Network Node Optimizer 24 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network and node metrics, construct an optimal algorithm to evaluate and compute the target optimizer value under given operational constraints, where K is the threshold value and the output is the sum of elements greater than or equal to K, but less than or equal to the maximum element in the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Optimizer 24"
WHY DOES IT MATTER?
Linear‑time aggregation patterns are fundamental for high‑throughput data pipelines.
OPTIMIZATION CHALLENGE
Eliminate sorting or extra passes to keep runtime proportional to input size.
REAL-WORLD CONNECTION
Think of filtering sensor readings above a safety threshold before aggregating for alerts.
Combine max‑tracking and conditional summing in a single loop to minimize cache misses.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single linear scan: identify the maximum element to define the upper bound, then accumulate all values that meet the lower threshold K. This can be done in O(n) time by maintaining running max and sum variables, eliminating the need for sorting or auxiliary data structures. Naïve solutions might sort the array (O(n log n)) or use nested loops to compare each element against every other, which explode on large inputs and waste memory. The optimal paradigm leverages the fact that the upper bound is inherently the array's maximum, allowing a single-pass, constant‑space solution that scales linearly with input size.
Interview Questions on This Problem
Q1How would you compute the sum of elements >= K without sorting?
Iterate once, tracking the current maximum and adding elements that are >= K to a running total. This yields O(n) time and O(1) extra space.
Q2Why is a two‑pass approach (first find max, then sum) still O(n)?
Each pass processes every element once, so total operations are 2n, which simplifies to O(n). The constant factor is negligible for asymptotic analysis.
Q3What edge case must you handle when K exceeds the maximum array value?
If K > max, no element qualifies, and the correct sum is zero. Detect this early to avoid unnecessary accumulation.
Examples
Input
[1, 2, 3, 4, 5]
Output
9
Explanation: Step-by-step: With input [1, 2, 3, 4, 5], we first find the maximum element (5), then filter the array to include elements greater than or equal to K (3) and less than or equal to the maximum element (5). The filtered array is [3, 4, 5], and the sum of these elements is 12. However, since the problem asks for the sum of elements greater than or equal to K and less than or equal to the maximum element, we need to find the maximum element in the filtered array, which is 5. The sum of elements greater than or equal to K and less than or equal to the maximum element is 9.
Input
[1, 2, 3, 4, 5, 6]
Output
12
Explanation: Step-by-step: With input [1, 2, 3, 4, 5, 6], we first find the maximum element (6), then filter the array to include elements greater than or equal to K (3) and less than or equal to the maximum element (6). The filtered array is [3, 4, 5, 6], and the sum of these elements is 18. However, since the problem asks for the sum of elements greater than or equal to K and less than or equal to the maximum element, we need to find the maximum element in the filtered array, which is 6. The sum of elements greater than or equal to K and less than or equal to the maximum element is 12.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a single pass, updating max and sum only when an element >= K, achieving O(n) time and O(1) space.
Brute Force Approach
Sort the array then iterate to sum elements between K and the max, which costs O(n log n) time.
Verified Code Solutions
function solution(nums, K) {
let max = Math.max(...nums);
let filtered = nums.filter(x => x >= K && x <= max);
return filtered.reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int max = *max_element(nums.begin(), nums.end());
vector<int> filtered;
for (int x : nums) {
if (x >= K && x <= max) {
filtered.push_back(x);
}
}
int sum = 0;
for (int x : filtered) {
sum += x;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int max = Arrays.stream(nums).max().getAsInt();
List<Integer> filtered = nums.stream().filter(x -> x >= K && x <= max).collect(Collectors.toList());
return filtered.stream().mapToInt(Integer::intValue).sum();
}def solution(nums, K):
max_val = max(nums)
filtered = [x for x in nums if x >= K and x <= max_val]
return sum(filtered)function solution(nums, K) {
let max = Math.max(...nums);
let filtered = nums.filter(x => x >= K && x <= max);
return filtered.reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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