Node Payload Detector 34 — Problem Statement & Solution Guide

QueueEasyGreedy Choice

TimeO(N)

SpaceO(1)

Problem Description

Given a sequence of data elements representing node and payload metrics, construct an optimal algorithm to evaluate and compute the target detector value under given operational constraints. The detector value is the maximum value in the array that is greater than K.

DSA Pattern Breakdown

"Node Payload Detector 34"

easy

WHY DOES IT MATTER?

Linear filtering and incremental aggregation form the backbone of real-time data pipelines, enabling systems to extract critical metrics without buffering entire datasets.

OPTIMIZATION CHALLENGE

The key insight is recognizing that sorting or heap maintenance is mathematically redundant; a single running variable updated conditionally achieves the same result with strictly lower complexity.

REAL-WORLD CONNECTION

This pattern mirrors network packet inspection engines and IoT sensor streams, where devices continuously filter telemetry against safety thresholds while maintaining minimal memory overhead.

Always clarify whether the input is a static array or a live queue before coding, and explicitly state that O(N) is the theoretical lower bound since every element must be evaluated at least once.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(1)

Core Theory — Why This Approach?

The Node Payload Detector problem fundamentally tests a candidate's ability to recognize when a linear scan suffices versus when heavier data structures are warranted. In queue-based or streaming architectures, data arrives sequentially, and the goal is to extract a specific aggregate metric without blocking the pipeline. A naive approach often involves sorting the entire dataset or maintaining a priority queue to track the maximum, which introduces unnecessary O(N log N) time overhead and O(N) space consumption. These methods fail on large-scale inputs because they require buffering the entire stream before computation can begin, violating real-time processing constraints and memory budgets.

The optimal paradigm relies on a single-pass linear evaluation that processes each element exactly once as it is dequeued. By maintaining a simple running maximum variable, the algorithm filters elements against the threshold K in constant time per item. This approach aligns with the theoretical lower bound for unstructured search problems, where every element must be inspected at least once to guarantee correctness. Consequently, the solution achieves O(N) time complexity and O(1) auxiliary space, making it ideal for high-throughput queue systems, log aggregators, and real-time metric dashboards.

From an engineering perspective, this pattern demonstrates the importance of constraint-aware algorithm design. Candidates must recognize that sorting or heap-based tracking introduces latency and memory pressure that scale poorly with input size. By leveraging the mathematical property that the maximum of a subset can be computed incrementally, the detector maintains pipeline throughput while minimizing resource allocation. This principle extends to distributed queue consumers, where stateless, single-pass workers can independently compute partial maximums and merge results without coordination overhead.

Interview Questions on This Problem

Q1How would you adapt this detector if the input arrives as an unbounded stream with strict memory limits?

I would process elements sequentially using a sliding window or stateless consumer pattern, maintaining only the running maximum variable in memory. Since the algorithm requires O(1) space, it naturally handles unbounded streams without buffering, ensuring constant memory footprint regardless of input volume.

Q2In a fintech trading system, K represents a dynamic volatility threshold that updates every millisecond. How do you handle this?

I would decouple the threshold evaluation from the scan by passing the current K value into each comparison step. If K changes mid-stream, the running maximum simply continues evaluating against the new threshold, requiring no data structure rebuilds or re-scans, preserving O(N) throughput.

Q3How would you parallelize this detector across multiple worker nodes in a distributed queue architecture?

I would partition the queue into disjoint chunks, assign each to a worker node, and have each compute its local maximum greater than K. A final reducer step would take the maximum of all local results, maintaining O(N) total time while leveraging horizontal scaling.

Examples

Example 1

Input

[1, 2, 3, 4, 5, 250, 6, 7, 8, 9]

Output

Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 250, 6, 7, 8, 9], we first find the maximum value greater than K. In this case, the maximum value greater than K is 250. However, all elements before 250 are less than or equal to K. Therefore, the maximum value greater than K is not 250, but 0.

Example 2

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9]

Output

Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9], we first find the maximum value greater than K. However, since all elements are less than or equal to K, the maximum value greater than K does not exist. Therefore, the output is 0.

Constraints

1 <= N <= 10^5
-10^4 <= metrics[i] <= 10^4
1 <= K <= N

Optimal Approach & Strategy

Iterate through the sequence exactly once, maintaining a running maximum of all elements that satisfy the threshold condition. This single-pass strategy guarantees O(N) time complexity and O(1) auxiliary space by eliminating unnecessary reordering or data structure overhead.

Brute Force Approach

Sort the entire sequence in descending order and iterate until locating the first element that exceeds K. This approach requires O(N log N) time due to the sorting step, making it inefficient and memory-heavy for large-scale inputs.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

function solution(nums, K) {
   let max = -Infinity;
   for (let num of nums) {
      if (num > K && num > max) {
         max = num;
      }
   }
   return max === -Infinity ? 0 : max;
}

class Solution {
public:
   int solution(vector<int>& nums, int K) {
      int max = INT_MIN;
      for (int num : nums) {
         if (num > K && num > max) {
            max = num;
         }
      }
      return max == INT_MIN ? 0 : max;
   }
};

class Solution {
   public int solution(int[] nums, int K) {
      int max = Integer.MIN_VALUE;
      for (int num : nums) {
         if (num > K && num > max) {
            max = num;
         }
      }
      return max == Integer.MIN_VALUE ? 0 : max;
   }
}

def solution(nums, K):
   max_val = float('-inf')
   for num in nums:
      if num > K and num > max_val:
         max_val = num
   return 0 if max_val == float('-inf') else max_val

function solution(nums, K) {
   let max = -Infinity;
   for (let num of nums) {
      if (num > K && num > max) {
         max = num;
      }
   }
   return max === -Infinity ? 0 : max;
}

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Node Payload Detector 34 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Queue

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Node Payload Detector 34 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor