Node Payload Consolidator 4 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a sequence of integer values representing data points in a distributed system. Given an array nums of integers and a threshold value K, your objective is to compute the aggregate sum of all elements in nums that strictly exceed K. If no element in the array satisfies this condition, the function must return 0. This operation simulates a consolidation step where only high-priority payloads (those above the threshold) are aggregated for further processing.
The input consists of a single array of integers and a single integer threshold. The output is a single integer representing the sum of the qualifying elements. The solution should efficiently iterate through the array once, accumulating the sum of elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Consolidator 4"
WHY DOES IT MATTER?
The filter‑and‑aggregate pattern is a foundational building block for analytics, monitoring, and real‑time decision making. Mastering it enables engineers to efficiently extract insights from raw streams without incurring heavy memory or compute costs.
OPTIMIZATION CHALLENGE
The key insight is recognizing that ordering does not affect the predicate, so we can avoid sorting or auxiliary data structures entirely. By maintaining a single accumulator and evaluating the condition on‑the‑fly, we achieve linear time and constant space.
REAL-WORLD CONNECTION
Imagine a distributed logging system where each node reports latency metrics. To trigger an alert, you need the total latency of all requests that exceed a latency SLA (K). Summing only the outliers in a single pass mirrors the Node Payload Consolidator scenario.
During an interview, write the loop first, then immediately add a comment like "if (num > K) sum += num;" – this shows you understand the predicate and aggregation without over‑engineering. Keep variable types wide enough (e.g., long long) to prevent overflow.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem belongs to the classic "filter‑and‑aggregate" pattern, where we must traverse a collection, apply a predicate (value > K) and combine the qualifying elements (sum). In algorithmic terms this is a single‑pass linear scan, which guarantees O(n) time because each element is examined exactly once. Naïve alternatives—such as sorting the array first and then performing a binary search to locate the first element greater than K—introduce an unnecessary O(n log n) overhead and extra space for the sorted copy, which becomes prohibitive for massive data streams typical in distributed systems. The optimal paradigm leverages the fact that the predicate is monotonic only with respect to the threshold, not the order of elements, allowing us to avoid any re‑ordering and achieve constant auxiliary space. This approach aligns with the streaming model where data points arrive continuously and must be aggregated on‑the‑fly without storing the entire dataset.
Interview Questions on This Problem
Q1How would you modify the solution if the array size could be up to 10^9 and the data is provided as a stream?
Use a running total variable and process each incoming integer one by one, adding it to the total only if it exceeds K. Since we never store the whole array, memory stays O(1) and time remains O(N) where N is the number of streamed elements.
Q2Can you compute the sum of elements greater than K in parallel using multiple threads? What challenges arise?
Yes, split the array into chunks, let each thread compute a local sum of values > K, then combine the partial sums with a thread‑safe reduction (e.g., atomic addition or a final aggregation step). Challenges include load balancing, avoiding false sharing, and ensuring that the threshold K is shared read‑only across threads.
Q3If the numbers can be negative and K is also negative, does the algorithm change?
No. The predicate "value > K" works uniformly for any integer sign. The algorithm still scans each element, checks the condition, and adds qualifying values, so no special handling is required beyond using a data type that can hold the possible sum range.
Examples
Input
nums = [12, 5, 8, 15, 3], K = 10
Output
27
Explanation: Iterate through the array: 12 > 10 (add 12, sum=12); 5 <= 10 (skip); 8 <= 10 (skip); 15 > 10 (add 15, sum=27); 3 <= 10 (skip). Final sum is 27.
Input
nums = [2, 4, 6, 8], K = 10
Output
0
Explanation: Iterate through the array: 2 <= 10 (skip); 4 <= 10 (skip); 6 <= 10 (skip); 8 <= 10 (skip). No elements exceed K, so return 0.
Input
nums = [100, -5, 200, 50], K = 75
Output
300
Explanation: Iterate through the array: 100 > 75 (add 100, sum=100); -5 <= 75 (skip); 200 > 75 (add 200, sum=300); 50 <= 75 (skip). Final sum is 300.
Input
nums = [1, 1, 1, 1], K = 0
Output
4
Explanation: Iterate through the array: 1 > 0 (add 1, sum=1); 1 > 0 (add 1, sum=2); 1 > 0 (add 1, sum=3); 1 > 0 (add 1, sum=4). Final sum is 4.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
The optimal solution scans the array once, adds qualifying elements to a running sum, achieving O(n) time and O(1) space.
Brute Force Approach
A naive method would be to use two nested loops, checking each element against every other to find those > K, which is O(n²).
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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