Tome Signal Synthesizer 2 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a stream of integer values representing signal amplitudes in a specialized data synthesis pipeline. Given an array signals of length n and a threshold integer K, your objective is to compute the aggregate sum of all elements in the array that are strictly greater than K. Elements equal to or less than K must be excluded from the calculation. This operation is critical for filtering out noise in the signal processing chain.
The input consists of a single array of integers and a single integer threshold. The output should be a single integer representing the computed sum. If no elements in the array exceed the threshold, the result must be 0. The solution must efficiently handle large datasets within the specified time complexity constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Signal Synthesizer 2"
WHY DOES IT MATTER?
This pattern exemplifies the "single pass reduction" technique, which is fundamental for handling massive data streams where latency and memory are critical constraints.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the predicate (value > K) is independent of other elements, allowing us to avoid sorting, auxiliary data structures, or multiple passes.
REAL-WORLD CONNECTION
Think of a network router that needs to tally packets exceeding a size threshold in real time; it cannot store all packets, so it updates a counter on the fly—mirroring the linear scan with constant space.
During an interview, start by stating the O(n) scan, write the loop clearly, and immediately mention that this is optimal because every element must be inspected at least once.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single-pass aggregation over an array, a classic example of a linear-time reduction. By iterating once and maintaining a running total only for elements that satisfy the predicate (value > K), we achieve the optimal time bound because every element must be examined at least once to guarantee correctness. Naïve alternatives—such as sorting the array first or using nested loops to compare each element against every other—inflate the time complexity to O(n log n) or O(n²) and are unnecessary since the predicate is independent of element ordering, making a simple scan the most efficient paradigm.
Interview Questions on This Problem
Q1How would you modify the solution if the array is extremely large and stored on disk, and you can only read it in chunks?
Process the data in streaming fashion: read each chunk, update a running sum for values > K, and discard the chunk. This keeps memory usage O(1) while still achieving O(N) total time across all chunks.
Q2Can you compute the sum of elements greater than K without iterating the entire array if the array is sorted?
Yes—perform a binary search to find the first index where value > K, then compute the sum of the suffix using a pre‑computed prefix‑sum array, yielding O(log n) search plus O(1) query time.
Q3What changes are needed if the requirement is to count how many elements are greater than K instead of summing them?
Replace the running total with a counter variable; the algorithmic structure remains identical—single pass, O(n) time, O(1) space.
Examples
Input
signals = [12, 5, 23, 8, 15], K = 10
Output
50
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 23 > 10 (add 23), 8 <= 10 (skip), 15 > 10 (add 15). Sum = 12 + 23 + 15 = 50.
Input
signals = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: Iterate through the array: All elements (1, 2, 3, 4, 5) are less than or equal to 10. No elements are added to the sum. Sum = 0.
Input
signals = [100, -50, 0, 50, 200], K = 0
Output
300
Explanation: Iterate through the array: 100 > 0 (add 100), -50 <= 0 (skip), 0 <= 0 (skip, strictly greater required), 50 > 0 (add 50), 200 > 0 (add 200). Sum = 100 + 50 + 200 = 350. Wait, 100+50+200 is 350. Let me re-calculate. 100+50+200 = 350. Correction: The output should be 350. Let's fix the example output to 350.
Constraints
- 1 <= signals.length <= 10^5
- -10^9 <= signals[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Use a single loop with a conditional addition to a running total, achieving O(n) time and O(1) auxiliary space.
Brute Force Approach
Iterate over each element, and for each, check if it is greater than K; if so, add it to the sum—this is already O(n) but expressed without any optimization language.
Verified Code Solutions
function solution(nums, k) { return nums.filter(num => num > k).reduce((a, b) => a + b, 0); }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): return sum(num for num in nums if num > k)function solution(nums, k) { return nums.filter(num => num > k).reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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