Vault Registry Consolidator 45 — Problem Statement & Solution Guide
Problem Description
Vault Registry Consolidator 45
You are given an array of integers, nums, that represents a series of vault and registry metrics collected over time, and a single integer K. Your task is to compute the *consolidator value*, defined as the sum of all elements in nums that are strictly greater than K.
The input consists of two lines: the first line contains the space‑separated integers of nums; the second line contains the integer K. The output is a single integer, the consolidator value.
Although the problem can be solved in linear time by a simple scan, an efficient solution can also be implemented using a monotonic stack to maintain a decreasing sequence of values, which allows early pruning of elements that cannot contribute to the sum. The stack approach is optional; any correct algorithm that runs within the given constraints is acceptable.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Registry Consolidator 45"
WHY DOES IT MATTER?
Filtering and aggregating data is a foundational pattern for analytics and real‑time monitoring.
OPTIMIZATION CHALLENGE
Avoid sorting or extra data structures; the key is a single pass with constant extra memory.
REAL-WORLD CONNECTION
Think of summing sales above a target threshold in a streaming transaction log.
Initialize the accumulator to 0 and update it only when the condition holds to minimize branch mispredictions.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a linear scan where each element is compared against a threshold K and conditionally accumulated. This is a classic example of a filter‑then‑aggregate pattern that can be solved in O(n) time because each element is visited exactly once.
A naive approach might attempt nested loops or sorting before summation, which inflates time complexity to O(n log n) or O(n^2) and adds unnecessary overhead. The optimal paradigm leverages the associative property of addition and the constant‑time comparison to achieve linear time with O(1) auxiliary space.
Interview Questions on This Problem
Q1How would you handle the case where K is larger than all elements in the array?
The sum would be zero because no element satisfies the >K condition. You can return 0 directly after the scan.
Q2Can this problem be solved without an explicit loop in a language like Python?
Yes, using built‑in functions like filter and sum or a generator expression. However, the underlying iteration still runs in O(n) time.
Q3What is the impact on complexity if the array is extremely large and stored on disk?
You must stream the data to keep memory usage O(1), preserving linear time overall. Random access is unnecessary; sequential reads suffice.
Examples
Input
3 7 2 9 5 5
Output
16
Explanation: Elements greater than 5 are 7 and 9. Their sum is 7+9=16.
Input
-1 0 5 10 15 10
Output
15
Explanation: Only 15 is greater than 10, so the sum is 15.
Input
100 200 300 400 250
Output
700
Explanation: Elements greater than 250 are 300 and 400. Their sum is 300+400=700.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- -1000000000 <= K <= 1000000000
- The result fits in a 64‑bit signed integer.
Optimal Approach & Strategy
Perform a single linear pass, adding elements to the sum only when they exceed K.
Brute Force Approach
Sort the array then iterate from the first element greater than K, summing the rest.
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} K
* @return {number}
*/
var consolidatorValue = function(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
};class Solution {
public:
int consolidatorValue(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int consolidatorValue(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
}class Solution:
def consolidatorValue(self, nums: List[int], K: int) -> int:
return sum(num for num in nums if num > K)/**
* @param {number[]} nums
* @param {number} K
* @return {number}
*/
var consolidatorValue = function(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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