LeetCode #3562 — HARD

Maximum Profit from Trading Stocks with Discounts

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer n, representing the number of employees in a company. Each employee is assigned a unique ID from 1 to n, and employee 1 is the CEO, is the direct or indirect boss of every employee. You are given two 1-based integer arrays, present and future, each of length n, where:

present[i] represents the current price at which the i^th employee can buy a stock today.
future[i] represents the expected price at which the i^th employee can sell the stock tomorrow.

The company's hierarchy is represented by a 2D integer array hierarchy, where hierarchy[i] = [u_i, v_i] means that employee u_i is the direct boss of employee v_i.

Additionally, you have an integer budget representing the total funds available for investment.

However, the company has a discount policy: if an employee's direct boss purchases their own stock, then the employee can buy their stock at half the original price (floor(present[v] / 2)).

Return the maximum profit that can be achieved without exceeding the given budget.

Note:

You may buy each stock at most once.
You cannot use any profit earned from future stock prices to fund additional investments and must buy only from budget.

Example 1:

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

Output: 5

Explanation:

Employee 1 buys the stock at price 1 and earns a profit of 4 - 1 = 3.
Since Employee 1 is the direct boss of Employee 2, Employee 2 gets a discounted price of floor(2 / 2) = 1.
Employee 2 buys the stock at price 1 and earns a profit of 3 - 1 = 2.
The total buying cost is 1 + 1 = 2 <= budget. Thus, the maximum total profit achieved is 3 + 2 = 5.

Example 2:

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

Output: 4

Explanation:

Employee 2 buys the stock at price 4 and earns a profit of 8 - 4 = 4.
Since both employees cannot buy together, the maximum profit is 4.

Example 3:

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

Output: 10

Explanation:

Employee 1 buys the stock at price 4 and earns a profit of 7 - 4 = 3.
Employee 3 would get a discounted price of floor(8 / 2) = 4 and earns a profit of 11 - 4 = 7.
Employee 1 and Employee 3 buy their stocks at a total cost of 4 + 4 = 8 <= budget. Thus, the maximum total profit achieved is 3 + 7 = 10.

Example 4:

Input: n = 3, present = [5,2,3], future = [8,5,6], hierarchy = [[1,2],[2,3]], budget = 7

Output: 12

Explanation:

Employee 1 buys the stock at price 5 and earns a profit of 8 - 5 = 3.
Employee 2 would get a discounted price of floor(2 / 2) = 1 and earns a profit of 5 - 1 = 4.
Employee 3 would get a discounted price of floor(3 / 2) = 1 and earns a profit of 6 - 1 = 5.
The total cost becomes 5 + 1 + 1 = 7 <= budget. Thus, the maximum total profit achieved is 3 + 4 + 5 = 12.

Constraints:

1 <= n <= 160
present.length, future.length == n
1 <= present[i], future[i] <= 50
hierarchy.length == n - 1
hierarchy[i] == [u_i, v_i]
1 <= u_i, v_i <= n
u_i != v_i
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.

Step 01

Brute Force Baseline

Problem summary: You are given an integer n, representing the number of employees in a company. Each employee is assigned a unique ID from 1 to n, and employee 1 is the CEO, is the direct or indirect boss of every employee. You are given two 1-based integer arrays, present and future, each of length n, where: present[i] represents the current price at which the ith employee can buy a stock today. future[i] represents the expected price at which the ith employee can sell the stock tomorrow. The company's hierarchy is represented by a 2D integer array hierarchy, where hierarchy[i] = [ui, vi] means that employee ui is the direct boss of employee vi. Additionally, you have an integer budget representing the total funds available for investment. However, the company has a discount policy: if an employee's direct boss purchases their own stock, then the employee can buy their stock at half the original price

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Tree

Example 1

2
[1,2]
[4,3]
[[1,2]]
3

Example 2

2
[3,4]
[5,8]
[[1,2]]
4

Example 3

3
[4,6,8]
[7,9,11]
[[1,2],[1,3]]
10

Step 02

Core Insight

What unlocks the optimal approach

- Compute <code>max_profit[u]</code> and <code>max_profit1[u]</code> for each node <code>u</code>
- <code>max_profit[u]</code> = maximum profit in the subtree of <code>u</code> assuming the parent of <code>u</code> has not bought the stock
- <code>max_profit1[u]</code> = maximum profit in the subtree of <code>u</code> assuming the parent of <code>u</code> has bought the stock
For each node <code>u</code>, consider two cases:
Buy the stock for <code>u</code> (at <code>present[u]</code> price if parent did not buy, or at <code>floor(present[u]/2)</code> if parent bought), then add the best <code>max_profit1</code> values of its children
Skip buying for <code>u</code>, then add the best <code>max_profit</code> values of its children

