Node Payload Architect 9 — Problem Statement & Solution Guide
Problem Description
In a distributed data processing pipeline, a specific node is responsible for aggregating payload sizes from a stream of incoming packets. You are provided with an array payloads containing the integer sizes of these packets and a capacity limit limit. Your task is to calculate the total size of all packets that can be safely processed by the node, defined as those with a size less than or equal to limit. Packets exceeding this capacity must be discarded and excluded from the total. Return the sum of the valid packet sizes. If no packets meet the criteria, return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Architect 9"
WHY DOES IT MATTER?
Linear scans are the backbone of many real-time analytics pipelines; they provide deterministic performance and minimal memory footprint, which is critical when processing millions of packets per second.
OPTIMIZATION CHALLENGE
The key insight is that each packet’s eligibility depends only on its own size, not on any combination of other packets. This allows us to avoid sorting or nested comparisons, reducing time from O(n log n) or O(n^2) to O(n).
REAL-WORLD CONNECTION
Think of a network router that must decide which packets to forward based on size constraints. It performs a single pass over the packet queue, summing sizes until the buffer limit is reached, mirroring this algorithm.
When explaining this in an interview, emphasize the independence of decisions and the importance of early exit conditions to avoid unnecessary work.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a classic linear scan where each packet size is evaluated against a capacity threshold. A naive approach might sort the array or use nested loops to compare each pair, leading to O(n log n) or O(n^2) time, which is unnecessary and costly for large streams. The optimal paradigm is a single-pass accumulation: iterate through the array once, add the size to a running total only if it does not exceed the limit. This leverages the fact that the decision for each packet is independent of others, eliminating the need for complex data structures or dynamic programming tables.
Interview Questions on This Problem
Q1How would you modify this algorithm if the node could process packets in any order but had a strict total size limit?
You would sort the packet sizes in ascending order and then greedily add them until the next packet would exceed the limit. This ensures the maximum number of packets is processed, similar to the classic knapsack problem with unit weights.
Q2In a distributed system, why is it important to keep the per-node processing time O(n) rather than O(n log n) or O(n^2)?
Distributed nodes often handle high-throughput streams; any additional logarithmic or quadratic factor can become a bottleneck, leading to increased latency and resource contention across the cluster. Linear time guarantees predictable scaling with data volume.
Q3What edge case would you test to ensure your implementation handles integer overflow correctly?
Test with an array containing the maximum integer values and a limit that is also near the maximum. Verify that the sum does not wrap around and that the algorithm correctly identifies packets that exceed the limit before adding them.
Examples
Input
payloads = [10, 20, 30, 40], limit = 25
Output
30
Explanation: Iterate through the array: 10 <= 25 (include, sum=10); 20 <= 25 (include, sum=30); 30 > 25 (exclude); 40 > 25 (exclude). Final sum is 30.
Input
payloads = [5, 15, 25], limit = 10
Output
5
Explanation: Iterate through the array: 5 <= 10 (include, sum=5); 15 > 10 (exclude); 25 > 10 (exclude). Final sum is 5.
Input
payloads = [100, 200, 300], limit = 50
Output
0
Explanation: Iterate through the array: 100 > 50 (exclude); 200 > 50 (exclude); 300 > 50 (exclude). No elements included, final sum is 0.
Input
payloads = [1, 2, 3, 4, 5], limit = 100
Output
15
Explanation: All elements are <= 100. Sum = 1 + 2 + 3 + 4 + 5 = 15.
Constraints
- 1 <= payloads.length <= 10^5
- 1 <= payloads[i] <= 10^9
- 1 <= limit <= 10^9
Optimal Approach & Strategy
Traverse the array once, adding each packet’s size to a total only if it’s less than or equal to the limit. This yields O(n) time and O(1) space.
Brute Force Approach
A naive solution might sort the array and then try every possible subset of packets, leading to O(n log n) or O(2^n) time. This is impractical for large inputs.
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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