Protocol Sensor Validator 46 — Problem Statement & Solution Guide
Problem Description
Protocol Sensor Validator 46
You are given an array of integers nums that represent a series of protocol and sensor measurements, and an integer threshold k. Your task is to compute the sum of all elements in nums that are strictly greater than k. The algorithm should run in linear time relative to the size of the input and use only constant extra space.
**Input**
- The first line contains a single integer n, the number of measurements.
- The second line contains n space‑separated integers, the elements of nums.
- The third line contains the integer k.
**Output**
- Output a single integer, the sum of all nums[i] such that nums[i] > k. If no element satisfies the condition, output 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Validator 46"
WHY DOES IT MATTER?
Linear‑time aggregation of filtered data is a foundational pattern for high‑throughput data pipelines.
OPTIMIZATION CHALLENGE
Eliminating sorting or nested loops reduces the complexity from O(n log n) or O(n²) to O(n).
REAL-WORLD CONNECTION
Think of a sensor hub that only records readings above a safety threshold, summing them to assess risk exposure.
Keep the accumulator in a primitive type and avoid temporary collections; this minimizes cache pressure and GC overhead.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The straightforward solution is to iterate once over the input array, accumulating the sum of values that exceed the threshold k. This single-pass linear algorithm leverages the fact that each element can be evaluated independently, guaranteeing O(n) time where n is the array length. Naïve alternatives, such as sorting the array first (O(n log n)) or using nested loops to compare each element against every other, inflate the runtime dramatically and are unnecessary because the condition depends only on a constant k, not on relative ordering. The optimal paradigm is a greedy linear scan combined with constant‑space accumulation, which directly exploits the problem’s additive and threshold nature without extra data structures.
Interview Questions on This Problem
Q1How would you modify the solution to handle very large integer sums that might overflow a 32‑bit type?
Use a 64‑bit integer type (e.g., long long in C++ or long in Java) for the accumulator, or employ arbitrary‑precision libraries if needed.
Q2Can this algorithm be parallelized, and if so, what would be the approach?
Yes, split the array into chunks, compute partial sums of elements > k in each chunk concurrently, then combine the partial results in a final reduction step.
Q3What is the time‑space trade‑off if you pre‑process the array into a prefix‑sum structure?
Prefix sums allow O(1) range queries but require O(n) extra space and O(n) preprocessing, which is unnecessary for a single global sum query.
Examples
Input
5 5 12 7 3 9 6
Output
28
Explanation: Elements greater than 6 are 12, 7 and 9. Their sum is 12 + 7 + 9 = 28.
Input
5 -2 -1 0 1 2 -1
Output
3
Explanation: Elements greater than -1 are 0, 1 and 2. Their sum is 0 + 1 + 2 = 3.
Input
1 100 50
Output
100
Explanation: The only element 100 is greater than 50, so the sum equals 100.
Constraints
- 1 <= n <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= k <= 10^9
- The solution must run in O(n) time and O(1) auxiliary space.
Optimal Approach & Strategy
Perform a single linear scan, adding each element that exceeds k to a running sum.
Brute Force Approach
Sort the array then iterate from the first element greater than k, summing the rest.
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var sumGreaterThanK = function(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
};class Solution {
public:
int sumGreaterThanK(vector<int>& nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
};class Solution {
public int sumGreaterThanK(int[] nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
}class Solution:
def sumGreaterThanK(self, nums: List[int], k: int) -> int:
return sum(num for num in nums if num > k)/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var sumGreaterThanK = function(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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