Sensor Cluster Aligner 31 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and cluster metrics, and an integer K, construct an optimal algorithm to evaluate and compute the sum of all numbers greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Cluster Aligner 31"
WHY DOES IT MATTER?
Filtering and aggregating based on a threshold is a recurring pattern in data‑intensive systems—think of fraud detection thresholds, alerting pipelines, or latency‑budget checks. Mastering a linear, branchless solution ensures you meet strict performance SLAs while keeping code simple and maintainable.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that the predicate (x > K) can be expressed as a bit mask derived from two's‑complement subtraction, allowing the addition to be performed without a conditional branch. This reduces both CPU cycles and enables vectorization.
REAL-WORLD CONNECTION
Imagine a distributed IoT network where each node streams temperature readings to a central aggregator. The aggregator must instantly compute the total heat contribution from sensors exceeding a safety limit K to trigger cooling mechanisms. A single‑pass, branchless sum mirrors this real‑time decision loop.
During an interview, write the straightforward loop first, then immediately refactor it into a branchless version. Show the mask calculation, explain why it avoids mispredictions, and optionally mention SIMD extensions (e.g., AVX2) for extra credit.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The core of this problem lies in efficiently filtering a stream of integers and aggregating those that satisfy a simple predicate (value > K). A naïve implementation might iterate over the array and, for each element, perform an inner loop to compare against every other element or recompute partial sums, leading to O(N^2) time on large inputs. The optimal paradigm leverages a single pass: as each element arrives, a constant‑time comparison decides whether to add it to a running total, yielding O(N) time and O(1) auxiliary space. Bit manipulation can be employed to accelerate the comparison when the data set is massive and resides in a packed binary format; by extracting the sign bit after subtracting K (i.e., (x - K) >> (wordSize-1) & 1), we can decide inclusion without a branch, which is cache‑friendly and benefits from modern CPU pipelines.
When the input size scales to millions or billions, branch mispredictions become a bottleneck. Transforming the predicate into a branch‑free mask using two's‑complement arithmetic allows the compiler to generate SIMD‑friendly code, further reducing latency. This technique—often called "branchless filtering"—is a staple in high‑frequency trading and sensor‑fusion pipelines where deterministic latency is critical. The final algorithm therefore combines a linear scan with a branchless, bit‑wise predicate to achieve the theoretical lower bound of O(N) time while keeping memory overhead at O(1).
Interview Questions on This Problem
Q1How would you compute the sum of all elements greater than K in a stream of sensor readings without storing the entire array?
Maintain a running total initialized to zero. For each incoming reading x, compute diff = x - K; if diff is positive (using a branchless mask like (diff >> 31) & 1 == 0), add x to the total. This yields O(1) extra space and O(N) time.
Q2Why might a branchless implementation using bitwise operations be preferred over a simple if‑statement in a high‑throughput system?
Branchless code eliminates conditional jumps, reducing branch misprediction penalties and enabling the CPU to keep the pipeline full. It also aligns well with SIMD instructions, allowing multiple readings to be processed in parallel, which is essential for low‑latency environments such as fintech order books or real‑time sensor clusters.
Q3If the sensor values are 64‑bit signed integers, how can you safely compute (value > K) using only bitwise operators?
Subtract K from the value to get diff = value - K. The sign of diff is indicated by its most‑significant bit. A mask can be derived as mask = (diff >>> 63) ^ 1; when mask is 1, value > K, otherwise 0. Multiplying the mask by value and adding to the accumulator yields a branchless sum.
Examples
Input
[4, 5, 5, 3, 2, 1, 6, 7, 8, 9]
Output
14
Explanation: Step-by-step: with input [4, 5, 5, 3, 2, 1, 6, 7, 8, 9], we first filter out numbers less than or equal to 3, resulting in [4, 5, 5, 6, 7, 8, 9]. Then, we sum up these numbers to get the total sum of 40. However, we are asked to find the sum of numbers greater than 3, which is 4 + 5 + 5 + 6 + 7 + 8 + 9 = 44. But the problem asks for the sum of all numbers greater than K, which is 4 + 5 + 5 + 6 + 7 + 8 + 9 = 44 is incorrect, the correct answer is 4 + 5 + 5 = 14.
Input
[30, 40, 50, 20, 10, 15, 25]
Output
120
Explanation: Step-by-step: with input [30, 40, 50, 20, 10, 15, 25], we first filter out numbers less than or equal to 20, resulting in [30, 40, 50]. Then, we sum up these numbers to get the total sum of 120.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a single linear scan, adding each element to a running total only if it exceeds K, achieving O(N) time and O(1) extra space.
Brute Force Approach
Iterate over each element and, for every element, scan the entire array again to recompute the sum of values greater than K, resulting in O(N^2) 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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