Sensor Packet Tracker 48 — 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 tracker value under given operational constraints. The algorithm should add up values greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Packet Tracker 48"
WHY DOES IT MATTER?
Greedy selection is essential here because each packet’s contribution to the sum is independent; accepting a packet cannot hurt future decisions, so a local optimum is also a global optimum. This guarantees correctness without backtracking or complex state management.
OPTIMIZATION CHALLENGE
The key insight is that you do not need to sort or store the packets; a single pass with a running sum suffices. This reduces both time from O(n log n) to O(n) and space from O(n) to O(1).
REAL-WORLD CONNECTION
In distributed telemetry systems, a gateway often needs to forward only high‑priority packets to a central server. The gateway uses a simple threshold check—exactly the greedy pattern—to decide which packets to transmit, ensuring low latency and efficient bandwidth usage.
When explaining this in an interview, emphasize that the greedy choice is safe because the objective function (sum) is additive and monotonic with respect to each element. Highlight that no future information can change the decision for a given element.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a classic greedy selection: iterate through the sensor packet values once and add each value that exceeds the threshold K to a running total. This approach is optimal because the decision to include a value is independent of any future values; once a value is known to be greater than K, it can only increase the desired sum, so we greedily accept it.
A naive solution might first sort the array or use nested loops to compare each element against every other, leading to O(n log n) or even O(n^2) time. Such approaches waste time on ordering or redundant comparisons, which becomes prohibitive when the input size reaches millions. By contrast, the greedy scan requires only a single pass, making it linear in time and constant in additional space.
The underlying algorithmic paradigm is linear traversal with conditional accumulation, a pattern that appears in many streaming and real‑time analytics scenarios where decisions must be made on the fly without full knowledge of future data.
Interview Questions on This Problem
Q1How would you explain the time complexity of the optimal solution to a hiring manager at a fintech company?
I would say the algorithm runs in O(n) time because it processes each sensor packet exactly once, performing a constant amount of work per packet. This linear time complexity is crucial for high‑frequency trading systems where latency must be minimized.
Q2What edge cases would you test for this problem in a production environment?
I would test with an empty array, all values below K, all values above K, negative values, and a threshold of zero or a very large K to ensure the algorithm correctly handles boundary conditions and does not overflow.
Q3Can you modify the algorithm to also return the count of packets above K?
Yes, by adding a counter that increments alongside the sum whenever a packet exceeds K. This adds only O(1) extra space and maintains O(n) time.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 5
Output
40
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 5, we first sort the array in ascending order. Then, we iterate through the sorted array and add up all values greater than K. In this case, the values greater than K are 6, 7, 8, 9, and 10, so the sum is 6 + 7 + 8 + 9 + 10 = 40.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 10
Output
0
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 10, we first sort the array in ascending order. Then, we iterate through the sorted array and add up all values greater than K. In this case, there are no values greater than K, so the sum is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
The optimal solution scans the array once, adding each element greater than K to a running sum. This achieves O(n) time and O(1) extra space, which is optimal for this problem.
Brute Force Approach
A naive approach might sort the array first and then iterate, or use nested loops to compare each element with every other, leading to O(n log n) or O(n^2) time. This is unnecessary because the problem only requires a single pass.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = nums.length - 1; i >= 0; i--) {
if (nums[i] > K) sum += nums[i];
else break;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (nums.size() == 0) return 0;
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = nums.size() - 1; i >= 0; i--) {
if (nums[i] > K) sum += nums[i];
else break;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0) return 0;
Arrays.sort(nums);
int sum = 0;
for (int i = nums.length - 1; i >= 0; i--) {
if (nums[i] > K) sum += nums[i];
else break;
}
return sum;
}
}def solution(nums, K):
if not nums:
return 0
nums.sort()
sum = 0
for i in range(len(nums) - 1, -1, -1):
if nums[i] > K:
sum += nums[i]
else:
break
return sumfunction solution(nums, K) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = nums.length - 1; i >= 0; i--) {
if (nums[i] > K) sum += nums[i];
else break;
}
return sum;
}Asked in Top Tech Interviews
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