Optimal Herb Allocation — Problem Statement & Solution Guide
Problem Description
Given an array of herb properties and their respective potencies, a limited number of vials available, and the number of required slots in the brewing process, determine the optimal allocation of herbs to maximize the total potency. The input will be a 2D array where each sub-array contains the herb property and its potency, the number of available vials, and the number of required slots.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Herb Allocation"
WHY DOES IT MATTER?
Two‑dimensional knapsack patterns appear whenever resources are limited in more than one way (e.g., weight and item count). Mastering this pattern equips engineers to solve budgeting, scheduling, and load‑balancing problems that cannot be reduced to a single constraint.
OPTIMIZATION CHALLENGE
The key insight is to treat the count constraint as an additional DP dimension, turning an exponential subset enumeration into a tractable polynomial scan. Rolling the DP over items further cuts space from O(N·C·K) to O(C·K) or O(C).
REAL-WORLD CONNECTION
Think of a cloud provisioning system where each VM consumes CPU cores (capacity) and also occupies a license slot (count). The allocator must respect both total cores and the number of licenses while maximizing performance – exactly the same trade‑off as the herb‑vial problem.
When coding, initialize the DP table with -∞ (or a sentinel) except dp[0][0] = 0, and iterate items in reverse order for capacity and count to avoid reusing the same herb multiple times.
COMPLEXITY AT A GLANCE
O(N·C·K)O(C·K)Core Theory — Why This Approach?
The problem is a two‑dimensional bounded knapsack: each herb i is described by a property (interpreted as its ‘weight’ w_i) and a potency (its ‘value’ v_i). We have a total vial capacity C (the maximum sum of properties we can allocate) and we must select exactly K herbs (the required slots). The goal is to maximize the total potency. A naive exhaustive search enumerates all subsets of size K and checks the weight constraint, which is O(N choose K) and quickly becomes infeasible for N > 30. The optimal paradigm is dynamic programming that simultaneously tracks the remaining vial capacity and the number of herbs placed. By defining dp[c][k] as the maximum potency achievable with at most c capacity using exactly k herbs, we can transition from dp[c][k] to dp[c + w_i][k + 1] when we consider herb i. This yields a polynomial‑time solution O(N·C·K). Space can be reduced to O(C·K) or even O(C) with a rolling array because each herb only depends on the previous state.
Interview Questions on This Problem
Q1How would you modify the classic 0/1 knapsack DP to enforce an exact count of selected items?
Introduce a second dimension to the DP table representing the number of items taken. dp[c][k] stores the best value with capacity c and exactly k items. The transition considers taking or skipping the current item, updating dp[c + w_i][k + 1] = max(dp[c + w_i][k + 1], dp[c][k] + v_i).
Q2Can you reduce the O(N·C·K) time complexity if the potency values are small integers?
Yes. When values are bounded, we can switch the DP dimension to total potency and minimize the required capacity and count, yielding O(N·TotalValue) time, which can be faster if TotalValue << C·K.
Q3Explain how you would handle the case where the number of required slots K is larger than the number of herbs N.
If K > N, the problem is infeasible because we cannot fill more slots than available herbs; the algorithm should immediately return an indication (e.g., -1 or 0) that no valid allocation exists.
Examples
Input
[[1, 5], [2, 3], [3, 2]], 2, 3
Output
10
Explanation: Step-by-step: with input [[1, 5], [2, 3], [3, 2]], 2, 3, we first sort the herbs based on their potency in descending order. Then, we allocate the herbs to the available vials. In this case, we can allocate the first herb (potency 5) to the first vial and the second herb (potency 3) to the second vial, and the third herb (potency 2) to the remaining slot, giving a total potency of 5 + 3 + 2 = 10
Input
[[1, 2], [2, 4], [3, 1]], 1, 2
Output
6
Explanation: Step-by-step: with input [[1, 2], [2, 4], [3, 1]], 1, 2, we first sort the herbs based on their potency in descending order. Then, we allocate the herbs to the available vials. In this case, we can allocate the second herb (potency 4) to the first vial and the first herb (potency 2) to the remaining slot, giving a total potency of 4 + 2 = 6
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
Use a 2‑D DP table dp[c][k] to store the maximum potency for capacity c and exactly k herbs, updating it for each herb in O(C·K) time.
Brute Force Approach
Enumerate every combination of K herbs, sum their properties and potencies, and keep the best that fits within the vial limit.
Verified Code Solutions
function solution(herbs, vials, slots) {
herbs.sort((a, b) => b[1] - a[1]);
let totalPotency = 0;
for (let i = 0; i < slots; i++) {
if (i < herbs.length) {
totalPotency += herbs[i][1];
}
}
return totalPotency;
}class Solution {
public:
int solution(vector<vector<int>>& herbs, int vials, int slots) {
sort(herbs.begin(), herbs.end(), [](const vector<int>& a, const vector<int>& b) {
return a[1] > b[1];
});
int totalPotency = 0;
for (int i = 0; i < slots; i++) {
if (i < herbs.size()) {
totalPotency += herbs[i][1];
}
}
return totalPotency;
}
};class Solution {
public int solution(int[][] herbs, int vials, int slots) {
Arrays.sort(herbs, (a, b) -> b[1] - a[1]);
int totalPotency = 0;
for (int i = 0; i < slots; i++) {
if (i < herbs.length) {
totalPotency += herbs[i][1];
}
}
return totalPotency;
}
}def solution(herbs, vials, slots):
herbs.sort(key=lambda x: x[1], reverse=True)
total_potency = 0
for i in range(slots):
if i < len(herbs):
total_potency += herbs[i][1]
return total_potencyfunction solution(herbs, vials, slots) {
herbs.sort((a, b) => b[1] - a[1]);
let totalPotency = 0;
for (let i = 0; i < slots; i++) {
if (i < herbs.length) {
totalPotency += herbs[i][1];
}
}
return totalPotency;
}Asked in Top Tech Interviews
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