Vault Interval Detector 17 — 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 detector value under given operational constraints. The target detector value is the maximum element in the array that is greater than the threshold K, or 0 if no such element exists.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Detector 17"
WHY DOES IT MATTER?
Selecting a maximum under a constraint is a common reduction that appears in many real‑time decision engines.
OPTIMIZATION CHALLENGE
The key is to avoid full sorting or nested checks, reducing the problem to a single linear pass.
REAL-WORLD CONNECTION
Think of a security system scanning sensor readings to trigger the highest alert above a safety threshold.
Keep the candidate variable immutable until a better value appears; avoid unnecessary array copies or extra data structures.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a simple selection query: find the maximum element that satisfies a predicate (value > K). A linear scan maintains a candidate maximum, updating it only when a larger qualifying element appears, guaranteeing optimality because each element must be examined at least once to ensure no larger qualifying value is missed. Naïve solutions such as sorting the entire array (O(N log N)) or using nested loops to compare each pair (O(N^2)) waste time and memory, especially when N can reach 10^7, causing time‑outs and excessive cache pressure. The optimal paradigm leverages the greedy single‑pass technique, which is both time‑optimal (Ω(N) lower bound) and space‑optimal (O(1) auxiliary space), making it ideal for high‑throughput systems.
Interview Questions on This Problem
Q1What is the time and space complexity of the optimal solution?
The optimal solution runs in O(N) time and uses O(1) extra space. It scans the array once, keeping only a single variable for the current maximum.
Q2How would you modify the algorithm if the requirement changed to the smallest element greater than K?
Initialize the candidate with a sentinel value like INF and update it when a smaller qualifying element is found. The rest of the logic stays the same, still O(N) time and O(1) space.
Q3Why is sorting not a suitable approach for this problem at scale?
Sorting incurs O(N log N) time, which is unnecessary because we only need one element, not a fully ordered list. The extra overhead can cause time‑outs on large inputs.
Examples
Input
[100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 1100, 1200, 1300, 1400]
Output
0
Explanation: Step-by-step: Given the input array, we iterate through each element. Since no element is greater than 1000, the target detector value is 0.
Input
[100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 1100, 1200, 1300, 1401]
Output
1401
Explanation: Step-by-step: Given the input array, we iterate through each element. Since 1401 is greater than 1000, the target detector value is 1401.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a single pass, tracking the maximum element that satisfies > K, returning 0 if none is found.
Brute Force Approach
Sort the array then scan from the end until you find a value > K, or use a double loop to compare every pair, both costing more than linear time.
Verified Code Solutions
function solution(nums, K) {
let max = -Infinity;
for (let num of nums) {
if (num > K) {
max = Math.max(max, num);
}
}
return max === -Infinity ? 0 : max;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int max = INT_MIN;
for (int num : nums) {
if (num > K) {
max = std::max(max, num);
}
}
return max == INT_MIN ? 0 : max;
}
};class Solution {
public int solution(int[] nums, int K) {
int max = Integer.MIN_VALUE;
for (int num : nums) {
if (num > K) {
max = Math.max(max, num);
}
}
return max == Integer.MIN_VALUE ? 0 : max;
}
}def solution(nums, K):
max_val = float('-inf')
for num in nums:
if num > K:
max_val = max(max_val, num)
return max_val if max_val != float('-inf') else 0function solution(nums, K) {
let max = -Infinity;
for (let num of nums) {
if (num > K) {
max = Math.max(max, num);
}
}
return max === -Infinity ? 0 : max;
}Asked in Top Tech Interviews
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