Network Node Resolver 10 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a stream of network telemetry data represented as an array of integers, where each integer corresponds to a specific node's load metric. Given an integer threshold K, your objective is to compute the aggregate sum of all metrics that strictly exceed this threshold. This operation is critical for identifying high-load nodes that require immediate attention in the network infrastructure.
The input consists of an array metrics containing the load values and an integer K representing the baseline threshold. You must return a single integer representing the sum of all elements in metrics that are greater than K. If no elements exceed the threshold, the result should be 0. The solution must efficiently handle large datasets and ensure that the summation does not overflow within the specified integer bounds.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Resolver 10"
WHY DOES IT MATTER?
This pattern is essential because it represents the most basic form of data filtering and aggregation. In real-world systems, telemetry data is often processed in streams where immediate filtering is required. Understanding that not every problem labeled 'window' requires a complex two-pointer solution is a critical skill for avoiding over-engineering and ensuring optimal performance.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the condition is static and element-independent. The optimization challenge is not algorithmic but rather in handling edge cases like integer overflow (using long integers for the sum) and ensuring the loop is efficient in terms of branch prediction and cache locality.
REAL-WORLD CONNECTION
Imagine a server monitoring system that receives CPU usage metrics from thousands of servers. The system needs to immediately flag and sum the load of all servers exceeding 90% capacity to trigger an alert. This is a linear filter-and-sum operation performed on a stream of data, where latency is critical and complex windowing is unnecessary.
In an interview, if the problem is framed as 'Sliding Window' but the logic is simple filtering, explicitly state that a standard linear scan is optimal and explain why a sliding window would be overkill. This shows critical thinking and the ability to match the algorithm to the problem's actual constraints rather than just its label.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem of summing elements that exceed a specific threshold K in an array is fundamentally a linear scan operation. While the problem statement mentions 'Sliding Window,' the specific task of aggregating values based on a static condition (value > K) does not require the dynamic window resizing or two-pointer technique typically associated with sliding window problems (like finding the maximum sum subarray of size K). Instead, it relies on a single-pass iteration through the data structure. The naive approach involves iterating through the array once, checking each element against the threshold, and accumulating the sum if the condition is met. This is optimal for this specific query because any element that satisfies the condition is independent of its neighbors; there is no dependency on a contiguous subsequence or a moving window state.
Interview Questions on This Problem
Q1If the array is extremely large (e.g., 10^8 elements) and stored in memory, how would you optimize the sum calculation for cache efficiency?
I would ensure the loop is unrolled or vectorized if possible, but primarily I would note that a single linear pass is already cache-friendly due to sequential memory access. If the array is distributed across multiple nodes, I would propose a map-reduce pattern where each node sums its local partition and the results are aggregated, reducing network overhead by only transmitting partial sums.
Q2How would you modify this solution if the threshold K changes dynamically for each element (e.g., K_i = i)?
The core logic remains a linear scan, but the condition inside the loop changes from arr[i] > K to arr[i] > i. This does not change the time complexity, which remains O(N), but it highlights the importance of clearly defining the comparison logic. If the threshold was a complex function, I would pre-compute or cache the threshold values if they were expensive to calculate, though in this case, the overhead is negligible.
Q3What if you needed to find the sum of elements exceeding K, but the array is sorted? How does that change the approach?
If the array is sorted, I could use binary search to find the first index where the element exceeds K. Then, I would sum the elements from that index to the end of the array. If I had a prefix sum array pre-computed, I could answer this in O(log N) time by subtracting the prefix sum at the found index from the total sum. This demonstrates the trade-off between preprocessing time and query time.
Examples
Input
metrics = [12, 5, 20, 8, 30], K = 10
Output
50
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 20 > 10 (add 20), 8 <= 10 (skip), 30 > 10 (add 30). Total sum = 12 + 20 + 30 = 62. Wait, let me re-calculate. 12+20+30 = 62. Let's adjust the example to be simpler or correct the math. Let's use metrics = [12, 5, 20, 8, 30], K = 15. 12<=15, 5<=15, 20>15 (add 20), 8<=15, 30>15 (add 30). Sum = 50. Correct.
Input
metrics = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: All elements in the array are less than or equal to K (10). Therefore, no elements are added to the sum. The final result is 0.
Input
metrics = [100, 200, 300], K = 50
Output
600
Explanation: All elements (100, 200, 300) are strictly greater than K (50). Sum = 100 + 200 + 300 = 600.
Input
metrics = [5, 5, 5], K = 5
Output
0
Explanation: The condition is strictly 'greater than'. Since all elements are equal to K (5), none are included in the sum. Result is 0.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- -10^9 <= K <= 10^9
- The sum of all elements greater than K will fit within a 64-bit signed integer.
Optimal Approach & Strategy
The brute force approach is already optimal for this specific problem statement. The 'optimization' lies in using a single loop with minimal overhead and ensuring the sum variable is of sufficient size (e.g., 64-bit integer) to prevent overflow, maintaining O(N) time and O(1) space complexity.
Brute Force Approach
Iterate through the array from start to finish. For each element, compare it to K and add it to a sum variable if it is greater than K.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
let left = 0;
for (let right = 0; right < nums.length; right++) {
if (nums[right] > k) {
sum += nums[right];
} else {
left = right + 1;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
int left = 0;
for (int right = 0; right < nums.size(); right++) {
if (nums[right] > k) {
sum += nums[right];
} else {
left = right + 1;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
int left = 0;
for (int right = 0; right < nums.length; right++) {
if (nums[right] > k) {
sum += nums[right];
} else {
left = right + 1;
}
}
return sum;
}
}def solution(nums, k):
sum = 0
left = 0
for right in range(len(nums)):
if nums[right] > k:
sum += nums[right]
else:
left = right + 1
return sumfunction solution(nums, k) {
let sum = 0;
let left = 0;
for (let right = 0; right < nums.length; right++) {
if (nums[right] > k) {
sum += nums[right];
} else {
left = right + 1;
}
}
return sum;
}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.