?

Yes, without DP, merging profits from subtrees under budget constraints would be inefficient and likely incorrect.

terminal

Solution

Solution 1: Tree Dynamic Programming

For each node $u$, we maintain a 2D array $f_u[j][pre]$, representing the maximum profit that can be obtained in the subtree rooted at $u$ with a budget not exceeding $j$ and whether $u$'s manager purchased stocks (where $pre=1$ means purchased, and $pre=0$ means not purchased). The answer is $f_1[\text{budget}][0]$.

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class Solution:
    def maxProfit(
        self,
        n: int,
        present: List[int],
        future: List[int],
        hierarchy: List[List[int]],
        budget: int,
    ) -> int:
        max = lambda a, b: a if a > b else b
        g = [[] for _ in range(n + 1)]
        for u, v in hierarchy:
            g[u].append(v)

        def dfs(u: int):
            nxt = [[0, 0] for _ in range(budget + 1)]
            for v in g[u]:
                fv = dfs(v)
                for j in range(budget, -1, -1):
                    for jv in range(j + 1):
                        for pre in (0, 1):
                            val = nxt[j - jv][pre] + fv[jv][pre]
                            if val > nxt[j][pre]:
                                nxt[j][pre] = val

            f = [[0, 0] for _ in range(budget + 1)]
            price = future[u - 1]

            for j in range(budget + 1):
                for pre in (0, 1):
                    cost = present[u - 1] // (pre + 1)
                    if j >= cost:
                        f[j][pre] = max(nxt[j][0], nxt[j - cost][1] + (price - cost))
                    else:
                        f[j][pre] = nxt[j][0]

            return f

        return dfs(1)[budget][0]

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