Node Vault Analyzer 34 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a stream of integer metrics derived from a distributed node-vault system. Given an array metrics of length n and a threshold integer K, your objective is to compute the cumulative sum of all elements in metrics that are strictly greater than K. Elements less than or equal to K are considered noise and must be excluded from the final aggregation. The solution must efficiently traverse the sequence to determine this target analyzer value.
The input consists of a single array of integers representing the raw data points and a single integer representing the filtering threshold. The output is a single integer representing the sum of the qualifying elements. If no elements in the array exceed the threshold, the result must be zero. This problem tests the ability to perform linear-time filtering and accumulation under strict numerical constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Vault Analyzer 34"
WHY DOES IT MATTER?
Filtering and aggregating streams in O(n) is a foundational pattern for high‑throughput data pipelines.
OPTIMIZATION CHALLENGE
The key is to avoid nested loops and extra data structures, reducing the problem to a simple linear scan.
REAL-WORLD CONNECTION
Think of a monitoring system that only records metrics exceeding a critical threshold to trigger alerts.
Keep the loop tight, use early continue for non‑qualifying values, and avoid unnecessary casts or function calls.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The naive solution iterates over every possible sub‑array, checking each element against K, which leads to O(n^2) time and quickly exceeds limits for large streams. By recognizing that the condition "greater than K" is independent of neighboring elements, we can reduce the problem to a single linear pass, accumulating only qualifying values, which is the optimal paradigm for this class of filter‑and‑aggregate tasks.
Sliding‑window techniques excel when the property to evaluate depends on a contiguous segment, but here the window size is effectively 1 for each element, allowing us to treat the array as a stream and maintain a running total without extra storage. This shift from quadratic to linear complexity is crucial for handling massive metric logs in real‑time systems.
Interview Questions on This Problem
Q1How would you compute the sum of elements greater than K in a single pass?
Initialize a sum variable to zero and iterate through the array, adding each element to the sum only if it exceeds K.
Q2What is the time and space complexity of the optimal solution?
Time complexity is O(n) and auxiliary space is O(1) because we only use a few scalar variables.
Q3Can this approach be adapted for a sliding‑window sum of elements > K within a fixed window size?
Yes, maintain a queue or two pointers to add new qualifying elements and subtract those exiting the window, still achieving O(n).
Examples
Input
metrics = [12, 5, 8, 20, 3], K = 10
Output
32
Explanation: Iterate through the array: 12 > 10 (add 12, sum=12); 5 <= 10 (skip); 8 <= 10 (skip); 20 > 10 (add 20, sum=32); 3 <= 10 (skip). Final sum is 32.
Input
metrics = [1, 2, 3, 4, 5], K = 100
Output
0
Explanation: Iterate through the array: 1 <= 100 (skip); 2 <= 100 (skip); 3 <= 100 (skip); 4 <= 100 (skip); 5 <= 100 (skip). No elements exceed the threshold, so the sum remains 0.
Input
metrics = [100, 100, 100], K = 99
Output
300
Explanation: Iterate through the array: 100 > 99 (add 100, sum=100); 100 > 99 (add 100, sum=200); 100 > 99 (add 100, sum=300). Final sum is 300.
Input
metrics = [-5, -1, 0, 2, 7], K = 1
Output
9
Explanation: Iterate through the array: -5 <= 1 (skip); -1 <= 1 (skip); 0 <= 1 (skip); 2 > 1 (add 2, sum=2); 7 > 1 (add 7, sum=9). Final sum is 9.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Perform a single pass, adding only elements > K to a running sum, achieving O(n) time and O(1) space.
Brute Force Approach
Check every sub‑array or element with nested loops, leading to O(n^2) time.
Verified Code Solutions
/**
* @param {number[]} metrics
* @param {number} K
* @return {number}
*/
var analyzeVault = function(metrics, K) {
let sum = 0;
for (let val of metrics) {
if (val > K) {
sum += val;
}
}
return sum;
};#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
int analyzeVault(const vector<int>& metrics, int K) {
int sum = 0;
for (int val : metrics) {
if (val > K) {
sum += val;
}
}
return sum;
}
};import java.util.*;
class Solution {
public int analyzeVault(int[] metrics, int K) {
int sum = 0;
for (int val : metrics) {
if (val > K) {
sum += val;
}
}
return sum;
}
}from typing import List
class Solution:
def analyzeVault(self, metrics: List[int], K: int) -> int:
return sum(val for val in metrics if val > K)/**
* @param {number[]} metrics
* @param {number} K
* @return {number}
*/
var analyzeVault = function(metrics, K) {
let sum = 0;
for (let val of metrics) {
if (val > K) {
sum += val;
}
}
return sum;
};Asked in Top Tech Interviews
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