Protocol Sensor Partition 49 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing protocol and sensor metrics, and an integer K, construct an optimal algorithm to compute the sum of elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Partition 49"
WHY DOES IT MATTER?
The filter‑and‑aggregate pattern appears in analytics, monitoring, and financial risk calculations where thresholds dictate inclusion. Mastery of this pattern ensures you can handle massive streams efficiently without unnecessary sorting or extra storage.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the predicate does not depend on other elements, allowing us to avoid any ordering or auxiliary data structures and simply accumulate on the fly.
REAL-WORLD CONNECTION
Think of a network router that only forwards packets exceeding a certain priority score; it scans incoming packets once, discarding low‑priority ones and aggregating metrics for the high‑priority traffic, mirroring the single‑pass sum > K.
During an interview, write the loop first, then immediately add the conditional check (if arr[i] > K) before updating the sum—this keeps the code concise and avoids off‑by‑one or missing‑element bugs.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a single linear scan over the input array, accumulating the values that satisfy the predicate (value > K). This is a classic example of a filter‑and‑aggregate pattern, where each element is examined exactly once, making the algorithm optimal in terms of time complexity. Naïve approaches that sort the array first or use nested loops inflate the runtime to O(N log N) or O(N²), which becomes prohibitive for large N (e.g., N > 10⁶) because the extra passes or comparisons dominate the execution time and memory usage. By leveraging the fact that the predicate is monotonic (greater than a constant) and that we only need the sum—not the order or positions of qualifying elements—the optimal paradigm is a single-pass, constant‑space accumulation, which is both cache‑friendly and trivially parallelizable if needed.
Interview Questions on This Problem
Q1How would you modify the solution if the requirement changed to compute the sum of the top‑M largest elements greater than K?
Maintain a min‑heap of size M while scanning; push elements > K into the heap and pop when size exceeds M. After the scan, sum the heap contents. This yields O(N log M) time and O(M) space.
Q2What is the time and space complexity if the input is a linked list instead of an array?
The algorithm remains O(N) time because we still need to visit each node once, and O(1) auxiliary space since we only keep a running total and a pointer to the next node.
Q3Explain how you could compute the sum of elements greater than K in a distributed setting where the array is sharded across multiple machines.
Each shard independently computes the local sum of values > K; a final reduction step aggregates these partial sums across machines, resulting in overall O(N/p) local work per machine (p = number of shards) and O(p) communication overhead.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
40
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we first identify the elements greater than 5, which are 6, 7, 8, 9, and 10. Then, we sum these elements to get the final output: 6 + 7 + 8 + 9 + 10 = 40.
Input
[30, 40, 50, 60, 70, 80, 90]
Output
390
Explanation: Step-by-step: Given the input [30, 40, 50, 60, 70, 80, 90], we first identify the elements greater than 30, which are 40, 50, 60, 70, 80, and 90. Then, we sum these elements to get the final output: 40 + 50 + 60 + 70 + 80 + 90 = 390.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Traverse the array once, checking each element against K and adding qualifying values to a running total; this yields O(N) time and O(1) extra space.
Brute Force Approach
Sort the array and then iterate from the first element greater than K, summing the rest; or use a nested loop to compare each element with every other, both O(N log N) or O(N²).
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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