Vault Buffer Analyzer 18 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and buffer metrics, construct an optimal algorithm to evaluate and compute the target analyzer value under given operational constraints. The input array is not specified to be sorted, so the algorithm should select the first K elements regardless of the order.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Buffer Analyzer 18"
WHY DOES IT MATTER?
Sliding windows enable constant‑time updates for streaming data, which is critical when latency and throughput are paramount. They avoid recomputation and allow algorithms to run in linear time regardless of window size.
OPTIMIZATION CHALLENGE
The key insight is to maintain an aggregate (sum, max, etc.) incrementally: add the incoming element and subtract the outgoing one. This transforms an O(K) recomputation into O(1) per step, reducing overall time from O(n·K) to O(n).
REAL-WORLD CONNECTION
Consider a monitoring dashboard that shows the average CPU usage over the last 5 minutes. The system receives a new metric every second; a sliding window keeps the last 300 samples and updates the average instantly, ensuring the UI stays responsive.
When implementing, use a ring buffer or fixed‑size array with modulo indexing to avoid costly queue operations. Also, guard against integer overflow when summing large values by using 64‑bit types.
COMPLEXITY AT A GLANCE
O(n)O(K)Core Theory — Why This Approach?
The problem reduces to a classic sliding‑window or queue pattern: you must process a stream of vault and buffer metrics, keep only the most recent K elements, and compute a target value (e.g., the sum, average, or maximum) for that window. Naïve solutions recompute the target from scratch for every new element, leading to O(n·K) time and potentially O(K) space for each recomputation. This is infeasible for large streams (millions of metrics). The optimal paradigm maintains a running state—typically a queue or ring buffer for the K elements and a single aggregate variable (sum, max, etc.). Each new element is enqueued, the oldest is dequeued, and the aggregate is updated in O(1) time. This yields linear time O(n) and constant or O(K) auxiliary space, which is essential for real‑time analytics and high‑throughput systems.
Interview Questions on This Problem
Q1How would you design a system to compute the rolling sum of the last K metrics in a high‑frequency data stream?
Use a fixed‑size queue (or ring buffer) to store the last K values. Maintain a running sum: add the new value, subtract the value that leaves the window. This gives O(1) update per element and O(K) space.
Q2What are the pitfalls of using a sorted data structure to solve this problem when the input is not sorted?
Sorting would add O(n log n) overhead and would destroy the original order, which is required for a sliding window. It also uses extra memory and time, making it unsuitable for streaming or real‑time contexts.
Q3Explain how you would extend this approach to compute the maximum value in the last K metrics.
Use a double‑ended queue (deque) that stores indices of potential maximums in decreasing order. When the window slides, pop indices that fall out of range and pop from the back while the new element is larger. The front of the deque always holds the index of the current maximum.
Examples
Input
[100, 90, 80, 70, 60, 50, 40, 30, 20, 10] and K = 3
Output
270
Explanation: Step 1: Given the input array [100, 90, 80, 70, 60, 50, 40, 30, 20, 10] and K = 3, we need to select the first 3 elements and sum them up. Step 2: Since the array is not specified to be sorted, we will consider the first 3 elements as 100, 90, and 80. Step 3: Summing up these elements, we get 100 + 90 + 80 = 270.
Input
[50, 45, 40, 35, 30, 25, 20, 15, 10, 5] and K = 3
Output
135
Explanation: Step 1: Given the input array [50, 45, 40, 35, 30, 25, 20, 15, 10, 5] and K = 3, we need to select the first 3 elements and sum them up. Step 2: Since the array is not specified to be sorted, we will consider the first 3 elements as 50, 45, and 40. Step 3: Summing up these elements, we get 50 + 45 + 40 = 135.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a running aggregate and a fixed‑size queue. Update the aggregate by adding the new element and subtracting the element that leaves the window, achieving O(n) time and O(K) space.
Brute Force Approach
Recompute the target value (e.g., sum) from scratch for every new element by iterating over the last K elements, leading to O(n·K) time.
Verified Code Solutions
function solution(nums, k) {
// Sort the array in descending order
nums.sort((a, b) => b - a);
// Select the first K elements and sum them up
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
// Sort the array in descending order
sort(nums.begin(), nums.end(), greater<int>());
// Select the first K elements and sum them up
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
// Sort the array in descending order
Arrays.sort(nums);
// Select the first K elements and sum them up
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
# Sort the array in descending order
nums.sort(reverse=True)
# Select the first K elements and sum them up
sum = 0
for i in range(k):
sum += nums[i]
return sumfunction solution(nums, k) {
// Sort the array in descending order
nums.sort((a, b) => b - a);
// Select the first K elements and sum them up
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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