Sensor Cluster Detector 4 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and cluster metrics, and a threshold value K, construct an optimal algorithm to evaluate and compute the maximum sum of metrics that can be achieved by selecting a subset of the sequence such that the sum of the metrics does not exceed the threshold value K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Cluster Detector 4"
WHY DOES IT MATTER?
Subset‑sum/knapsack patterns appear in resource allocation, budgeting, and capacity planning problems where you must fit a set of items within a hard limit while extracting maximum value. Mastering this pattern equips engineers to design efficient schedulers, load balancers, and cost‑optimizers.
OPTIMIZATION CHALLENGE
The key insight is to replace exponential enumeration with a monotonic feasibility predicate. Binary searching the answer reduces the search space to log K steps, and the bitset DP provides an O(N·K/word) feasibility test by leveraging word‑level parallelism, turning a potentially O(N·K) DP into a fast, cache‑friendly operation.
REAL-WORLD CONNECTION
Imagine a distributed sensor network where each node reports a data payload size. The central aggregator can only ingest up to K bytes per cycle. Determining the maximal payload combination without overflow mirrors the subset‑sum problem, and the binary‑search‑plus‑bitset technique is analogous to probing the network with incremental load tests to find the saturation point.
During an interview, first articulate the monotonic property (if a sum S is achievable, any S' ≤ S is also achievable). Then propose binary search on S and immediately suggest a bitset DP for the feasibility check – this shows you understand both algorithmic theory and practical low‑level optimization.
COMPLEXITY AT A GLANCE
O(N * K / 64 * log K)O(K / 64)Core Theory — Why This Approach?
The problem is a classic variant of the 0/1 knapsack where we must pick a subset of values whose total does not exceed a capacity K while maximizing the achieved sum. A naïve exhaustive search enumerates all 2^N subsets, which explodes even for moderate N (e.g., N=30 yields over a billion possibilities) and is therefore infeasible for the input sizes typical in interview settings. The optimal paradigm combines binary search on the answer space with a fast feasibility test: we binary‑search the maximum achievable sum S (0 ≤ S ≤ K) and, for each mid‑point, run a subset‑sum feasibility check using a bitset DP that runs in O(N·K/word) time. The bitset shifts represent adding each element to all previously reachable sums, and checking whether any reachable sum ≥ mid confirms that S is attainable. This reduces the exponential blow‑up to a pseudo‑polynomial solution that scales to N≈10^5 and K≈10^5, comfortably fitting within typical time limits.
Interview Questions on This Problem
Q1How would you adapt the solution if the sensor metrics could be negative as well as positive?
Negative values break the monotonicity needed for binary search on the answer because adding a negative can increase the feasible sum range. The fix is to separate positives and negatives: run the standard DP on positives to compute all reachable positive sums, then for each reachable sum, try to add any combination of negatives (which can be pre‑processed with a separate DP or greedy if their total magnitude is small). Alternatively, shift all values by a constant to make them non‑negative, adjust K accordingly, and run the original algorithm.
Q2Explain why a simple greedy approach that picks the largest metrics first fails for this problem.
Greedy selection assumes that larger items always contribute more to the optimal sum, but a large metric may overshoot K while a combination of smaller metrics can fill the remaining capacity more tightly. Counter‑examples (e.g., K=10, metrics=[9,6,5]) show greedy picks 9 (sum=9) whereas optimal is 6+5=11>10 not allowed, but 6+5 exceeds K, so best is 6+5? Actually best is 6+5=11 >10, so 9 is optimal; need a better example: K=12, metrics=[8,7,6,5]; greedy picks 8 then cannot add any other (sum=8) while optimal is 7+5=12. Hence greedy is not optimal.
Q3What modifications are needed to answer the variant: "Find the number of subsets whose sum is exactly K"?
Instead of a boolean bitset, maintain a count bitset (or an array of 64‑bit integers) where each position stores the number of ways to achieve that sum modulo a large prime (to avoid overflow). For each metric, shift the count array left by the metric and add it to the original counts. After processing all elements, the entry at index K holds the answer. This runs in O(N·K/word) time and O(K) space, similar to the feasibility DP but with additive updates.
Examples
Input
[10, 20, 30, 40, 50], 100
Output
90
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and threshold 100, we start by adding the largest metric 50, then add the next largest metric 40, giving a sum of 90 which does not exceed the threshold value 100.
Input
[5, 10, 15, 20, 25], 30
Output
25
Explanation: Step-by-step: with input [5, 10, 15, 20, 25] and threshold 30, we start by adding the largest metric 25, which does not exceed the threshold value 30, giving a sum of 25.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Binary‑search the answer and, for each candidate, run a bitset DP that updates reachable sums in O(N·K/word) time, yielding O(N·K/word · log K) overall.
Brute Force Approach
Enumerate every subset, compute its sum, and keep the largest sum ≤ K; this runs in O(2^N) time.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let num of nums) {
if (sum + num <= k) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.rbegin(), nums.rend());
int sum = 0;
for (int num : nums) {
if (sum + num <= k) {
sum += num;
}
}
return sum;
}
}class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = nums.length - 1; i >= 0; i--) {
if (sum + nums[i] <= k) {
sum += nums[i];
}
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
sum = 0
for num in nums:
if sum + num <= k:
sum += num
return sumfunction solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let num of nums) {
if (sum + num <= k) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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