Tome Signal Detector 8 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a stream of sensor readings from a specialized detection array. The system requires you to identify specific high-intensity signals that exceed a baseline threshold. Given an array of integers representing the signal magnitudes and an integer k, your objective is to compute the cumulative sum of the first k elements that are strictly greater than 3. If fewer than k elements meet this criterion, return the sum of all qualifying elements found. This operation simulates a greedy selection process where you traverse the data in its original order, accumulating values only when they satisfy the intensity condition, until the quota k is fulfilled or the data stream is exhausted.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Signal Detector 8"
WHY DOES IT MATTER?
The filter‑and‑aggregate pattern is fundamental because many real‑world analytics tasks involve extracting a subset of data that meets a condition and then summarizing it. Mastering this pattern teaches candidates to avoid unnecessary data structures and to think in terms of single‑pass, constant‑space solutions.
OPTIMIZATION CHALLENGE
The key insight is early termination: once k qualifying elements are found, the scan can stop, preventing traversal of the remaining (potentially huge) tail of the array. This reduces the effective runtime from O(n) to O(m) where m is the position of the k‑th qualifying element.
REAL-WORLD CONNECTION
Consider a network intrusion detection system that flags packets exceeding a risk score. It must quickly sum the severity of the first k critical alerts to decide whether to trigger an alarm, mirroring the need to process high‑priority events in a streaming fashion.
During an interview, write the loop that checks the predicate first, then increment the counter and accumulator; avoid checking the counter before the predicate, which can lead to off‑by‑one errors when the k‑th element is exactly at the boundary.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The task reduces to a single linear scan of the input sequence while maintaining a counter of qualifying elements and an accumulator for their sum. In algorithmic terms this is a classic "filter‑and‑aggregate" pattern where we only consider elements that satisfy a predicate (value > 3) and stop once we have collected k of them. A naive solution might repeatedly search the array for the next qualifying element using nested loops, leading to O(n·k) time in the worst case, which quickly becomes prohibitive for large n (e.g., n up to 10^7). The optimal paradigm leverages the fact that the predicate is monotonic and does not require any re‑ordering or random access; a single pass suffices, giving O(n) time and O(1) auxiliary space. This approach also aligns with streaming models where the data may arrive incrementally, allowing early termination as soon as k qualifying values are observed.
Interview Questions on This Problem
Q1How would you modify the solution if the threshold (3) and the required count k were provided at runtime for each query on the same static array?
Pre‑process the array by storing the indices of all elements > threshold in a separate list. For each query, perform a binary search to locate the first index >= 0 and then sum the next k values using a prefix‑sum array built over the filtered list. This yields O(log n) per query after O(n) preprocessing.
Q2Explain why a two‑pointer technique is unnecessary for this problem, unlike typical sliding‑window problems.
Two‑pointer methods rely on a movable window whose boundaries are adjusted based on a condition that can both increase and decrease (e.g., sum ≤ X). Here we only need to count elements that satisfy a static predicate and stop after k matches; there is no need to shrink or expand a window, so a simple counter suffices.
Q3If the input stream is infinite and you must output the running sum after each qualifying element, what changes to the algorithm are required?
Maintain a running total and a count of qualifying elements seen so far. After each new element, if it > 3, increment the count and add to the total; once count reaches k, you can optionally stop processing or continue reporting the same total for subsequent elements.
Examples
Input
nums = [1, 4, 2, 5, 3, 6], k = 2
Output
9
Explanation: Traverse the array: 1 is <= 3 (skip). 4 is > 3 (take, count=1, sum=4). 2 is <= 3 (skip). 5 is > 3 (take, count=2, sum=9). Since count equals k (2), stop. Return 9.
Input
nums = [0, 1, 2, 3], k = 5
Output
0
Explanation: Traverse the array: 0, 1, 2, and 3 are all <= 3. No elements are taken. The loop ends without reaching k=5. Return the accumulated sum, which is 0.
Input
nums = [10, 11, 12, 13], k = 3
Output
33
Explanation: Traverse the array: 10 is > 3 (take, count=1, sum=10). 11 is > 3 (take, count=2, sum=21). 12 is > 3 (take, count=3, sum=33). Since count equals k (3), stop. Return 33.
Input
nums = [3, 4, 3, 5, 3, 6], k = 1
Output
4
Explanation: Traverse the array: 3 is <= 3 (skip). 4 is > 3 (take, count=1, sum=4). Since count equals k (1), stop immediately. Return 4.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- 1 <= k <= 10^5
Optimal Approach & Strategy
Perform a single linear pass, incrementing a counter and sum whenever an element >3 is seen, and break when the counter reaches k.
Brute Force Approach
Repeatedly search the array for the next element >3 using a nested loop, resulting in O(n·k) time.
Verified Code Solutions
function solution(nums, k) {
let count = 0;
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > 3) {
count++;
sum += nums[i];
if (count === k) {
return sum;
}
}
}
return -1; // Return -1 if input array has less than k elements
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int count = 0;
int sum = 0;
for (int num : nums) {
if (num > 3) {
count++;
sum += num;
if (count == k) {
return sum;
}
}
}
return -1; // Return -1 if input array has less than k elements
}
};class Solution {
public int solution(int[] nums, int k) {
int count = 0;
int sum = 0;
for (int num : nums) {
if (num > 3) {
count++;
sum += num;
if (count == k) {
return sum;
}
}
}
return -1; // Return -1 if input array has less than k elements
}
}def solution(nums, k):
count = 0
sum = 0
for num in nums:
if num > 3:
count += 1
sum += num
if count == k:
return sum
return -1 # Return -1 if input array has less than k elementsfunction solution(nums, k) {
let count = 0;
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > 3) {
count++;
sum += nums[i];
if (count === k) {
return sum;
}
}
}
return -1; // Return -1 if input array has less than k elements
}Asked in Top Tech Interviews
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