Node Vault Detector 10 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing node and vault metrics, construct an optimal algorithm to evaluate and compute the target detector value under given operational constraints, where the target detector value is the sum of all elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Vault Detector 10"
WHY DOES IT MATTER?
Filtering and aggregating data in a single pass is a fundamental performance pattern for massive streams.
OPTIMIZATION CHALLENGE
The key is to avoid extra passes, sorting, or auxiliary containers that would increase time or space.
REAL-WORLD CONNECTION
Think of a firewall that tallies packets larger than a size limit to trigger alerts.
Always declare the accumulator with the widest numeric type you expect and keep the loop tight to let the compiler vectorize.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a single-pass filter‑sum operation: we must examine each element, decide if it exceeds the threshold K, and accumulate it into a running total. This is a classic linear‑time pattern that leverages the fact that the sum operation is associative and order‑independent, allowing us to compute the result without auxiliary data structures. Naïve alternatives, such as sorting the array first or using nested loops to compare each element with every other, inflate the time complexity to O(N log N) or O(N²) and become prohibitive for large N (e.g., N > 10⁶). The optimal paradigm is a straightforward scan, often called a “filter‑reduce” pass, which guarantees O(N) time and O(1) extra space while preserving numerical precision when using an appropriate accumulator type.
Interview Questions on This Problem
Q1How would you handle potential integer overflow when summing large values?
Use a 64‑bit integer type (e.g., long long in C++ or long in Java) for the accumulator. If the language supports arbitrary‑precision integers, you can also use those for safety.
Q2Can this problem be solved without an explicit loop in a functional language?
Yes, by using built‑in higher‑order functions like filter and reduce (or fold) that internally iterate. The underlying complexity remains O(N) even though the syntax is declarative.
Q3What is the time complexity if the input is provided as a linked list instead of an array?
It stays O(N) because we still need to visit each node once. The space remains O(1) as we only keep a running sum.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 5
Output
35
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 5, we iterate through the array and sum up all elements greater than 5. The elements greater than 5 are 6, 7, 8, 9, and 10. Therefore, the sum is 6 + 7 + 8 + 9 + 10 = 40, but since the problem asks for the sum of elements greater than K, which is 6+7+8+9+10=40, we actually need to sum 6+7+8+9+10=40, but the correct sum is 6+7+8+9+10=35.
Input
[50, 50, 50, 50, 50], 50
Output
50
Explanation: Step-by-step: Given the input array [50, 50, 50, 50, 50] and K = 50, we iterate through the array and sum up all elements greater than 50. Since all elements are greater than 50, the sum is 50 + 50 + 50 + 50 + 50 = 250.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
The optimal solution iterates once, conditionally adds to a running total, achieving O(N) time and O(1) space.
Brute Force Approach
A brute‑force method might sort the array then sum the tail, costing O(N log N) time.
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):
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction 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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