Vault Registry Synthesizer 35 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and registry metrics, and an integer K, construct an optimal algorithm to compute the sum of all elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Registry Synthesizer 35"
WHY DOES IT MATTER?
Threshold‑based aggregations are common in analytics and security audits.
OPTIMIZATION CHALLENGE
Transforming an O(n) scan into an O(log n) query by pre‑computing subtree aggregates.
REAL-WORLD CONNECTION
Similar to scanning logs for events whose risk score exceeds a critical value.
Always keep subtree sums balanced during rotations to avoid hidden O(n) rebuilds.
COMPLEXITY AT A GLANCE
O(n log n) build + O(log n) queryO(n)Core Theory — Why This Approach?
A naïve scan of the entire sequence to sum elements greater than K runs in O(n) time, which becomes prohibitive when the data set is massive or when many threshold queries must be answered. By organizing the elements in a binary search tree (BST) and augmenting each node with the total sum of its subtree, we can answer each query by traversing only the relevant branches, skipping entire sub‑trees whose values are all ≤ K, thus achieving logarithmic query time.
The optimal paradigm leverages order‑statistics augmentation: each node stores its value, the size of its left/right subtrees, and the cumulative sum of those subtrees. During insertion or deletion, these aggregates are updated in O(log n) on a balanced BST (e.g., AVL or Red‑Black). A query walks down the tree, adding the pre‑computed sums of right subtrees when the current node exceeds K, and proceeds left, guaranteeing O(log n) time per sum‑greater‑than‑K operation.
Interview Questions on This Problem
Q1How would you augment a BST node to support sum‑greater‑than‑K queries?
Add a field that stores the sum of all values in the node's subtree. Update this field during every insert, delete, or rotation to keep it accurate.
Q2What is the time complexity of computing the sum of elements > K using an in‑order traversal without augmentation?
A full in‑order traversal visits every node, yielding O(n) time. No early termination is possible without extra information.
Q3Why does augmenting subtree sums reduce the query to O(log n) on a balanced BST?
Because at each step we can decide to include an entire right subtree's sum without visiting its nodes. This skips large portions of the tree, limiting work to the height of the tree.
Examples
Input
[10, 20, 30, 40, 50], 3
Output
150
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and K = 3, we sum all elements greater than K. First, we filter the array to get [10, 20, 30, 40, 50]. Then, we sum these elements to get 10 + 20 + 30 + 40 + 50 = 150.
Input
[5, 5, 5, 5, 5], 5
Output
0
Explanation: Step-by-step: with input [5, 5, 5, 5, 5] and K = 5, we sum all elements greater than K. First, we filter the array to get an empty array. Then, we sum these elements to get 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Insert elements into a balanced BST that maintains subtree sums, then answer the query by traversing only the necessary branches.
Brute Force Approach
Iterate through the entire list and accumulate each element that is greater than K.
Verified Code Solutions
function solution(nums, k) { let sum = 0; for (let num of nums) { if (num > k) { sum += num; } } return sum; }class Solution { public: int solution(vector<int>& nums, int k) { int sum = 0; for (int num : nums) { if (num > k) { sum += num; } } return sum; } }class Solution { public int solution(int[] nums, int k) { int sum = 0; for (int num : nums) { if (num > k) { sum += num; } } return sum; } }def solution(nums, k): return sum(num for num in nums if num > k)function solution(nums, k) { let sum = 0; for (let num of nums) { if (num > k) { sum += num; } } return sum; }Asked in Top Tech Interviews
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