Vault Interval Validator 33 — 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 validator value under given operational constraints. The target validator value is the sum of elements greater than K in each window of size W.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Validator 33"
WHY DOES IT MATTER?
Sliding windows turn quadratic scans into linear passes, essential for real‑time analytics.
OPTIMIZATION CHALLENGE
The key is to avoid recomputing the sum for overlapping windows by updating only the changed elements.
REAL-WORLD CONNECTION
Think of monitoring vault transaction streams where you need the sum of high‑value events over the last W seconds.
Keep a running total and conditionally adjust it as elements enter or leave; avoid extra containers.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The sliding‑window technique transforms a problem that naïvely requires recomputing a metric for every subarray into a linear‑time scan by reusing information from the previous window. In the Vault Interval Validator, a naïve solution would iterate over each of the N‑W+1 windows and, for each, sum all elements greater than K, resulting in O(N·W) time which explodes for large N and W.
The optimal paradigm maintains a running sum of only those elements that satisfy the >K condition. As the window slides one position, the element exiting the window is subtracted from the sum if it was >K, and the new entering element is added if it exceeds K. This constant‑time update per step yields an overall O(N) time algorithm with O(1) auxiliary space, perfectly scaling to massive input streams.
Interview Questions on This Problem
Q1How does the sliding‑window approach reduce the time complexity for this problem?
It updates the answer incrementally instead of recomputing from scratch for each window. Each slide costs O(1), giving O(N) total.
Q2What edge cases must you handle when implementing the validator?
When the array length is smaller than the window size, no full window exists. Also, K can be negative, so every element might qualify.
Q3Can you compute the result using a deque or other data structure?
A deque is unnecessary because we only need the sum of qualifying elements, not their order. Simple variables suffice for O(1) updates.
Examples
Input
[1, 2, 3, 4, 5], 10, 3
Output
0
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], K = 10, and window size W = 3, we calculate the sum of elements in each window. For the first window [1, 2, 3], the sum is 6, which is less than K. For the second window [2, 3, 4], the sum is 9, which is less than K. For the third window [3, 4, 5], the sum is 12, which is also less than K. Since there are no elements greater than K in any window, the output is 0.
Input
[10, 20, 30, 40, 50], 20, 3
Output
150
Explanation: Step-by-step: with input [10, 20, 30, 40, 50], K = 20, and window size W = 3, we calculate the sum of elements greater than K in each window. For the first window [10, 20, 30], the sum of elements greater than K is 30 + 20 = 50. For the second window [20, 30, 40], the sum of elements greater than K is 30 + 40 = 70. For the third window [30, 40, 50], the sum of elements greater than K is 30 + 40 + 50 = 120. The total sum of elements greater than K is 50 + 70 + 30 = 150.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a running sum of elements > K; when the window moves, subtract the outgoing element if it was > K and add the incoming one if it exceeds K, achieving O(N) time.
Brute Force Approach
Iterate over each possible window and, for each, scan all W elements to sum those > K, leading to O(N·W) time.
Verified Code Solutions
function solution(nums, k, w) {
let sum = 0;
for (let i = 0; i <= nums.length - w; i++) {
let windowSum = 0;
for (let j = i; j < i + w; j++) {
if (nums[j] > k) {
windowSum += nums[j];
}
}
sum += windowSum;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k, int w) {
int sum = 0;
for (int i = 0; i <= nums.size() - w; i++) {
int windowSum = 0;
for (int j = i; j < i + w; j++) {
if (nums[j] > k) {
windowSum += nums[j];
}
}
sum += windowSum;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k, int w) {
int sum = 0;
for (int i = 0; i <= nums.length - w; i++) {
int windowSum = 0;
for (int j = i; j < i + w; j++) {
if (nums[j] > k) {
windowSum += nums[j];
}
}
sum += windowSum;
}
return sum;
}
}def solution(nums, k, w):
total_sum = 0
for i in range(len(nums) - w + 1):
window_sum = 0
for j in range(i, i + w):
if nums[j] > k:
window_sum += nums[j]
total_sum += window_sum
return total_sumfunction solution(nums, k, w) {
let sum = 0;
for (let i = 0; i <= nums.length - w; i++) {
let windowSum = 0;
for (let j = i; j < i + w; j++) {
if (nums[j] > k) {
windowSum += nums[j];
}
}
sum += windowSum;
}
return sum;
}Asked in Top Tech Interviews
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