Vault Interval Extractor 30 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and interval metrics, construct an optimal algorithm to evaluate and compute the target extractor value under given operational constraints. The algorithm should sum the elements in the array that are greater than or equal to the threshold K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Extractor 30"
WHY DOES IT MATTER?
The linear‑scan greedy pattern is fundamental for any problem where decisions are independent and additive, enabling constant‑space, linear‑time solutions that scale to massive data volumes.
OPTIMIZATION CHALLENGE
The key insight is recognizing that sorting or complex data structures are unnecessary because the inclusion criterion is a simple comparison against a constant K, allowing the problem to be reduced to a single pass.
REAL-WORLD CONNECTION
Think of a financial audit system that needs to total all transactions above a compliance threshold; the auditor can process logs line‑by‑line without loading the entire dataset, mirroring the streaming variant of this algorithm.
During an interview, write the loop first, then immediately add the conditional check (value >= K) and the accumulation; this demonstrates clarity of thought and avoids over‑engineering.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem asks for the sum of all array elements that meet or exceed a given threshold K. This is a classic greedy scenario where each element can be evaluated independently; there is no need for combinatorial decisions or ordering because the contribution of each qualifying element to the final sum is additive and monotonic. A naive solution might attempt to sort the array or use nested loops to compare each element with every other, which inflates the time complexity to O(N log N) or O(N^2) and quickly becomes infeasible for large N (e.g., N > 10^6). The optimal paradigm leverages the fact that the decision to include an element depends solely on its value relative to K, allowing a single linear pass that accumulates the answer in O(N) time and O(1) auxiliary space. This linear‑time greedy scan is provably optimal because any algorithm must inspect each element at least once to determine whether it satisfies the threshold, establishing a lower bound of Ω(N).
Interview Questions on This Problem
Q1How would you modify the solution if the input array is streamed and cannot be stored entirely in memory?
Maintain a running total and a simple counter while reading each element from the stream; for each value, if it is >= K, add it to the accumulator. Since only the sum and K are needed, the memory footprint stays O(1) regardless of stream length.
Q2What changes are required if the problem asks for the count of elements >= K instead of their sum?
Replace the accumulator with a count variable and increment it each time an element meets the threshold. The algorithmic structure remains identical—still a single pass with O(N) time and O(1) space.
Q3Can you extend the solution to handle multiple queries, each with a different K, efficiently?
Sort the array once (O(N log N)) and compute a prefix‑sum array. For each query K, binary‑search the first index where value >= K, then retrieve the sum of the suffix using the prefix sums in O(log N) per query, achieving O(N log N + Q log N) total time.
Examples
Input
[3, 4, 5, 2, 1, 6]
Output
9
Explanation: Step-by-step: Given the input array [3, 4, 5, 2, 1, 6] and threshold K = 3, we iterate through the array and sum the elements that are greater than or equal to 3. The elements 3, 4, and 5 meet this condition, so the output is 3 + 4 + 5 = 12. However, we should handle edge cases where the input array is empty or contains only one element.
Input
[30, 40, 50, 20, 10, 60]
Output
150
Explanation: Step-by-step: Given the input array [30, 40, 50, 20, 10, 60] and threshold K = 25, we iterate through the array and sum the elements that are greater than or equal to 25. The elements 30, 40, and 50 meet this condition, so the output is 30 + 40 + 50 = 120. However, we should handle edge cases where the input array is empty or contains only one element.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
The optimal solution makes a single pass, checking each element against K and accumulating the sum, achieving O(N) time and O(1) auxiliary space.
Brute Force Approach
A naive method would sort the array and then sum elements, costing O(N log N) time, or use nested loops to compare each element with every other, leading to O(N^2).
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
if (nums.length === 0) return 0;
for (let num of nums) {
if (num >= K) sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (nums.empty()) return 0;
int sum = 0;
for (int num : nums) {
if (num >= K) sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0) return 0;
int sum = 0;
for (int num : nums) {
if (num >= K) sum += num;
}
return sum;
}
}def solution(nums, K):
if not nums:
return 0
return sum(num for num in nums if num >= K)function solution(nums, K) {
let sum = 0;
if (nums.length === 0) return 0;
for (let num of nums) {
if (num >= K) sum += num;
}
return sum;
}Asked in Top Tech Interviews
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