Resource Allocation Maximization — Problem Statement & Solution Guide
Problem Description
You are given a total of 927 units of resources and a list of 13 sectors with varying resource requirements. Determine the maximum number of sectors that can be allocated resources without exceeding the available resources. If multiple allocations are possible, choose the one with the most sectors.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Resource Allocation Maximization"
WHY DOES IT MATTER?
The 0/1 Knapsack pattern is fundamental in optimization problems where resources are limited and items are indivisible. It appears in scheduling, budgeting, and resource allocation scenarios. Understanding this pattern allows engineers to recognize when greedy approaches fail (e.g., when item values are not proportional to weights) and when DP is the correct tool.
OPTIMIZATION CHALLENGE
The key insight is that the value of each item is uniform (1), which simplifies the DP transition. Instead of tracking the maximum value, we track the maximum count. This allows for potential optimizations such as using a 1D array for space efficiency and early termination if the remaining capacity is insufficient for any further items.
REAL-WORLD CONNECTION
Consider a cloud provider allocating virtual machines (VMs) to physical servers. Each VM has a CPU/memory footprint (weight) and a uniform billing unit (value). The goal is to maximize the number of VMs hosted on a server without exceeding its capacity. This is directly analogous to the problem, where sectors are VMs and resources are server capacity.
In interviews, always clarify if the 'value' is uniform. If it is, you can mention that the problem is a special case of knapsack. Also, be prepared to discuss space optimization from $O(n \times W)$ to $O(W)$ by using a 1D DP array and iterating backwards.
COMPLEXITY AT A GLANCE
O(n * W)O(W)Core Theory — Why This Approach?
This problem is a classic instance of the 0/1 Knapsack problem, specifically optimized for maximizing the count of items rather than their total value. In the standard knapsack formulation, each item has a weight (resource requirement) and a value (here, value is uniformly 1 for every sector). The goal is to select a subset of items such that the sum of their weights does not exceed the capacity (927 units) and the sum of their values is maximized. A naive recursive approach would explore all $2^n$ subsets, which is computationally infeasible for $n=13$ in a tight loop but becomes catastrophic for larger $n$. The optimal paradigm is Dynamic Programming (DP), which exploits the principle of optimality: the optimal solution to the problem depends on the optimal solutions to its subproblems. By defining a state $dp[i][w]$ as the maximum number of sectors that can be allocated using the first $i$ sectors and a resource budget of $w$, we can build the solution iteratively from smaller subproblems.
Interview Questions on This Problem
Q1At a fintech platform, you need to allocate a fixed marketing budget across multiple campaigns. Each campaign has a cost and a guaranteed ROI. If the ROI is uniform across all campaigns, how would you maximize the number of campaigns funded? How does this differ from maximizing total ROI?
If ROI is uniform, the problem reduces to maximizing the count of items within a weight constraint, which is a 0/1 Knapsack problem where value is 1 for all items. This differs from maximizing total ROI (standard knapsack) where values vary. The DP state remains similar, but the transition logic simplifies because we are adding 1 to the count rather than a variable value. The time complexity is $O(n \times W)$, where $W$ is the budget.
Q2In a distributed systems startup, you have a limited number of CPU cores and a list of microservices with varying core requirements. How would you determine the maximum number of services that can be deployed simultaneously? What if the number of services is 1000 and the core limit is 10,000?
This is a 0/1 Knapsack problem where the 'value' of each service is 1. For $n=1000$ and $W=10,000$, the DP table would be $1000 \times 10,000 = 10^7$ entries, which is feasible in memory and time. The key is to use a 1D DP array to optimize space to $O(W)$, iterating through services and updating the DP array from high to low capacity to avoid reusing the same service multiple times.
Q3At a global product company, you are designing a resource allocation algorithm for cloud infrastructure. If the resource requirements are very large (e.g., up to $10^9$) but the number of sectors is small (e.g., 30), how would you adapt your approach?
