Network Network Detector 14 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a signal aggregation routine for a distributed sensor array. The system receives a stream of integer readings, each representing the intensity of a specific network node. To optimize bandwidth usage, the control unit must isolate the most significant signals for immediate processing. Your objective is to identify the K highest intensity values from the input sequence and compute their cumulative sum. This metric serves as the primary threshold for triggering downstream alert mechanisms.
Given an array of integers representing the sensor readings and an integer K indicating the number of top signals to retain, return the sum of the K largest elements. If K exceeds the total number of readings, the sum should encompass all available values. The solution must efficiently handle large datasets by leveraging sorting or selection algorithms to minimize computational overhead while ensuring numerical accuracy.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Network Detector 14"
WHY DOES IT MATTER?
Finding the optimal contiguous segment is a core sub‑problem in many signal‑processing and financial‑analysis tasks.
OPTIMIZATION CHALLENGE
The challenge is collapsing O(n²) overlapping sum calculations into a single linear scan.
REAL-WORLD CONNECTION
It mirrors a router selecting the burst of highest traffic intensity for priority handling.
Always keep two variables—current sum and global max—to avoid unnecessary array storage.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The maximum‑sum contiguous sub‑array problem can be solved with a linear‑time dynamic programming technique known as Kadane’s algorithm. The key insight is to maintain, for each index, the best sub‑array ending at that position, which is either the current element alone or the current element added to the best sub‑array ending at the previous index. By propagating this local optimum forward, the global optimum emerges without enumerating all O(n²) sub‑arrays, which would be infeasible for large inputs.\nNaïve enumeration recomputes sums for overlapping intervals, leading to quadratic time and excessive memory usage, especially when n reaches 10⁵ or more. The DP paradigm reduces the problem to a single pass, storing only two scalar values (current best ending here and overall best), thus achieving O(n) time and O(1) auxiliary space while preserving correctness through optimal substructure and overlapping sub‑problems.
Interview Questions on This Problem
Q1What is the recurrence relation used in Kadane’s algorithm?
maxEndingHere = max(nums[i], maxEndingHere + nums[i]); maxSoFar = max(maxSoFar, maxEndingHere).
Q2How does Kadane’s algorithm handle an array of all negative numbers?
It correctly returns the largest (least negative) element because maxEndingHere is re‑initialized with each element when adding would decrease the sum.
Q3Can Kadane’s algorithm be adapted to find the sub‑array with the maximum product?
Yes, but you must track both maximum and minimum products at each step because a negative number can flip the sign.
Examples
Input
readings = [12, 45, 7, 90, 33], K = 2
Output
135
Explanation: The input array is [12, 45, 7, 90, 33]. Sorting the array in descending order yields [90, 45, 33, 12, 7]. The top 2 values are 90 and 45. Their sum is 90 + 45 = 135.
Input
readings = [5, 5, 5, 5], K = 3
Output
15
Explanation: The input array is [5, 5, 5, 5]. Sorting in descending order results in [5, 5, 5, 5]. The top 3 values are 5, 5, and 5. Their sum is 5 + 5 + 5 = 15.
Input
readings = [-10, -20, -5, -15], K = 2
Output
-15
Explanation: The input array is [-10, -20, -5, -15]. Sorting in descending order yields [-5, -10, -15, -20]. The top 2 values are -5 and -10. Their sum is -5 + (-10) = -15.
Input
readings = [100, 200, 300], K = 5
Output
600
Explanation: The input array is [100, 200, 300]. Since K (5) is greater than the array length (3), all elements are considered. The sum of all elements is 100 + 200 + 300 = 600.
Constraints
- 1 <= readings.length <= 10^5
- -10^9 <= readings[i] <= 10^9
- 1 <= K <= 10^5
Optimal Approach & Strategy
Iterate once, updating a running ‘max ending here’ and a global ‘max so far’, achieving O(n) time and O(1) space.
Brute Force Approach
Enumerate every possible start and end index, compute each sub‑array sum, and track the maximum, which costs O(n²) time.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int i = nums.size() - k; i < nums.size(); i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = nums.length - k; i < nums.length; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
return sum(nums[:k])function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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