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Difficulty: Hard
Topics: Dynamic Programming, Tree DP, Knapsack, Graph Theory
You are given an integer n, representing the number of employees in a company.
Each employee has a unique ID from 1 to n, and employee 1 is the CEO.
You are given two 1-based arrays:
You are also given a hierarchy array of edges:
hierarchy[i] = [u, v] means employee u is the direct boss of employee vThere is a total budget representing the amount available for buying stocks.
If a boss buys stock, then each of their direct subordinates can buy stock at half price:
You may buy each employee’s stock at most once, and you cannot reuse the profit from selling to buy more stocks.
Return the maximum total profit achievable without exceeding the budget.
Profit for employee i:
Input:
n = 2
present = [1, 2]
future = [4, 3]
hierarchy = [[1, 2]]
budget = 3
Output:
5
Explanation:
Input:
n = 2
present = [3, 4]
future = [5, 8]
hierarchy = [[1, 2]]
budget = 4
Output:
4
Explanation:
Only Employee 2 can buy within budget → profit = 4.
Input:
n = 3
present = [4, 6, 8]
future = [7, 9, 11]
hierarchy = [[1, 2], [1, 3]]
budget = 10
Output:
10
Explanation:
Input:
n = 3
present = [5, 2, 3]
future = [8, 5, 6]
hierarchy = [[1, 2], [2, 3]]
budget = 7
Output:
12
Explanation:
1 <= n <= 160
1 <= present[i], future[i] <= 50
1 <= budget <= 160
hierarchy.length = n - 1
Employee 1 is root of the hierarchy tree
No cycles in the hierarchy
import java.util.*;
public class Main { public static class Result { int[] dp0; int[] dp1; int size;
Result(int[] dp0, int[] dp1, int size) {
this.dp0 = dp0;
this.dp1 = dp1;
this.size = size;
}
} public static void main(String[] args) { // Example Input int n = 3; int[] present = {5, 2, 3}; int[] future = {8, 5, 6}; int[][] hierarchy = { {1, 2}, {2, 3} }; int budget = 7;
int ans = maxProfit(n, present, future, hierarchy, budget);
System.out.println("Maximum Profit = " + ans);
}
public static int maxProfit(
int n,
int[] present,
int[] future,
int[][] hierarchy,
int budget
) {
List<Integer>[] g = new ArrayList[n];
for (int i = 0; i < n; i++) {
g[i] = new ArrayList<>();
}
for (int[] e : hierarchy) {
g[e[0] - 1].add(e[1] - 1);
}
return dfs(0, present, future, g, budget).dp0[budget];
}
private static Result dfs(
int u,
int[] present,
int[] future,
List<Integer>[] g,
int budget
) {
int cost = present[u];
int dCost = present[u] / 2;
int[] dp0 = new int[budget + 1];
int[] dp1 = new int[budget + 1];
int[] subProfit0 = new int[budget + 1];
int[] subProfit1 = new int[budget + 1];
int uSize = cost;
for (int v : g[u]) {
Result childResult = dfs(v, present, future, g, budget);
uSize += childResult.size;
for (int i = budget; i >= 0; i--) {
for (int sub = 0; sub <= Math.min(childResult.size, i); sub++) {
if (i - sub >= 0) {
subProfit0[i] = Math.max(
subProfit0[i],
subProfit0[i - sub] + childResult.dp0[sub]
);
subProfit1[i] = Math.max(
subProfit1[i],
subProfit1[i - sub] + childResult.dp1[sub]
);
}
}
}
}
for (int i = 0; i <= budget; i++) {
dp0[i] = subProfit0[i];
dp1[i] = subProfit0[i];
if (i >= dCost) {
dp1[i] = Math.max(
dp1[i],
subProfit1[i - dCost] + future[u] - dCost
);
}
if (i >= cost) {
dp0[i] = Math.max(
dp0[i],
subProfit1[i - cost] + future[u] - cost
);
}
}
return new Result(dp0, dp1, uSize);
}
}
— Written by Saurabh Patil • B.Tech CSE • Software Developer🎯 Select Challenge to activate
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