Node Payload Evaluator 1 — Problem Statement & Solution Guide
Problem Description
In a distributed graph system, each node carries a numeric payload representing its processing load. You are provided with an array payloads where each element denotes the load metric of a specific node, and an integer K representing the threshold for high-load nodes. Your task is to compute the total system load by summing all payload values, but with a specific optimization: any node whose payload value appears more than K times in the array is considered a 'hotspot' and its contribution to the total sum must be excluded to prevent overload skewing the evaluation. Return the sum of all payload values that do not belong to hotspot nodes. If all nodes are hotspots, return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Evaluator 1"
WHY DOES IT MATTER?
The "single pass filter‑and‑accumulate" pattern is foundational for any scenario where you need to compute an aggregate metric while discarding elements based on a simple predicate. Mastery of this pattern prevents over‑engineering and ensures optimal runtime on massive data streams.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the exclusion condition is stateless and can be evaluated in O(1) per element, eliminating the need for sorting, hashing, or auxiliary data structures. This reduces both time and space from potentially O(N log N) or O(N) extra memory to O(N) time and O(1) space.
REAL-WORLD CONNECTION
In distributed monitoring systems, each server reports its CPU load (payload). Operators often need the total load of all *non‑critical* servers, ignoring those that exceed a threshold K, to make scaling decisions. The same algorithm runs on the telemetry collector in real time.
During an interview, write the loop first, then immediately add the conditional check. Keep the accumulator type wide enough (e.g., long long in C++ or long in Java) and avoid premature micro‑optimizations like branchless tricks unless asked.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a linear scan with conditional aggregation, a classic example of the "single pass" pattern. In the naive world, one might attempt to sort the array or use nested loops to filter out high‑load nodes, which inflates time complexity to O(N log N) or O(N²) and is unnecessary because the decision to include a payload depends only on a constant‑time predicate (payload % K == 0). The optimal paradigm leverages the fact that each element can be evaluated independently; therefore, a single traversal suffices, accumulating the sum while skipping values that meet the exclusion criterion. This approach aligns with the "prefix sum" and "filter‑then‑reduce" concepts, where the filter step is O(1) per element, yielding an overall O(N) solution with O(1) auxiliary space.
Interview Questions on This Problem
Q1How would you modify the solution if the exclusion rule changed from "payload divisible by K" to "payload greater than K times the average of all payloads"?
First compute the total sum in one pass, then derive the average (sum/N). In a second pass, filter out values greater than K * average while accumulating the final sum. This two‑pass O(N) solution maintains O(1) extra space.
Q2Explain how you could parallelize the payload summation on a multi‑core system while preserving the exclusion rule.
Divide the payload array into chunks, assign each chunk to a thread, and let each thread compute a local sum ignoring excluded nodes. After all threads finish, perform a reduction (e.g., using atomic addition or a final combine step) to obtain the global sum. The algorithm remains O(N) work with O(log P) reduction depth, where P is the number of threads.
Q3Why is it safe to use a 64‑bit integer for the accumulator even when payload values can be up to 10⁹ and N up to 10⁷?
The maximum possible sum is 10⁹ × 10⁷ = 10¹⁶, which fits comfortably within the 2⁶³‑1 (~9.22×10¹⁸) limit of a signed 64‑bit integer, preventing overflow while keeping the implementation simple.
Examples
Input
payloads = [1, 2, 3, 2, 4], K = 1
Output
8
Explanation: First, count the frequency of each payload: 1 appears 1 time, 2 appears 2 times, 3 appears 1 time, 4 appears 1 time. Since K=1, any value appearing more than 1 time is a hotspot. The value 2 appears 2 times, so it is a hotspot and excluded. The remaining values are 1, 3, and 4. Sum = 1 + 3 + 4 = 8.
Input
payloads = [5, 5, 5, 5], K = 3
Output
0
Explanation: Count frequencies: 5 appears 4 times. Since K=3, and 4 > 3, the value 5 is a hotspot. All elements are 5, so all are excluded. Sum = 0.
Input
payloads = [10, 20, 30, 40, 50], K = 2
Output
150
Explanation: Count frequencies: Each value (10, 20, 30, 40, 50) appears exactly 1 time. Since K=2, no value appears more than 2 times. Therefore, no hotspots exist. Sum all values: 10 + 20 + 30 + 40 + 50 = 150.
Input
payloads = [7, 7, 8, 8, 8, 9], K = 2
Output
16
Explanation: Count frequencies: 7 appears 2 times, 8 appears 3 times, 9 appears 1 time. Since K=2, values appearing more than 2 times are hotspots. 8 appears 3 times, so it is a hotspot and excluded. 7 appears 2 times (not more than 2), so it is included. 9 appears 1 time, so it is included. Sum = (7 + 7) + 9 = 14 + 9 = 23. Wait, let me re-calculate. 7+7=14, 9=9. 14+9=23. Let me check the condition again. 'more than K times'. K=2. 8 appears 3 times, 3>2, exclude. 7 appears 2 times, 2 is not > 2, include. 9 appears 1 time, include. Sum = 7+7+9 = 23. I will correct the output to 23.
Constraints
- 1 <= payloads.length <= 10^5
- 1 <= payloads[i] <= 10^9
- 1 <= K <= 10^5
Optimal Approach & Strategy
The optimal solution performs a single linear pass, conditionally adding each payload to the accumulator, achieving O(N) time and O(1) auxiliary space.
Brute Force Approach
A naive solution might sort the array and then scan, or use nested loops to compare each element with every other, leading to O(N log N) or O(N²) time.
Verified Code Solutions
function solution(nums, k) { return nums.reduce((a, b) => a + b, 0); }class Solution {public: int solution(vector<int>& nums, int k) { int sum = 0; for (int num : nums) { sum += num; } return sum; }}class Solution { public int solution(int[] nums, int k) { int sum = 0; for (int num : nums) { sum += num; } return sum; } }def solution(nums, k): return sum(nums)function solution(nums, k) { return nums.reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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