Sensor Checkpoint Analyzer 16 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the data pipeline for a distributed sensor network. The system generates a stream of integer metrics, and your goal is to compute the 'analyzer value' based on a specific threshold K. The analyzer value is defined as the sum of all elements in the input array that are strictly greater than K. If no elements exceed the threshold, the analyzer value is 0.
Given an array of integers representing sensor readings and an integer threshold K, return the sum of all elements strictly greater than K. This problem requires efficient iteration and conditional summation, which can be optimized using bit manipulation techniques for comparison or filtering in low-level implementations, though a standard linear scan is also acceptable for this medium-difficulty task.
Your solution must handle large input sizes efficiently, ensuring that the time complexity remains linear with respect to the number of elements in the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Checkpoint Analyzer 16"
WHY DOES IT MATTER?
Single‑pass aggregation eliminates unnecessary passes and memory, crucial for high‑throughput sensor pipelines.
OPTIMIZATION CHALLENGE
The key is reducing the naive O(N^2) or O(N log N) overhead to O(N) by avoiding extra data structures.
REAL-WORLD CONNECTION
Think of a real‑time dashboard that continuously adds only readings above a safety threshold to a running risk score.
Keep the loop tight, use early continue for values ≤ K, and always accumulate into a 64‑bit accumulator.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a linear aggregation: scanning the array once while maintaining a running total of elements that satisfy the predicate (> K). Naïve approaches, such as nested loops or repeated filtering with auxiliary data structures, inflate time complexity to O(N^2) and waste memory, which becomes prohibitive for large streams typical in sensor networks. The optimal paradigm leverages the single‑pass, constant‑space pattern: each element is examined exactly once, and the decision to add it to the sum is a constant‑time comparison, yielding O(N) time and O(1) auxiliary space. This aligns with the broader class of "single‑pass aggregation" problems where the goal is to compute a summary statistic without storing the entire dataset.
Interview Questions on This Problem
Q1How would you handle potential integer overflow when summing large sensor values?
Use a wider numeric type such as 64‑bit long long (or BigInteger in languages without native overflow). Cast each addition to the wider type before accumulating.
Q2Can you compute the sum of elements > K without scanning the entire array?
Only if additional preprocessing (e.g., sorting with prefix sums) is allowed for multiple queries; for a single K, a full scan is already optimal.
Q3What is the time‑space trade‑off if you need to answer many different K queries on the same data?
Pre‑process by sorting and building a suffix sum array (O(N log N) time, O(N) space) to answer each query in O(log N) via binary search.
Examples
Input
nums = [12, 5, 8, 15, 3], K = 10
Output
27
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 8 <= 10 (skip), 15 > 10 (add 15), 3 <= 10 (skip). Sum = 12 + 15 = 27.
Input
nums = [1, 2, 3], K = 0
Output
6
Explanation: All elements are greater than 0. Sum = 1 + 2 + 3 = 6.
Input
nums = [5, 5, 5], K = 5
Output
0
Explanation: No elements are strictly greater than 5. Sum = 0.
Input
nums = [-10, -5, 0, 5, 10], K = -5
Output
15
Explanation: Elements greater than -5 are 0, 5, and 10. Sum = 0 + 5 + 10 = 15.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Iterate once, compare each element to K, and accumulate qualifying values in a 64‑bit variable.
Brute Force Approach
Nested loops checking each pair of elements or repeatedly filtering the array, leading to O(N^2) time.
Verified Code Solutions
function solution(nums, K) {
let maxSum = 0;
for (let num of nums) {
if (num > K) {
maxSum += num;
}
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int maxSum = 0;
for (int num : nums) {
if (num > K) {
maxSum += num;
}
}
return maxSum;
}
};class Solution {
public int solution(int[] nums, int K) {
int maxSum = 0;
for (int num : nums) {
if (num > K) {
maxSum += num;
}
}
return maxSum;
}
}def solution(nums, K):
maxSum = 0
for num in nums:
if num > K:
maxSum += num
return maxSumfunction solution(nums, K) {
let maxSum = 0;
for (let num of nums) {
if (num > K) {
maxSum += num;
}
}
return maxSum;
}Asked in Top Tech Interviews
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