Vault Buffer Validator 9 — 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 validator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Buffer Validator 9"
WHY DOES IT MATTER?
Monotonic stacks turn quadratic range‑search problems into linear scans, essential for high‑throughput metric validation.
OPTIMIZATION CHALLENGE
The key is reducing repeated boundary checks from O(n²) to a single pass by maintaining order invariants.
REAL-WORLD CONNECTION
They mirror how browsers manage nested tags or how operating systems track resource allocation windows.
Push indices, not values, and compute contributions at pop time to avoid extra passes.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem reduces to finding, for each element, the nearest larger (or smaller) metric on its left and right to determine its contribution to the validator value. A monotonic stack maintains a strictly decreasing (or increasing) sequence of indices, allowing constant‑time access to the next boundary for each element, which yields an overall linear scan. Naïve double loops recompute these boundaries for every element, leading to O(n²) time that explodes on large input sizes typical of vault‑buffer logs. By leveraging the stack’s LIFO property, each element is pushed and popped at most once, guaranteeing O(n) time and O(n) auxiliary space, the optimal paradigm for this class of range‑query problems.
Interview Questions on This Problem
Q1How does a monotonic stack help compute nearest greater elements in linear time?
It keeps indices in decreasing order so the top always holds the next greater candidate. When a new element arrives, you pop until the stack top is larger, establishing the nearest greater on the left.
Q2What is the worst‑case space usage of the stack in this algorithm?
The stack can hold at most all n indices when the sequence is strictly monotonic. Hence the space complexity is O(n).
Q3How would you modify the algorithm to handle equal values correctly?
Treat equal values as either strictly greater or not based on problem rules; typically you pop only while the top is strictly less to keep the first occurrence as the boundary. This ensures duplicates are processed without breaking the monotonic invariant.
Examples
Input
[10, 20, 30, 40, 50], 3
Output
150
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50] and k = 3, we first sort the array in descending order. Then, we select the top 3 elements (30, 40, 50) and sum them up to get the target validator value, which is 120. However, this is not the correct answer because the problem statement asks for the target validator value under given operational constraints, which are not specified.
Input
[5, 5, 5, 5, 5], 3
Output
15
Explanation: Step-by-step: Given the input [5, 5, 5, 5, 5] and k = 3, we first sort the array in descending order. Then, we select the top 3 elements (5, 5, 5) and sum them up to get the target validator value, which is 15.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a decreasing monotonic stack to determine nearest larger boundaries in a single O(n) traversal.
Brute Force Approach
For each element, scan left and right to find the first larger metric, resulting in O(n²) time.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
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(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
return sum(nums[:k])function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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