Protocol Pipeline Analyzer 42 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a sequence of integer packets in a data pipeline. Given an array nums representing the packet values and an integer threshold, identify all packets whose value strictly exceeds the threshold. Compute and return the sum of these qualifying packet values.
If no packets exceed the threshold, the sum is 0. The solution must efficiently iterate through the array to accumulate the sum of valid elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Pipeline Analyzer 42"
WHY DOES IT MATTER?
This pattern exemplifies the filter‑then‑aggregate technique, a fundamental building block for data‑stream processing, analytics, and real‑time monitoring where you must extract and summarize information on the fly.
OPTIMIZATION CHALLENGE
The key insight is to merge the filtering condition and accumulation into one loop, eliminating the need for an intermediate collection and thus reducing both time (no extra pass) and space (no auxiliary array).
REAL-WORLD CONNECTION
Think of a network router that forwards packets only if their size exceeds a certain limit; the router tallies the total bytes of forwarded packets without storing each packet, mirroring the single‑pass sum of qualifying values.
During an interview, write the loop first, then immediately add the conditional check and accumulator; avoid premature optimization like building a filtered list—focus on clarity and constant‑space reasoning.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single-pass filter‑and‑aggregate operation on an integer array. In algorithmic terms, we are asked to compute the sum of a subset of elements defined by a simple predicate (value > threshold). A naïve solution might attempt nested loops or repeated scanning, but because the predicate is stateless and each element can be evaluated independently, a linear scan suffices. This is a classic example of the "single‑pass" or "streaming" paradigm, where we process each input exactly once, maintaining only constant‑size state (the running total). The optimal approach leverages this property to achieve O(n) time and O(1) auxiliary space, which scales gracefully even for very large inputs where quadratic or multi‑pass strategies would become prohibitive.
On large inputs, any algorithm that revisits elements or constructs auxiliary structures proportional to the input size (e.g., building a filtered list before summing) incurs unnecessary overhead both in time and memory. By folding the filter and aggregation steps into a single loop, we eliminate the extra pass and avoid extra allocations. This aligns with the principle of "in‑place" computation, a cornerstone of efficient algorithm design for streaming data, real‑time analytics, and low‑latency pipelines.
Interview Questions on This Problem
Q1How would you modify the solution if the threshold could be negative and the array contains both positive and negative integers?
The same linear scan works unchanged because the predicate (value > threshold) remains valid for any integer range; just ensure the accumulator is initialized to 0 (or a long type) to correctly handle negative sums.
Q2What changes are needed if the problem asks for the count of elements greater than the threshold instead of their sum?
Replace the running total with a counter variable and increment it each time an element satisfies the predicate; the overall structure and complexity remain O(n) time and O(1) space.
Q3Can you compute the sum of elements greater than the threshold in a distributed setting where the array is sharded across multiple machines?
Each shard independently runs the linear scan to compute a local sum of qualifying elements; a final reduction step aggregates these local sums, preserving O(n) total work and O(1) per‑node extra space.
Examples
Input
nums = [12, 5, 8, 20, 3], threshold = 10
Output
32
Explanation: Iterate through nums: 12 > 10 (add 12), 5 <= 10 (skip), 8 <= 10 (skip), 20 > 10 (add 20), 3 <= 10 (skip). Sum = 12 + 20 = 32.
Input
nums = [1, 2, 3, 4, 5], threshold = 10
Output
0
Explanation: No element in nums is greater than 10. Sum remains 0.
Input
nums = [100, 200, 300], threshold = 150
Output
500
Explanation: 100 <= 150 (skip), 200 > 150 (add 200), 300 > 150 (add 300). Sum = 200 + 300 = 500.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= threshold <= 10^9
Optimal Approach & Strategy
Iterate once, checking each element against the threshold and adding qualifying values directly to a running total, using only constant extra space.
Brute Force Approach
Create a new list of all elements > threshold, then iterate over that list to compute the sum; this uses two passes and extra O(n) space.
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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