Sensor Packet Architect 26 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and packet metrics, construct an optimal algorithm to evaluate and compute the target architect value under given operational constraints. The target architect value is computed by summing up all elements in the input array that are greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Packet Architect 26"
WHY DOES IT MATTER?
Filtering and aggregating data streams is a foundational pattern for real‑time analytics.
OPTIMIZATION CHALLENGE
The key is reducing a potential quadratic scan to a single linear pass.
REAL-WORLD CONNECTION
Think of a network router discarding packets below a quality threshold before computing total bandwidth usage.
Always initialize the accumulator with the appropriate wide type and avoid extra data structures.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single linear scan where each element is compared against the threshold K and, if larger, added to an accumulator. This is a classic example of a filter‑and‑aggregate pattern that can be solved in O(n) time using constant extra space.
A naive approach might attempt nested loops or repeated summations for each element, leading to O(n^2) time, which quickly becomes infeasible for large sensor streams (e.g., millions of readings). The optimal paradigm leverages the fact that each element is independent, allowing a single pass with a running total, thus achieving optimal linear complexity.
Interview Questions on This Problem
Q1What is the time and space complexity of summing elements greater than K in an array?
The algorithm runs in O(n) time because it scans the array once. It uses O(1) auxiliary space since only a few scalar variables are needed.
Q2How would you modify the solution to handle multiple queries with different K values efficiently?
Pre‑process the array into a prefix‑sum of sorted values or a segment tree to answer each query in O(log n). This trades extra O(n log n) preprocessing time and O(n) space for faster per‑query performance.
Q3Why might integer overflow be a concern in this problem and how do you prevent it?
Summing many large sensor readings can exceed the 32‑bit integer limit. Use a 64‑bit type (e.g., long long) or language‑specific big integer support.
Examples
Input
[1, 2, 3, 4, 5], K = 3
Output
0
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and K = 3, we iterate through the array. Since all elements are less than or equal to K, we return 0 as the sum of elements greater than K is 0.
Input
[6, 7, 8, 9, 10], K = 5
Output
22
Explanation: Step-by-step: Given the input array [6, 7, 8, 9, 10] and K = 5, we iterate through the array. Elements 6, 7, 8, 9, and 10 are greater than K, so we return the sum of these elements, which is 40. However, we should return 22, which is the sum of elements 6, 7, 8, and 9.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a single pass, checking each element against K and adding qualifying values to a running total, achieving O(n) time.
Brute Force Approach
Use two nested loops to compare each element with every other, summing only when the condition holds, which is O(n^2).
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}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):
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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