Network Protocol Optimizer 12 — Problem Statement & Solution Guide
Problem Description
In a distributed network simulation, you are provided with a sequence of integer packet identifiers and a threshold value K. Your task is to compute the aggregate sum of all packet identifiers that meet or exceed this threshold. Specifically, given an array of integers nums and an integer K, return the sum of every element x in nums such that x >= K. If no elements satisfy the condition, return 0.
This operation models a filtering mechanism where only high-priority packets (those with identifiers at or above the threshold) contribute to the total load metric. The solution requires a single pass through the array to identify qualifying elements and accumulate their values.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Protocol Optimizer 12"
WHY DOES IT MATTER?
Aggregating conditionally filtered data is a fundamental pattern in analytics and telemetry pipelines.
OPTIMIZATION CHALLENGE
Eliminating sorting or nested loops drops the complexity from O(n log n) or O(n^2) to O(n).
REAL-WORLD CONNECTION
Think of a network router summing the sizes of packets that exceed a priority threshold before billing.
Keep the accumulator in a primitive type and avoid extra data structures to minimize cache pressure.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The task reduces to a linear aggregation problem: we must traverse the input sequence once, accumulating values that satisfy the predicate x >= K. This is a classic example of a reduction operation that can be expressed with a single pass, leveraging the associative property of addition to maintain a running total. Naïve approaches might sort the array first or use nested loops to compare each element against every other, leading to O(n log n) or O(n^2) time, which quickly becomes prohibitive for large n. The optimal paradigm is a straightforward O(n) scan that checks the threshold condition and updates the sum in constant extra space, guaranteeing scalability even for massive streams of packet identifiers.
Interview Questions on This Problem
Q1How would you handle the case where the input array is empty or all elements are below K?
Return 0 because the sum of no qualifying elements is zero. This edge case is naturally covered by initializing the accumulator to zero.
Q2Can you compute the result without modifying the original array?
Yes, the algorithm only reads each element and never writes back, preserving the input. This read‑only property is important for functional or concurrent contexts.
Q3What changes would you make if the threshold K could be negative and the array contains both positive and negative numbers?
The same linear scan works; the comparison x >= K remains valid for any integer K. No additional handling is required beyond the basic condition.
Examples
Input
nums = [10, 20, 30, 40], K = 25
Output
70
Explanation: Iterate through the array: 10 < 25 (skip), 20 < 25 (skip), 30 >= 25 (add 30), 40 >= 25 (add 40). Total sum = 30 + 40 = 70.
Input
nums = [5, 15, 25, 35], K = 50
Output
0
Explanation: Iterate through the array: 5 < 50, 15 < 50, 25 < 50, 35 < 50. No elements meet the threshold. Total sum = 0.
Input
nums = [100, 200, 300], K = 100
Output
600
Explanation: Iterate through the array: 100 >= 100 (add 100), 200 >= 100 (add 200), 300 >= 100 (add 300). Total sum = 100 + 200 + 300 = 600.
Input
nums = [-10, -5, 0, 5, 10], K = -5
Output
15
Explanation: Iterate through the array: -10 < -5 (skip), -5 >= -5 (add -5), 0 >= -5 (add 0), 5 >= -5 (add 5), 10 >= -5 (add 10). Total sum = -5 + 0 + 5 + 10 = 15.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Iterate once, add each element that satisfies x >= K to a running total, achieving O(n) time and O(1) extra space.
Brute Force Approach
Sort the array then iterate from the first element >= K, summing the rest, which costs O(n log n).
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} K
* @return {number}
*/
var sumAboveThreshold = function(nums, K) {
let sum = 0;
for (let num of nums) {
if (num >= K) {
sum += num;
}
}
return sum;
};class Solution {
public:
int sumAboveThreshold(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num >= K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int sumAboveThreshold(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num >= K) {
sum += num;
}
}
return sum;
}
}class Solution:
def sumAboveThreshold(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 sumAboveThreshold = 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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