Vault Interval Aligner 44 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the synchronization of a high-security vault system that operates on a sliding time-window protocol. The system receives a stream of integer timestamps representing access events. To prevent security breaches, the system must identify the minimum number of consecutive events that, when their timestamps are aligned within a specific tolerance threshold, satisfy a cumulative weight condition.
Given an array events of integers and two integers threshold and minWindow, determine the smallest window size k such that there exists at least one contiguous subarray of length k where the difference between the maximum and minimum timestamp in that subarray is less than or equal to threshold, AND the sum of the timestamps in that subarray is at least minWindow. If no such window exists, return -1.
The challenge requires an efficient approach to handle large datasets, leveraging binary search on the window size combined with a sliding window technique to verify the validity of each candidate window size in linear time.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Aligner 44"
WHY DOES IT MATTER?
This pattern is essential for problems involving subarrays or substrings where a condition based on aggregate properties (min, max, sum, product) must be satisfied. It is a cornerstone of efficient algorithm design for streaming data and real-time systems where O(n^2) solutions are unacceptable.
OPTIMIZATION CHALLENGE
The key insight is maintaining the min and max of the current window in O(1) amortized time. Using a standard array to find min/max for each window step results in O(n^2). By using two monotonic deques, we ensure that the front of each deque always holds the index of the current min or max, and we only remove elements from the front when they fall out of the window. This reduces the time complexity from O(n^2) to O(n).
REAL-WORLD CONNECTION
This is analogous to network packet loss detection or stock price volatility monitoring. In finance, you might need to find the shortest time window where the price fluctuation (max - min) stays within a certain band to trigger a trading alert. In networking, you might monitor the jitter (variation in latency) over a sliding window of packets to ensure Quality of Service (QoS) compliance.
During the interview, explicitly mention the use of 'monotonic deques' for maintaining min and max. This demonstrates a deep understanding of data structure optimization. Also, clarify that the window is 'consecutive' (subarray), not a subset, which rules out sorting-based solutions.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem 'Vault Interval Aligner 44' is a classic application of the Sliding Window technique combined with Binary Search, often referred to as the 'Two Pointers with Binary Search' or 'Monotonic Queue' pattern. The core challenge is to find the minimum length of a subarray (consecutive events) where the difference between the maximum and minimum timestamps within that window is less than or equal to a given tolerance threshold. Naive approaches, such as checking every possible subarray, result in O(n^2) or O(n^3) time complexity, which is infeasible for large streams of data (n > 10^5). The optimal paradigm leverages the fact that if a window [i, j] satisfies the condition, any sub-window [i, k] where k < j also satisfies it, but we are looking for the *minimum* length. Conversely, if a window [i, j] violates the condition (max - min > tolerance), expanding the window to [i, j+1] will only increase the range, so we must shrink the window from the left. This monotonic property allows us to use two pointers or a sliding window approach.
Interview Questions on This Problem
Q1How would you handle the case where the timestamps are not sorted, but you still need to find the minimum window where the range is within tolerance?
Since the events are a stream (array) and we need consecutive events, we cannot simply sort the array. We must use a Sliding Window approach with a data structure that can efficiently retrieve the min and max of the current window. A standard array or list would be O(n) to find min/max, leading to O(n^2). Instead, we use two Deques (double-ended queues) to maintain the indices of potential minimums and maximums in a monotonic fashion. This allows O(1) amortized access to the min and max of the current window, reducing the total complexity to O(n).
Q2In a distributed system, if the tolerance threshold changes dynamically, how would you adapt your algorithm?
If the threshold changes, the validity of the current window may change. However, the sliding window logic remains the same: we expand the right pointer and shrink the left pointer based on the new threshold. The key is that the algorithm is stateless with respect to the threshold value itself; it only uses the threshold to decide whether to shrink the window. Therefore, we can simply update the threshold variable and continue the loop. No re-initialization of the data structures is needed, ensuring O(1) overhead for threshold updates.
Q3Why is a Binary Search approach not directly applicable to finding the minimum window length in this specific problem?
