Vault Interval Optimizer 3 — Problem Statement & Solution Guide
Problem Description
In a high-frequency trading environment, a monitoring system tracks a stream of integer-valued vault interval metrics. You are provided with an array metrics of length n and a threshold integer K. Your task is to compute the aggregate sum of all elements in metrics that are strictly greater than K. The solution must be efficient enough to handle large-scale data streams, ensuring linear time complexity relative to the input size. You may utilize direct scanning or frequency-based aggregation strategies to achieve optimal performance. The output should be a single integer representing the total sum of qualifying elements. If no elements exceed the threshold, return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Optimizer 3"
WHY DOES IT MATTER?
Filtering and aggregating over a stream is a foundational pattern in real‑time analytics, enabling systems to compute KPIs on the fly without persisting the entire dataset.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that the predicate is independent of other elements, allowing a single pass with a running accumulator instead of costly multi‑pass or auxiliary‑structure solutions.
REAL-WORLD CONNECTION
Think of a network router that counts packets exceeding a size threshold; each packet is examined once, and qualifying sizes are summed to monitor bandwidth spikes.
During an interview, write the loop that both dequeues (or iterates) and updates the sum in the same line – it demonstrates awareness of constant‑space streaming and avoids unnecessary temporary containers.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single pass filter‑and‑aggregate operation over a stream of integers. In the naive world, one might repeatedly scan the array for each query or maintain auxiliary structures that incur extra overhead, which quickly becomes prohibitive when n reaches millions or when the data arrives as a live stream. The optimal paradigm leverages the linearity of addition and the fact that the predicate (value > K) is stateless – it depends only on the current element, not on any history. By iterating once, checking the condition, and accumulating the qualifying values, we achieve O(n) time with O(1) auxiliary space, which is the theoretical lower bound for any algorithm that must inspect each element at least once.
When the input is presented as a queue (or any FIFO container), the same principle applies: dequeue each metric, evaluate it, and add it to a running total if it exceeds K. This approach respects the streaming nature of high‑frequency trading data, allowing the algorithm to work in an online fashion without storing the entire array. The key insight is that no sorting, heap, or segment tree is required because the aggregation does not depend on ordering or range queries – a simple linear scan suffices.
Interview Questions on This Problem
Q1How would you modify the solution if you needed the sum of the top‑M values greater than K instead of all values?
Maintain a min‑heap of size M while scanning. For each element > K, push it onto the heap; if the heap exceeds M, pop the smallest. After the scan, sum the heap contents. This runs in O(n log M) time and O(M) space.
Q2Explain how you could compute the same sum in a distributed system where the array is partitioned across multiple nodes.
Each node independently computes the local sum of elements > K using the linear scan. Then a reduce step aggregates the local sums (e.g., via a map‑reduce shuffle) to produce the global total. The overall complexity remains O(n) distributed, with only O(1) communication per node.
Q3Why is a queue a suitable data structure for processing a live stream of metrics in this problem?
A queue naturally models FIFO arrival of metrics, allowing constant‑time enqueue for new data and constant‑time dequeue for processing. Since the aggregation only needs the current element, we can dequeue, evaluate, and discard without extra storage, preserving O(1) space per element.
Examples
Input
metrics = [12, 5, 18, 3, 22, 7], K = 10
Output
40
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 18 > 10 (add 18), 3 <= 10 (skip), 22 > 10 (add 22), 7 <= 10 (skip). Sum = 12 + 18 + 22 = 40.
Input
metrics = [1, 2, 3, 4, 5], K = 100
Output
0
Explanation: No element in the array is strictly greater than 100. Therefore, the aggregate sum is 0.
Input
metrics = [100, 200, 300, 400], K = 150
Output
900
Explanation: 100 <= 150 (skip), 200 > 150 (add 200), 300 > 150 (add 300), 400 > 150 (add 400). Sum = 200 + 300 + 400 = 900.
Input
metrics = [-5, -10, 0, 5, 10], K = -3
Output
15
Explanation: -5 <= -3 (skip), -10 <= -3 (skip), 0 > -3 (add 0), 5 > -3 (add 5), 10 > -3 (add 10). Sum = 0 + 5 + 10 = 15.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- -10^9 <= K <= 10^9
- The sum of all elements exceeding K is guaranteed to fit within a 64-bit signed integer.
Optimal Approach & Strategy
Perform a single linear pass, checking each element against K and accumulating qualifying values, achieving O(n) time and O(1) extra space.
Brute Force Approach
Repeatedly scan the entire array for each element or query, leading to O(n^2) time.
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
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.