Network Network Tracker 46 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network and network metrics, and an integer K, construct an optimal algorithm to evaluate and compute the sum of all elements greater than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Network Tracker 46"
WHY DOES IT MATTER?
Threshold‑based aggregation appears in many analytics pipelines where only values above a cutoff matter.
OPTIMIZATION CHALLENGE
Eliminating sorting reduces the algorithm from O(N log N) to linear time, crucial for high‑throughput streams.
REAL-WORLD CONNECTION
Network monitoring systems often sum traffic volumes exceeding a danger threshold to trigger alerts.
Implement the scan with early exit for empty inputs and use a 64‑bit accumulator to avoid overflow.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
A naive solution would sort the array or insert every element into a heap, then traverse to accumulate values >= K, leading to O(N log N) time and extra space. The optimal paradigm leverages the fact that the condition is a simple threshold, allowing a single linear scan that adds qualifying elements directly, achieving O(N) time and O(1) auxiliary space. This approach exploits the monotonic nature of the predicate (value >= K) and avoids unnecessary ordering operations, which become prohibitive on large inputs. By treating the problem as a filter‑and‑aggregate task rather than a selection or ranking problem, we reduce both time and memory footprints dramatically.
Interview Questions on This Problem
Q1Why is sorting or heap insertion unnecessary for computing the sum of elements >= K?
Sorting or heap insertion adds O(N log N) overhead, but the threshold check is order‑independent, so a single pass suffices.
Q2What is the time and space complexity of the optimal solution?
Time complexity is O(N) and auxiliary space is O(1) because we only maintain a running total.
Q3How would you handle potential integer overflow when summing large numbers?
Use a wider integer type (e.g., long long in C++ or Python's arbitrary‑precision int) or check for overflow before addition.
Examples
Input
[3, 4, 5], 3
Output
7
Explanation: Step-by-step: with input [3, 4, 5] and K = 3, we filter elements greater than or equal to K, giving us [3, 4]. Then, we sum these elements, resulting in 3 + 4 = 7.
Input
[30, 40, 50], 25
Output
70
Explanation: Step-by-step: with input [30, 40, 50] and K = 25, we filter elements greater than or equal to K, giving us [30, 40]. Then, we sum these elements, resulting in 30 + 40 = 70.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a single pass, adding elements >= K to a running sum, achieving O(N) time and O(1) space.
Brute Force Approach
Sort the array or push all elements into a heap, then sum the qualifying tail, costing O(N log N).
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
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