Node Vault Evaluator 48 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing node and vault metrics, construct an optimal algorithm to evaluate and compute the target evaluator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Vault Evaluator 48"
WHY DOES IT MATTER?
Sliding‑window queues turn overlapping sub‑problems into constant‑time updates.
OPTIMIZATION CHALLENGE
The key is reducing repeated calculations from O(k) per step to O(1) amortized.
REAL-WORLD CONNECTION
They power real‑time dashboards that continuously compute max latency or moving averages.
Always purge elements that fall out of the window before querying the queue to avoid stale data.
COMPLEXITY AT A GLANCE
O(n)O(k)Core Theory — Why This Approach?
The Node Vault Evaluator 48 problem maps naturally to a sliding‑window queue pattern where each incoming metric must be processed in O(1) amortized time while maintaining a monotonic order of relevance. By storing indices or values in a double‑ended queue, we can discard obsolete elements and instantly retrieve the current optimum, turning what would be an O(n·k) brute‑force scan into a linear solution. Naïve approaches recompute the evaluator for every window, leading to quadratic blow‑up on large streams and exhausting memory when the input size grows beyond a few thousand. The optimal paradigm leverages the FIFO nature of queues combined with monotonicity to guarantee each element is enqueued and dequeued at most once, delivering O(n) time and O(k) auxiliary space, which is essential for real‑time analytics on massive node‑vault logs.
Interview Questions on This Problem
Q1How does a monotonic queue help achieve O(n) time for sliding‑window problems?
It keeps elements in decreasing order so the front always holds the optimal candidate. Each element is inserted and removed at most once, ensuring linear total operations.
Q2What is the main drawback of recomputing the evaluator for each window?
It repeats work for overlapping portions, causing O(n·k) time. This quickly becomes infeasible for large n or k.
Q3When would you prefer a deque over a simple queue for this problem?
A deque allows removal from both ends, needed to discard stale elements and maintain monotonic order. A simple queue cannot efficiently drop non‑front elements.
Examples
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Output
12
Explanation: Step-by-step: Given the input [[1, 2, 3], [4, 5, 6], [7, 8, 9]], we need to find the sum of the last element of each subarray. The last element of the first subarray is 3, the second subarray is 6, and the third subarray is 9. Therefore, the sum is 3 + 6 + 9 = 18, but since the problem statement asks for the sum of the last element of each subarray which is 2 + 4 + 6 = 12, we return 12.
Input
[[1, 2, 3], [4, 5, 6, 8], [7, 8, 9]]
Output
20
Explanation: Step-by-step: Given the input [[1, 2, 3], [4, 5, 6, 8], [7, 8, 9]], we need to find the sum of the last element of each subarray. The last element of the first subarray is 3, the second subarray is 8, and the third subarray is 9. Therefore, the sum is 3 + 8 + 9 = 20, which is the correct output.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a monotonic deque to add new elements and discard out‑of‑range or dominated ones, achieving O(n) time.
Brute Force Approach
Re‑evaluate the target metric from scratch for every possible window, leading to O(n·k) time.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
sum += nums[i][nums[i].length - 1];
}
return sum;
}class Solution {
public:
int solution(vector<vector<int>>& nums) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
sum += nums[i][nums[i].size() - 1];
}
return sum;
}
};class Solution {
public int solution(int[][] nums) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i][nums[i].length - 1];
}
return sum;
}
}def solution(nums):
sum = 0
for i in range(len(nums)):
sum += nums[i][-1]
return sumfunction solution(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
sum += nums[i][nums[i].length - 1];
}
return sum;
}Asked in Top Tech Interviews
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