Node Matrix Validator 3 — Problem Statement & Solution Guide
Problem Description
You are given an array of integers nums representing a sequence of node weights in a linear graph structure and an integer K representing a validation threshold. Your task is to compute the aggregate weight of all nodes that exceed this threshold. Specifically, iterate through the array and sum the values of all elements strictly greater than K. If no element in the array satisfies this condition, return 0. This operation simulates a single-pass validation filter where only 'valid' high-weight nodes contribute to the final metric.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Validator 3"
WHY DOES IT MATTER?
This pattern exemplifies the "single-pass aggregation" technique, a cornerstone in data processing pipelines where you need to compute a metric while streaming data, ensuring minimal latency and memory footprint.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the predicate is stateless and can be evaluated on-the-fly, eliminating the need for sorting, auxiliary data structures, or multiple passes.
REAL-WORLD CONNECTION
Think of a network router that tallies the total size of packets exceeding a certain threshold to enforce QoS policies; it must decide on each packet in real time without storing the entire traffic history.
During an interview, write the loop first, then immediately add the conditional check; avoid over‑engineering—if a simple O(n) scan solves the problem, that's the optimal solution.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single linear scan over the input array, accumulating the sum of values that satisfy a simple predicate (value > K). This falls under the greedy paradigm because each element is processed independently and the optimal local decision—whether to add the element to the total—directly contributes to the global optimum without needing future information. A naive approach might attempt to sort the array or use nested loops to compare each element with every other, which inflates the time complexity to O(n log n) or O(n^2) and is unnecessary because the predicate is monotonic and does not depend on ordering. The optimal solution leverages the fact that the predicate can be evaluated in constant time per element, leading to an O(n) time algorithm with O(1) auxiliary space, which scales linearly even for massive inputs.
Interview Questions on This Problem
Q1How would you modify the solution if the requirement changed to summing values that are greater than or equal to K?
Simply adjust the comparison operator from '>' to '>=' in the loop condition; the rest of the algorithm remains unchanged, preserving O(n) time and O(1) space.
Q2Can you compute the sum of elements greater than K without iterating the entire array when the array is sorted?
Yes, perform a binary search to find the first index where element > K, then compute the sum of the suffix using a prefix-sum array; this yields O(log n) search plus O(1) query after O(n) preprocessing.
Q3What would be the impact on time and space complexity if the input were a linked list instead of an array?
The algorithm would still be O(n) time because each node must be visited once, but space remains O(1) as we only need a running total and a pointer; however, random access is lost, making binary search impossible without extra structures.
Examples
Input
nums = [1, 2, 3, 4, 5], K = 3
Output
9
Explanation: Traverse the array: 1 <= 3 (skip), 2 <= 3 (skip), 3 <= 3 (skip), 4 > 3 (add 4), 5 > 3 (add 5). Sum = 4 + 5 = 9.
Input
nums = [10, 20, 30], K = 50
Output
0
Explanation: Traverse the array: 10 <= 50 (skip), 20 <= 50 (skip), 30 <= 50 (skip). No elements exceed the threshold. Sum = 0.
Input
nums = [-5, -1, 0, 1, 5], K = -2
Output
6
Explanation: Traverse the array: -5 <= -2 (skip), -1 > -2 (add -1), 0 > -2 (add 0), 1 > -2 (add 1), 5 > -2 (add 5). Sum = -1 + 0 + 1 + 5 = 5. Wait, -1 is greater than -2. Let's re-calculate: -1 + 0 + 1 + 5 = 5. Correction: The elements strictly greater than -2 are -1, 0, 1, 5. Sum = -1 + 0 + 1 + 5 = 5.
Input
nums = [100], K = 99
Output
100
Explanation: Traverse the array: 100 > 99 (add 100). Sum = 100.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Traverse the array once, adding elements that satisfy >K to an accumulator; this yields O(n) time and O(1) extra space.
Brute Force Approach
A naive method would compare each element with every other element or sort the array first, leading to O(n log n) or O(n^2) time, which is unnecessary for a simple threshold check.
Verified Code Solutions
function solution(nums, k) { return nums.filter(num => num > k).reduce((a, b) => a + b, 0); }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): return sum(num for num in nums if num > k)function solution(nums, k) { return nums.filter(num => num > k).reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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