When the capacity $W$ is very large, the standard $O(n \times W)$ DP becomes infeasible. In this case, we can use a meet-in-the-middle approach or a DP based on the number of items selected. Since $n=30$ is small, we can split the items into two halves, enumerate all subsets of each half (2^15 = 32,768 subsets), calculate their total resource usage and count, and then combine the best subsets from both halves that fit within the total capacity. This reduces the complexity to $O(2^{n/2} \log(2^{n/2}))$.
Examples
Input
927, [[10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130]]
Output
4
Explanation: Step-by-step: Given 927 units of resources and a list of sectors with varying resource requirements, we sort the sectors in ascending order of their resource requirements. We then iterate over the sorted sectors and allocate resources to each sector until we run out of resources. In this case, we can allocate resources to the first 4 sectors (10, 20, 30, 40) without exceeding the available resources, giving us a total of 4 sectors.
Input
927, [[100, 200, 300, 500, 700, 900, 1000, 100, 50, 20, 10, 5, 1]]
Output
3
Explanation: Step-by-step: Given 927 units of resources and a list of sectors with varying resource requirements, we sort the sectors in ascending order of their resource requirements. We then iterate over the sorted sectors and allocate resources to each sector until we run out of resources. In this case, we can allocate resources to the first 3 sectors (100, 200, 300) without exceeding the available resources, giving us a total of 3 sectors.
Constraints
- 1 ≤ number of sectors ≤ 1000
- 1 ≤ resource requirement per sector ≤ 1000
- 1 ≤ total resources ≤ 1000000
Optimal Approach & Strategy
Use dynamic programming with a 1D array of size 928, where $dp[w]$ stores the maximum number of sectors that can be allocated with a resource budget of $w$. Iterate through each sector and update the DP array from high to low capacity to ensure each sector is used at most once. The final answer is $dp[927]$. This approach runs in $O(n \times W)$ time and $O(W)$ space.
Brute Force Approach
Generate all $2^{13}$ subsets of the 13 sectors, calculate the total resource usage and count for each subset, and keep track of the subset with the maximum count that does not exceed 927 units. This approach is exponential and becomes infeasible for larger inputs.
Verified Code Solutions
function solution(resources, sectors) {
sectors.sort((a, b) => a[0] - b[0]);
let allocated = 0;
let count = 0;
for (let i = 0; i < sectors.length; i++) {
if (resources >= sectors[i][0]) {
resources -= sectors[i][0];
allocated += sectors[i][0];
count++;
} else {
break;
}
}
return count;
}class Solution {
public:
int solution(int resources, vector<pair<int, int>> sectors) {
sort(sectors.begin(), sectors.end(), [](pair<int, int> a, pair<int, int> b) {
return a.first < b.first;
});
int allocated = 0;
int count = 0;
for (int i = 0; i < sectors.size(); i++) {
if (resources >= sectors[i].first) {
resources -= sectors[i].first;
allocated += sectors[i].first;
count++;
} else {
break;
}
}
return count;
}
};class Solution {
public int solution(int resources, int[][] sectors) {
Arrays.sort(sectors, (a, b) -> a[0] - b[0]);
int allocated = 0;
int count = 0;
for (int i = 0; i < sectors.length; i++) {
if (resources >= sectors[i][0]) {
resources -= sectors[i][0];
allocated += sectors[i][0];
count++;
} else {
break;
}
}
return count;
}
}def solution(resources, sectors):
sectors.sort(key=lambda x: x[0])
allocated = 0
count = 0
for sector in sectors:
if resources >= sector[0]:
resources -= sector[0]
allocated += sector[0]
count += 1
else:
break
return countfunction solution(resources, sectors) {
sectors.sort((a, b) => a[0] - b[0]);
let allocated = 0;
let count = 0;
for (let i = 0; i < sectors.length; i++) {
if (resources >= sectors[i][0]) {
resources -= sectors[i][0];
allocated += sectors[i][0];
count++;
} else {
break;
}
}
return count;
}Asked in Top Tech Interviews
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