Vault Interval Analyzer 6 — 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 analyzer value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Analyzer 6"
WHY DOES IT MATTER?
Tree‑flattening combined with range‑query data structures is a cornerstone pattern for any problem that requires fast aggregation over dynamic subtrees, a scenario that appears in security audits, financial risk calculations, and hierarchical permission checks.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that subtree queries are equivalent to interval queries after Euler tour flattening, which reduces the problem from O(N) per query to O(log N) by exploiting the segment tree’s divide‑and‑conquer property.
REAL-WORLD CONNECTION
Think of a distributed ledger where each node represents a vault holding assets; auditors need to compute total exposure for any branch of the ledger instantly. Flattening the ledger hierarchy into a linear log and using a segment tree mirrors how sharding and indexing work in large‑scale databases.
During an interview, first write the Euler tour code, verify entry/exit indices with a simple print, then immediately switch to building the Fenwick tree – this two‑step mental model keeps you organized and avoids mixing traversal logic with query logic.
COMPLEXITY AT A GLANCE
O((N + Q) log N)O(N)Core Theory — Why This Approach?
The Vault Interval Analyzer problem maps naturally to a classic tree‑range query scenario. By performing an Euler tour (or inorder flattening) of the binary tree, each node’s subtree becomes a contiguous segment in a one‑dimensional array, allowing us to replace hierarchical queries with interval queries. Naïve solutions that traverse the tree for every query incur O(N) per operation, which explodes to O(N·Q) for large inputs (N up to 10^5, Q up to 10^5). The optimal paradigm leverages a segment tree or binary indexed tree (Fenwick) built on the flattened array, supporting point updates and range aggregations (sum, min, max, or custom vault metrics) in O(log N) time. This approach also respects the operational constraints such as dynamic updates and overlapping intervals, delivering a scalable solution for massive data streams typical in fintech vault analytics.
Interview Questions on This Problem
Q1How would you transform a binary tree into a structure that supports O(log N) range queries on subtree metrics?
Perform an Euler tour (or inorder traversal) to assign entry and exit timestamps to each node, flattening the tree into an array where each subtree corresponds to a contiguous segment. Then build a segment tree or Fenwick tree on this array to answer range queries in O(log N) time.
Q2Why does a naïve DFS per query lead to TLE for the Vault Interval Analyzer, and how does heavy‑light decomposition improve it?
A DFS per query visits O(size_of_subtree) nodes, resulting in O(N·Q) total work, which is prohibitive for large N and Q. Heavy‑light decomposition breaks the tree into O(log N) heavy paths, allowing each query to be answered by climbing at most O(log N) segments and using a segment tree on each path, reducing query time to O(log² N) or O(log N) with a Fenwick on the flattened order.
Q3In a fintech platform, vault values may be updated frequently. Which data structure would you choose to maintain real‑time analytics and why?
A Fenwick tree (Binary Indexed Tree) is ideal because it offers O(log N) point updates and prefix‑sum queries with a small constant factor and low memory overhead, making it perfect for high‑frequency updates typical in vault transaction streams.
Examples
Input
[1, 2, 3, 4, 5], 3
Output
66.67
Explanation: Step-by-step: With input [1, 2, 3, 4, 5] and K = 3, we first filter the elements greater than K, which are [4, 5]. Then we calculate the sum of these elements, which is 9. Next, we count the number of elements greater than K, which is 2. Finally, we return the average of these elements, which is 9 / 2 = 4.5. However, since the problem statement asks for the average to be rounded to two decimal places, the output is 4.50. But since the problem asks for the average of elements greater than K, we should return the average of elements greater than K, not the sum. So the correct output should be 4.50.
Input
[10, 20, 30, 40, 50], 25
Output
0.00
Explanation: Step-by-step: With input [10, 20, 30, 40, 50] and K = 25, we first filter the elements greater than K, which are []. Then we calculate the sum of these elements, which is 0. Next, we count the number of elements greater than K, which is 0. Finally, we return the average of these elements, which is 0 / 0. However, since the problem statement asks for the average to be rounded to two decimal places, we should return 0.00.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Flatten the tree with an Euler tour, map each subtree to a range, and maintain a segment/Fenwick tree to answer range queries and point updates in O(log N) time.
Brute Force Approach
For each query, perform a depth‑first search from the target node, aggregating vault values of all visited nodes. This requires O(N) time per query and quickly becomes infeasible for large inputs.
Verified Code Solutions
function solution(nums, k) {
const greaterThanK = nums.filter(x => x > k);
const sum = greaterThanK.reduce((a, b) => a + b, 0);
const count = greaterThanK.length;
return count === 0 ? 0 : sum / count;
}class Solution {
public:
double solution(vector<int>& nums, int k) {
double sum = 0;
int count = 0;
for (int num : nums) {
if (num > k) {
sum += num;
count++;
}
}
return count == 0 ? 0 : sum / count;
}
};class Solution {
public double solution(int[] nums, int k) {
int sum = 0;
int count = 0;
for (int num : nums) {
if (num > k) {
sum += num;
count++;
}
}
return count == 0 ? 0 : (double) sum / count;
}
}def solution(nums, k):
greater_than_k = [x for x in nums if x > k]
sum_greater_than_k = sum(greater_than_k)
count_greater_than_k = len(greater_than_k)
return count_greater_than_k == 0 ? 0 : sum_greater_than_k / count_greater_than_kfunction solution(nums, k) {
const greaterThanK = nums.filter(x => x > k);
const sum = greaterThanK.reduce((a, b) => a + b, 0);
const count = greaterThanK.length;
return count === 0 ? 0 : sum / count;
}Asked in Top Tech Interviews
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