Network Protocol Architect 1 — Problem Statement & Solution Guide
Problem Description
In a distributed network monitoring system, a sequence of integer telemetry values is collected from various protocol nodes. Each value represents a specific metric such as latency, throughput, or error count. The system administrator defines a critical threshold K. Your task is to compute the 'Architect Value', which is defined as the sum of all telemetry values in the sequence that are strictly greater than K.
Given an array of integers representing the telemetry sequence and an integer K, return the sum of all elements that exceed K. If no elements exceed K, return 0.
Input: An array of integers telemetry and an integer K.
Output: A single integer representing the sum of all elements in telemetry that are strictly greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Protocol Architect 1"
WHY DOES IT MATTER?
This pattern—single pass with conditional accumulation—is a cornerstone for streaming analytics, enabling real‑time metric computation without storing the entire dataset.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the sum operation does not require any historical context beyond the current accumulator, allowing us to discard the array after a single read and achieve O(1) extra space.
REAL-WORLD CONNECTION
In distributed monitoring, each node often aggregates latency or error counts on‑the‑fly before sending a compact summary to a central dashboard, mirroring the exact logic of summing values that meet a threshold.
During an interview, write the loop first, then immediately add the conditional check; this reduces the chance of off‑by‑one errors and makes the code self‑documenting.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a single‑pass aggregation over a linear collection. The naive solution would iterate over every element and, for each, perform a secondary operation such as recomputing a prefix sum or invoking a nested loop, leading to O(N^2) time on large inputs. In contrast, the optimal paradigm leverages the fact that the sum operation is associative and commutative, allowing us to maintain a running total while scanning the array once. This approach eliminates redundant work and guarantees linear time regardless of input size.
When dealing with massive telemetry streams, memory footprints become critical. By using a constant‑size accumulator (e.g., a 64‑bit integer) we avoid auxiliary data structures, keeping the space complexity at O(1). This is especially important in distributed monitoring systems where each node may process millions of metrics per second. The algorithm’s simplicity also reduces cache misses and branch mispredictions, further improving real‑world throughput.
The optimal solution therefore follows the classic "single pass with accumulator" pattern: read each value, compare it to the threshold K, and add it to the result only if it satisfies the strict inequality. This yields an O(N) time, O(1) auxiliary space algorithm that scales gracefully to the constraints typical of production‑grade monitoring pipelines.
Interview Questions on This Problem
Q1How would you modify the solution if the requirement changed to sum values that are strictly greater than K?
The modification is trivial: replace the comparison operator from '<' to '>' while keeping the same single‑pass accumulator. The time and space complexities remain O(N) and O(1) respectively.
Q2Can you compute the sum of values less than K without iterating the entire array if the array is sorted?
Yes. In a sorted array you can locate the first element ≥ K using binary search (O(log N)) and then compute the prefix sum up to that index, either by maintaining a prefix‑sum array (O(N) preprocessing) or by summing on the fly from the start to the found index, yielding O(log N + M) where M is the number of qualifying elements.
Q3What potential overflow issues arise and how would you guard against them in a language like C++ or Java?
If the telemetry values and K are 32‑bit integers, the cumulative sum may exceed the 32‑bit range. Use a 64‑bit type (long long in C++, long in Java) for the accumulator, and optionally check for overflow before addition if the language does not automatically promote the type.
Examples
Input
telemetry = [12, 5, 18, 3, 22], K = 10
Output
52
Explanation: Iterate through the array: 1. 12 > 10: Add 12 to sum (sum = 12) 2. 5 <= 10: Skip 3. 18 > 10: Add 18 to sum (sum = 30) 4. 3 <= 10: Skip 5. 22 > 10: Add 22 to sum (sum = 52) Final result is 52.
Input
telemetry = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: Iterate through the array: 1. 1 <= 10: Skip 2. 2 <= 10: Skip 3. 3 <= 10: Skip 4. 4 <= 10: Skip 5. 5 <= 10: Skip No elements exceed K. Final result is 0.
Input
telemetry = [100, 200, 300], K = 150
Output
500
Explanation: Iterate through the array: 1. 100 <= 150: Skip 2. 200 > 150: Add 200 to sum (sum = 200) 3. 300 > 150: Add 300 to sum (sum = 500) Final result is 500.
Input
telemetry = [-5, -10, 0, 5, 10], K = -3
Output
15
Explanation: Iterate through the array: 1. -5 <= -3: Skip 2. -10 <= -3: Skip 3. 0 > -3: Add 0 to sum (sum = 0) 4. 5 > -3: Add 5 to sum (sum = 5) 5. 10 > -3: Add 10 to sum (sum = 15) Final result is 15.
Constraints
- 1 <= telemetry.length <= 10^5
- -10^9 <= telemetry[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Maintain a single accumulator while scanning the array once, adding only values that satisfy the condition, achieving O(N) time and O(1) space.
Brute Force Approach
Iterate over each element and, for each, recompute the sum of all previous qualifying elements, leading to O(N^2) time.
Verified Code Solutions
function solution(nums, k) {
return nums.filter(num => num > k).reduce((a, b) => a + b, 0);
}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):
return sum(num for num in nums if num > k)function solution(nums, k) {
return nums.filter(num => num > k).reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.