Protocol Pipeline Synthesizer 48 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a linear stream of integer metrics derived from a network protocol pipeline. Given an array of integers representing these metrics and a threshold value K, your objective is to compute the aggregate sum of all elements that strictly exceed the threshold K. Elements equal to or less than K are considered noise and must be excluded from the calculation. If no elements satisfy the condition, the result is zero.
The input consists of a single array of integers and an integer K. The output is a single integer representing the computed sum. This problem models a filtering and aggregation operation common in real-time data processing systems where only significant deviations above a baseline are of interest.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Pipeline Synthesizer 48"
WHY DOES IT MATTER?
Filtering and aggregation in a single pass is essential for high-throughput systems where latency and memory usage directly impact performance. It ensures that each data element is processed exactly once, minimizing CPU cycles and cache misses.
OPTIMIZATION CHALLENGE
The key insight is that the condition "value > K" can be evaluated on the fly, and the sum can be updated incrementally. This eliminates the need for auxiliary data structures or multiple traversals, reducing both time and space complexity.
REAL-WORLD CONNECTION
In network packet inspection, routers must quickly decide whether to forward or drop packets based on size or priority thresholds. A single-pass filter that aggregates statistics (e.g., total bytes above a threshold) is analogous to this algorithm.
When explaining this to an interviewer, emphasize the streaming nature of the solution and highlight that the algorithm is O(n) time, O(1) space, and can be extended to handle dynamic thresholds or distributed data with minimal changes.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a classic linear filtering and aggregation task. We must traverse the array once, compare each metric to the threshold K, and accumulate the sum of those strictly greater than K. A naive approach that uses nested loops or repeatedly scans the array for each element would be O(n^2) and is unnecessary; the optimal paradigm is a single-pass scan, which is O(n) time and O(1) auxiliary space. This pattern is fundamental in streaming data processing, where we cannot afford to store or reprocess data multiple times. By maintaining a running total and updating it only when the condition holds, we avoid any extra memory overhead and achieve the best possible performance for this class of problems.
Interview Questions on This Problem
Q1How would you adapt this algorithm if the threshold K changes frequently during runtime?
You would precompute a prefix sum array of the original metrics and then, for each new K, perform a binary search to find the first index where the metric exceeds K. The sum of elements greater than K can then be obtained by subtracting the prefix sum up to that index from the total sum. This allows each query to run in O(log n) time after an initial O(n) preprocessing step.
Q2If the array of metrics cannot fit into memory, how would you compute the sum of elements greater than K?
You would process the data in a streaming fashion: read chunks of the array from disk or a network stream, maintain a running sum of elements that exceed K, and discard the chunk after processing. This approach uses O(1) memory regardless of the array size and ensures the algorithm scales to arbitrarily large inputs.
Q3How can you parallelize this sum computation across multiple threads or machines?
Divide the array into disjoint segments, assign each segment to a thread or worker, and compute a local sum of elements greater than K. After all workers finish, perform a reduction (e.g., sum the local sums) to obtain the global result. Care must be taken to avoid race conditions and to handle the case where K is shared across workers.
Examples
Input
nums = [12, 5, 20, 8, 15], K = 10
Output
47
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 20 > 10 (add 20), 8 <= 10 (skip), 15 > 10 (add 15). Total sum = 12 + 20 + 15 = 47.
Input
nums = [3, 3, 3, 3], K = 3
Output
0
Explanation: All elements are equal to K. Since the condition is strictly greater than K, no elements are added. Total sum = 0.
Input
nums = [-5, -2, 0, 1, 100], K = -1
Output
101
Explanation: Check each element: -5 <= -1 (skip), -2 <= -1 (skip), 0 > -1 (add 0), 1 > -1 (add 1), 100 > -1 (add 100). Total sum = 0 + 1 + 100 = 101.
Input
nums = [1000000, 999999, 1000001], K = 1000000
Output
1000001
Explanation: 1000000 is not greater than 1000000 (skip). 999999 is less than 1000000 (skip). 1000001 is greater than 1000000 (add 1000001). Total sum = 1000001.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Use the same single-pass scan but emphasize that you avoid any extra data structures or repeated passes. The algorithm runs in O(n) time and O(1) space, which is optimal for this problem.
Brute Force Approach
Loop through each element, check if it’s greater than K, and add it to a sum variable. This is a single-pass O(n) algorithm, but it’s often described as the naive approach because it directly implements the problem statement without any optimization.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0 || nums.length === 1) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = nums.length - 1; i >= 0; i--) {
if (nums[i] > K) sum += nums[i];
else break;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (nums.size() == 0 || nums.size() == 1) return 0;
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = nums.size() - 1; i >= 0; i--) {
if (nums[i] > K) sum += nums[i];
else break;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0 || nums.length == 1) return 0;
Arrays.sort(nums);
int sum = 0;
for (int i = nums.length - 1; i >= 0; i--) {
if (nums[i] > K) sum += nums[i];
else break;
}
return sum;
}
}def solution(nums, K):
if len(nums) == 0 or len(nums) == 1:
return 0
nums.sort()
sum = 0
for i in range(len(nums) - 1, -1, -1):
if nums[i] > K:
sum += nums[i]
else:
break
return sumfunction solution(nums, K) {
if (nums.length === 0 || nums.length === 1) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = nums.length - 1; i >= 0; i--) {
if (nums[i] > K) sum += nums[i];
else break;
}
return sum;
}Asked in Top Tech Interviews
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