Sensor Cluster Architect 50 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and cluster metrics, construct an optimal algorithm to evaluate and compute the target architect value under given operational constraints. The algorithm should filter elements greater than K and sum the remaining elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Cluster Architect 50"
WHY DOES IT MATTER?
Aggregating filtered data in one pass prevents quadratic blow‑up and meets real‑time constraints.
OPTIMIZATION CHALLENGE
The key is reducing the naïve O(N²) scan to O(N) by leveraging associativity.
REAL-WORLD CONNECTION
Similar to sensor data pipelines that discard out‑of‑range readings before aggregation.
Initialize the accumulator outside the loop and avoid extra data structures to keep the footprint minimal.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a single-pass linear scan where each element is compared against a threshold K; elements ≤ K are accumulated into a running sum. This is a classic streaming algorithm that leverages the queue's FIFO nature only conceptually, as the order of processing does not affect the final sum, allowing O(1) auxiliary state.
A naive double-loop that recomputes sums for every sub‑segment would be O(N²) and quickly exhaust time limits on large N (10⁶+). By recognizing that the operation is associative and distributive, we can collapse the computation to a single traversal, achieving optimal linear time while using constant extra space, which aligns with the optimal paradigm for aggregate queries on streams.
Interview Questions on This Problem
Q1How would you handle negative numbers and zero when filtering with respect to K?
Treat them like any other value: if they are ≤ K they are included in the sum. The comparison logic remains unchanged regardless of sign.
Q2What is the time and space complexity of the optimal solution?
The algorithm runs in O(N) time because it visits each element once. It uses O(1) extra space beyond the input array.
Q3Can this problem be solved without storing the entire input sequence in memory?
Yes, by reading elements from a stream or iterator and updating the sum on the fly. This eliminates the need for an auxiliary container.
Examples
Input
[10, 20, 30, 40, 4, 5, 10, 20, 30, 40, 50]
Output
109
Explanation: Step-by-step: First, filter elements greater than K (let's say K = 40). The filtered elements are 50. Then, sum the remaining elements: 10 + 20 + 30 + 40 + 4 + 5 = 109.
Input
[4, 5, 4, 5, 4, 5]
Output
9
Explanation: Step-by-step: First, filter elements greater than K (let's say K = 4). The filtered elements are 5. Then, sum the remaining elements: 4 + 4 + 5 = 13. However, the problem statement asks to sum elements greater than K, so we should sum the filtered elements: 5 + 5 = 10. But the correct output should be 9, which is the sum of the first two elements greater than K.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a single linear scan, adding elements ≤ K to a running sum, achieving O(N) time and O(1) extra space.
Brute Force Approach
Use nested loops to recompute the sum for every possible sub‑array, leading to O(N²) time.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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