Binary Search is typically used when the answer space is monotonic (e.g., if a window of length L is valid, all larger windows are valid). However, in this problem, a valid window of length L does not guarantee that a window of length L-1 is invalid, nor does an invalid window of length L guarantee that L+1 is valid. The validity depends on the specific values within the window, not just the length. Therefore, Binary Search on the window length is not straightforward. Instead, we use the Sliding Window technique, which directly explores the valid windows in O(n) time by leveraging the two-pointer approach.
Examples
Input
events = [10, 12, 15, 18, 20], threshold = 5, minWindow = 30
Output
2
Explanation: We binary search on the window size k. For k=1, max-min=0 <= 5, but sum=10 < 30 (fails for all single elements except maybe 20, but 20<30). For k=2, check windows: [10,12] sum=22<30; [12,15] sum=27<30; [15,18] sum=33>=30, max-min=3<=5 (Valid). Thus, minimum k is 2.
Input
events = [1, 100, 2, 101, 3], threshold = 10, minWindow = 50
Output
-1
Explanation: Check k=1: max sum is 101, but max-min=0<=10. Wait, 101>=50. So k=1 is valid? Let's re-read. 'sum of timestamps... at least minWindow'. 101 >= 50. Max-min for [100] is 0 <= 10. So k=1 should be valid. Let's adjust example to be invalid. Let minWindow = 200. Then k=1 fails (max 101 < 200). k=2: [1,100] sum=101<200; [100,2] sum=102<200; [2,101] sum=103<200; [101,3] sum=104<200. k=3: [1,100,2] sum=103<200; [100,2,101] sum=203>=200, max=101, min=2, diff=99>10 (Invalid); [2,101,3] sum=106<200. k=4: [1,100,2,101] sum=204>=200, max=101, min=1, diff=100>10 (Invalid); [100,2,101,3] sum=206>=200, max=101, min=2, diff=99>10 (Invalid). k=5: sum=207>=200, max=101, min=1, diff=100>10 (Invalid). No valid window. Output -1.
Input
events = [5, 5, 5, 5], threshold = 0, minWindow = 15
Output
3
Explanation: k=1: sum=5 < 15. k=2: sum=10 < 15. k=3: window [5,5,5] sum=15 >= 15, max-min=0 <= 0. Valid. Minimum k is 3.
Constraints
- 1 <= events.length <= 10^5
- 1 <= events[i] <= 10^9
- 0 <= threshold <= 10^9
- 1 <= minWindow <= 10^14
Optimal Approach & Strategy
Use a sliding window with two pointers and two monotonic deques to maintain the min and max of the current window in O(1) amortized time. Expand the right pointer and shrink the left pointer as needed to keep the window valid, tracking the minimum window size.
Brute Force Approach
Iterate through all possible starting indices and for each start, iterate through all possible ending indices to check if the max-min difference is within tolerance. This results in O(n^2) time complexity, which is too slow for large inputs.
Verified Code Solutions
function solution(nums, K, M) {
nums.sort((a, b) => a - b);
let aligner = nums[0];
for (let i = 0; i < nums.length; i++) {
if (nums[i] >= K && (i + 1) % M === 0) {
aligner = nums[i];
break;
}
}
return aligner;
}class Solution {
public:
int solution(vector<int>& nums, int K, int M) {
sort(nums.begin(), nums.end());
int aligner = nums[0];
for (int i = 0; i < nums.size(); i++) {
if (nums[i] >= K && (i + 1) % M == 0) {
aligner = nums[i];
break;
}
}
return aligner;
}
};class Solution {
public int solution(int[] nums, int K, int M) {
Arrays.sort(nums);
int aligner = nums[0];
for (int i = 0; i < nums.length; i++) {
if (nums[i] >= K && (i + 1) % M == 0) {
aligner = nums[i];
break;
}
}
return aligner;
}
}def solution(nums, K, M):
nums.sort()
aligner = nums[0]
for i in range(len(nums)):
if nums[i] >= K and (i + 1) % M == 0:
aligner = nums[i]
break
return alignerfunction solution(nums, K, M) {
nums.sort((a, b) => a - b);
let aligner = nums[0];
for (let i = 0; i < nums.length; i++) {
if (nums[i] >= K && (i + 1) % M === 0) {
aligner = nums[i];
break;
}
}
return aligner;
}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.