Protocol Sensor Tracker 27 — Problem Statement & Solution Guide
Problem Description
In a distributed monitoring network, a sequence of integer readings is captured by a central protocol sensor. The system requires a specific aggregation metric known as the 'Tracker Value' to assess initial stability. Given an array of integers representing the sensor readings and an integer K, your task is to compute the sum of the first K elements in the sequence. This operation simulates a greedy selection strategy where the earliest data points are prioritized for immediate evaluation without considering subsequent values. The solution must efficiently process the input to return this cumulative sum, ensuring that the computation adheres to the operational constraints of the monitoring protocol.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Tracker 27"
WHY DOES IT MATTER?
Prefix‑sum (greedy) patterns turn repeated aggregation into a single pass, eliminating redundant work.
OPTIMIZATION CHALLENGE
Reducing from O(N*K) to O(K) cuts runtime dramatically for large N and small K.
REAL-WORLD CONNECTION
Network monitors often need the sum of recent packets to detect spikes, analogous to summing the first K readings.
Cache the running total as you read the stream; you can answer any K query instantly without re‑scanning.
COMPLEXITY AT A GLANCE
O(K)O(1)Core Theory — Why This Approach?
The problem reduces to computing a prefix sum, a classic greedy aggregation where each element contributes exactly once to the final metric. By iterating from the start and accumulating values, we guarantee the minimal work needed to obtain the sum of the first K readings.
A naĂŻve approach might recompute sums for every possible K or use nested loops, leading to O(N*K) time on large inputs, which quickly exceeds limits. The optimal paradigm leverages a single linear pass (or direct indexing when K is known) to achieve O(K) time and O(1) auxiliary space, making it scalable for massive sensor streams.
Interview Questions on This Problem
Q1How would you handle cases where K exceeds the array length?
Clamp K to the array size or return an error based on specification. This prevents out‑of‑bounds access and ensures correct results.
Q2Can you compute the sum of first K elements without an explicit loop?
Yes, by using built‑in language functions like slice and reduce or prefix sum arrays. These abstractions still run in O(K) time under the hood.
Q3What is the difference between a greedy prefix sum and a sliding‑window sum?
A prefix sum always starts at index 0, while a sliding window can start anywhere and moves across the array. Both are O(K) for a fixed window size, but their use‑cases differ.
Examples
Input
readings = [10, 20, 30, 40, 50], K = 3
Output
60
Explanation: The first 3 elements are 10, 20, and 30. Summing them yields 10 + 20 + 30 = 60.
Input
readings = [5, 15, 25], K = 2
Output
20
Explanation: The first 2 elements are 5 and 15. Summing them yields 5 + 15 = 20.
Input
readings = [100, 200, 300, 400], K = 4
Output
1000
Explanation: The first 4 elements are 100, 200, 300, and 400. Summing them yields 100 + 200 + 300 + 400 = 1000.
Input
readings = [7, 14, 21, 28], K = 1
Output
7
Explanation: The first 1 element is 7. The sum is simply 7.
Constraints
- 1 <= readings.length <= 10^5
- -10^9 <= readings[i] <= 10^9
- 1 <= K <= readings.length
Optimal Approach & Strategy
Traverse the array once, accumulating values until K elements are processed.
Brute Force Approach
Use a nested loop to recompute sums for each possible K, leading to O(N*K) time.
Verified Code Solutions
function solution(nums, k) {
let trackerValue = 0;
for (let i = 0; i < k; i++) {
trackerValue += nums[i];
}
return trackerValue;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int trackerValue = 0;
for (int i = 0; i < k; i++) {
trackerValue += nums[i];
}
return trackerValue;
}
};class Solution {
public int solution(int[] nums, int k) {
int trackerValue = 0;
for (int i = 0; i < k; i++) {
trackerValue += nums[i];
}
return trackerValue;
}
}def solution(nums, k):
tracker_value = 0
for i in range(k):
tracker_value += nums[i]
return tracker_valuefunction solution(nums, k) {
let trackerValue = 0;
for (let i = 0; i < k; i++) {
trackerValue += nums[i];
}
return trackerValue;
}Asked in Top Tech Interviews
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