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3562: Maximum Profit from Trading Stocks with Discounts
class Solution {
    private List<Integer>[] g;
    private int[] present;
    private int[] future;
    private int budget;

    public int maxProfit(int n, int[] present, int[] future, int[][] hierarchy, int budget) {
        this.present = present;
        this.future = future;
        this.budget = budget;

        g = new ArrayList[n + 1];
        Arrays.setAll(g, k -> new ArrayList<>());

        for (int[] e : hierarchy) {
            g[e[0]].add(e[1]);
        }

        return dfs(1)[budget][0];
    }

    private int[][] dfs(int u) {
        int[][] nxt = new int[budget + 1][2];

        for (int v : g[u]) {
            int[][] fv = dfs(v);
            for (int j = budget; j >= 0; j--) {
                for (int jv = 0; jv <= j; jv++) {
                    for (int pre = 0; pre < 2; pre++) {
                        int val = nxt[j - jv][pre] + fv[jv][pre];
                        if (val > nxt[j][pre]) {
                            nxt[j][pre] = val;
                        }
                    }
                }
            }
        }

        int[][] f = new int[budget + 1][2];
        int price = future[u - 1];

        for (int j = 0; j <= budget; j++) {
            for (int pre = 0; pre < 2; pre++) {
                int cost = present[u - 1] / (pre + 1);
                if (j >= cost) {
                    f[j][pre] = Math.max(nxt[j][0], nxt[j - cost][1] + (price - cost));
                } else {
                    f[j][pre] = nxt[j][0];
                }
            }
        }

        return f;
    }
}

// Accepted solution for LeetCode #3562: Maximum Profit from Trading Stocks with Discounts
func maxProfit(n int, present []int, future []int, hierarchy [][]int, budget int) int {
	g := make([][]int, n+1)
	for _, e := range hierarchy {
		u, v := e[0], e[1]
		g[u] = append(g[u], v)
	}

	var dfs func(u int) [][2]int
	dfs = func(u int) [][2]int {
		nxt := make([][2]int, budget+1)

		for _, v := range g[u] {
			fv := dfs(v)
			for j := budget; j >= 0; j-- {
				for jv := 0; jv <= j; jv++ {
					for pre := 0; pre < 2; pre++ {
						nxt[j][pre] = max(nxt[j][pre], nxt[j-jv][pre]+fv[jv][pre])
					}
				}
			}
		}

		f := make([][2]int, budget+1)
		price := future[u-1]

		for j := 0; j <= budget; j++ {
			for pre := 0; pre < 2; pre++ {
				cost := present[u-1] / (pre + 1)
				if j >= cost {
					buyProfit := nxt[j-cost][1] + (price - cost)
					f[j][pre] = max(nxt[j][0], buyProfit)
				} else {
					f[j][pre] = nxt[j][0]
				}
			}
		}
		return f
	}

	return dfs(1)[budget][0]
}

# Accepted solution for LeetCode #3562: Maximum Profit from Trading Stocks with Discounts
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]

// Accepted solution for LeetCode #3562: Maximum Profit from Trading Stocks with Discounts
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3562: Maximum Profit from Trading Stocks with Discounts
// class Solution {
//     private List<Integer>[] g;
//     private int[] present;
//     private int[] future;
//     private int budget;
// 
//     public int maxProfit(int n, int[] present, int[] future, int[][] hierarchy, int budget) {
//         this.present = present;
//         this.future = future;
//         this.budget = budget;
// 
//         g = new ArrayList[n + 1];
//         Arrays.setAll(g, k -> new ArrayList<>());
// 
//         for (int[] e : hierarchy) {
//             g[e[0]].add(e[1]);
//         }
// 
//         return dfs(1)[budget][0];
//     }
// 
//     private int[][] dfs(int u) {
//         int[][] nxt = new int[budget + 1][2];
// 
//         for (int v : g[u]) {
//             int[][] fv = dfs(v);
//             for (int j = budget; j >= 0; j--) {
//                 for (int jv = 0; jv <= j; jv++) {
//                     for (int pre = 0; pre < 2; pre++) {
//                         int val = nxt[j - jv][pre] + fv[jv][pre];
//                         if (val > nxt[j][pre]) {
//                             nxt[j][pre] = val;
//                         }
//                     }
//                 }
//             }
//         }
// 
//         int[][] f = new int[budget + 1][2];
//         int price = future[u - 1];
// 
//         for (int j = 0; j <= budget; j++) {
//             for (int pre = 0; pre < 2; pre++) {
//                 int cost = present[u - 1] / (pre + 1);
//                 if (j >= cost) {
//                     f[j][pre] = Math.max(nxt[j][0], nxt[j - cost][1] + (price - cost));
//                 } else {
//                     f[j][pre] = nxt[j][0];
//                 }
//             }
//         }
// 
//         return f;
//     }
// }

// Accepted solution for LeetCode #3562: Maximum Profit from Trading Stocks with Discounts
function maxProfit(
    n: number,
    present: number[],
    future: number[],
    hierarchy: number[][],
    budget: number,
): number {
    const g: number[][] = Array.from({ length: n + 1 }, () => []);

    for (const [u, v] of hierarchy) {
        g[u].push(v);
    }

    const dfs = (u: number): number[][] => {
        const nxt: number[][] = Array.from({ length: budget + 1 }, () => [0, 0]);

        for (const v of g[u]) {
            const fv = dfs(v);
            for (let j = budget; j >= 0; j--) {
                for (let jv = 0; jv <= j; jv++) {
                    for (let pre = 0; pre < 2; pre++) {
                        nxt[j][pre] = Math.max(nxt[j][pre], nxt[j - jv][pre] + fv[jv][pre]);
                    }
                }
            }
        }

        const f: number[][] = Array.from({ length: budget + 1 }, () => [0, 0]);
        const price = future[u - 1];

        for (let j = 0; j <= budget; j++) {
            for (let pre = 0; pre < 2; pre++) {
                const cost = Math.floor(present[u - 1] / (pre + 1));
                if (j >= cost) {
                    const profitIfBuy = nxt[j - cost][1] + (price - cost);
                    f[j][pre] = Math.max(nxt[j][0], profitIfBuy);
                } else {
                    f[j][pre] = nxt[j][0];
                }
            }
        }

        return f;
    };

    return dfs(1)[budget][0];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × \textbudget^2)

Space

O(n × \textbudget)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.