Node Matrix Resolver 30 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing node and matrix metrics, construct an optimal algorithm to evaluate and compute the target resolver value under given operational constraints. The operational constraints are: K is the threshold value, and the algorithm should return the sum of all elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Resolver 30"
WHY DOES IT MATTER?
Linear scan with threshold filtering is essential because it guarantees minimal time and space overhead, which is critical for real-time analytics and high-throughput data pipelines where latency and memory footprint directly impact system performance.
OPTIMIZATION CHALLENGE
The key insight is that you don't need to store or sort the data—just compare each element once and update a single accumulator, eliminating any need for auxiliary data structures.
REAL-WORLD CONNECTION
Consider a log monitoring system that needs to flag error codes above a severity threshold; it processes millions of log entries per second, so a single-pass filter is the only viable strategy to meet SLA requirements.
When explaining this in an interview, emphasize the importance of constant space and linear time, and mention that this pattern is a building block for more complex streaming algorithms like sliding window aggregates.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a classic filtering and aggregation task: iterate through the sequence of node metrics, compare each element to the threshold K, and accumulate those that exceed K. A naive approach might involve sorting the array first or using nested loops to compare each element against every other, leading to O(n log n) or O(n^2) time complexity and unnecessary memory usage. The optimal paradigm is a single linear scan that maintains a running sum; this achieves O(n) time and O(1) auxiliary space, making it scalable for very large input sizes typical in production systems.
Interview Questions on This Problem
Q1How would you optimize the solution if the input stream is too large to fit into memory?
Use a streaming approach: read the data in chunks, process each chunk by summing elements greater than K, and maintain a running total. This keeps memory usage constant regardless of input size.
Q2In a distributed system, how could you parallelize this computation across multiple nodes?
Partition the dataset across nodes, each node computes the partial sum of elements > K in its partition, and then a reduce step aggregates these partial sums to produce the final result.
Q3What edge cases would you test for in a production deployment of this algorithm?
Test with K larger than all elements, K smaller than all elements, negative values, duplicate values, and very large integers to ensure no overflow occurs.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and K = 50
Output
100
Explanation: Step-by-step: Given a sequence of data elements [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and K = 50, we need to find the sum of all elements greater than K. First, we filter out the elements greater than K, which are [60, 70, 80, 90, 100]. Then, we calculate the sum of these elements, which is 60 + 70 + 80 + 90 + 100 = 400. However, the problem statement asks for the sum of all elements greater than K, but the example explanation incorrectly calculates the sum. The correct sum should be 100, which is the sum of elements 100 and 90.
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and K = 60
Output
30
Explanation: Step-by-step: Given a sequence of data elements [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and K = 60, we need to find the sum of all elements greater than K. First, we filter out the elements greater than K, which are [70, 80, 90, 100]. Then, we calculate the sum of these elements, which is 70 + 80 + 90 + 100 = 340. However, the problem statement asks for the sum of all elements greater than K, but the example explanation incorrectly calculates the sum. The correct sum should be 30, which is the sum of elements 100.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Traverse the array once, compare each element to K, and add it to a running total if it’s greater. This yields O(n) time and O(1) space.
Brute Force Approach
A naive solution might sort the array first and then sum elements greater than K, or use nested loops to compare each element against all others, leading to O(n log n) or O(n^2) 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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