Protocol Tome Detector 41 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a lightweight validation routine for a legacy data stream. The input consists of an array of integers representing signal magnitudes and a single integer threshold, k. The objective is to determine the 'detector count', defined strictly as the number of elements in the array that are strictly greater than k.
This problem is categorized under Greedy strategies in the context of linear scanning, where the optimal approach involves a single pass through the data to minimize computational overhead. Although the pattern tag suggests Recursive Backtracking, the mathematical nature of counting elements above a threshold is inherently linear and deterministic, making recursion unnecessary and inefficient for this specific metric. The solution must efficiently traverse the sequence and accumulate the count without modifying the original data structure.
Your function should accept the array of integers and the threshold value, then return the integer count of values exceeding the threshold. Ensure that your implementation handles edge cases such as empty arrays or thresholds that are higher or lower than all elements in the sequence.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Detector 41"
WHY DOES IT MATTER?
The greedy pattern here guarantees optimality with minimal overhead, which is critical when processing high‑volume data streams where latency and memory usage directly impact system performance.
OPTIMIZATION CHALLENGE
The core insight is that each element’s contribution to the count is independent; thus, a single comparison per element suffices, eliminating the need for sorting or auxiliary data structures.
REAL-WORLD CONNECTION
In distributed log aggregation, a similar pattern is used to filter out logs above a severity threshold before forwarding them to a monitoring service, ensuring only relevant alerts consume bandwidth.
When explaining this in an interview, emphasize that the greedy choice is both necessary and sufficient, and that the algorithm’s simplicity is its strength—avoid overcomplicating with unnecessary data structures.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the number of array elements strictly greater than a threshold k. While a naive solution might sort the array or use nested loops to compare each element against every other, such approaches add unnecessary time complexity and memory overhead. The optimal greedy strategy is to perform a single linear scan, incrementing a counter whenever an element exceeds k. This yields an O(n) time complexity and O(1) auxiliary space, which is optimal because every element must be examined at least once to guarantee correctness.
Greedy algorithms are characterized by making a locally optimal choice at each step with the hope of finding a global optimum. In this context, the local choice is “count this element if it is greater than k.” Since the decision to count or skip an element does not affect future decisions, the greedy approach is both correct and efficient. The key insight is that the problem reduces to a simple comparison, eliminating the need for sorting, binary search, or additional data structures.
Large inputs (e.g., millions of signal magnitudes) make any approach that involves extra passes, sorting, or auxiliary arrays impractical. The linear scan ensures that the algorithm scales linearly with input size, making it suitable for real‑time data stream validation in embedded or high‑throughput systems.
Interview Questions on This Problem
Q1How would you explain the time complexity of this solution to a hiring manager at a fintech company?
I would state that the algorithm runs in linear time, O(n), because it processes each signal magnitude exactly once. This guarantees that even with millions of data points, the runtime grows proportionally, which is essential for real‑time fraud detection pipelines.
Q2What edge cases would you test for in a production environment?
I would test an empty array, all elements equal to k, all elements greater than k, all elements less than k, and a mix of negative and positive values to ensure the comparison logic handles sign correctly.
Q3Can you extend this greedy approach to count elements within a range [a, b]?
Yes, by modifying the condition to check if a < element <= b during the single pass. The time and space complexities remain O(n) and O(1), respectively, because the logic still requires only one traversal.
Examples
Input
nums = [12, 5, 8, 15, 3], k = 10
Output
2
Explanation: Iterate through the array: 12 > 10 (count=1), 5 <= 10, 8 <= 10, 15 > 10 (count=2), 3 <= 10. The final count is 2.
Input
nums = [1, 2, 3, 4, 5], k = 0
Output
5
Explanation: All elements (1, 2, 3, 4, 5) are strictly greater than 0. The count accumulates to 5.
Input
nums = [10, 20, 30], k = 30
Output
0
Explanation: 10 <= 30, 20 <= 30, 30 is not strictly greater than 30. No elements satisfy the condition, so the output is 0.
Input
nums = [], k = 5
Output
0
Explanation: The array is empty, so there are no elements to evaluate. The count remains 0.
Constraints
- 0 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= k <= 10^9
Optimal Approach & Strategy
The optimal solution scans the array once, incrementing a counter for each element greater than k, achieving O(n) time and O(1) space.
Brute Force Approach
A naive approach might sort the array and then perform a binary search to find the first element greater than k, or use nested loops to compare each element against k, both of which add unnecessary overhead.
Verified Code Solutions
function solution(nums, k) {
let count = 0;
for (let num of nums) {
if (num > k) {
count++;
}
}
return count;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int count = 0;
for (int num : nums) {
if (num > k) {
count++;
}
}
return count;
}
};class Solution {
public int solution(int[] nums, int k) {
int count = 0;
for (int num : nums) {
if (num > k) {
count++;
}
}
return count;
}
}def solution(nums, k):
count = 0
for num in nums:
if num > k:
count += 1
return countfunction solution(nums, k) {
let count = 0;
for (let num of nums) {
if (num > k) {
count++;
}
}
return count;
}Asked in Top Tech Interviews
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