Network Node Validator 19 — Problem Statement & Solution Guide

Linked ListEasyGreedy Choice

TimeO(N)

SpaceO(1)

Problem Description

You are tasked with processing a linear sequence of integer metrics collected from a distributed network. The system requires a specific validation score to be computed based on a threshold parameter K. Your objective is to iterate through the sequence and calculate the cumulative sum of all elements that strictly exceed the threshold K. Elements equal to or less than K are ignored in the final calculation. If no elements exceed the threshold, the validator value defaults to zero.

The input consists of a list of integers representing the node metrics and an integer K representing the validation threshold. The output is a single integer representing the computed validator value. This problem tests your ability to implement a straightforward linear scan with a conditional filter, ensuring efficient O(n) time complexity and O(1) space complexity.

DSA Pattern Breakdown

"Network Node Validator 19"

easy

WHY DOES IT MATTER?

Summation with a predicate is a fundamental streaming pattern; mastering it demonstrates the ability to process unbounded data efficiently, a skill crucial for real‑time analytics and low‑latency services.

OPTIMIZATION CHALLENGE

The key insight is recognizing that the predicate can be evaluated on-the-fly, allowing a single accumulator to replace any need for auxiliary collections, thereby collapsing both time and space complexity to their minima.

REAL-WORLD CONNECTION

Think of a network monitoring system that streams packet latencies; you need the total latency of packets exceeding a threshold to trigger alerts, and you must compute this on the fly without storing the entire stream.

During an interview, write the traversal loop first, then immediately add the conditional sum inside it. Avoid premature optimizations like building arrays; keep the code tight and explain that the list's sequential nature forces a one‑pass solution.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to a single-pass traversal of a singly linked list, accumulating values that satisfy a simple predicate (value > K). In a naive setting, one might repeatedly scan the list for each node or use auxiliary data structures to store intermediate results, leading to O(N^2) time or unnecessary memory overhead. The optimal paradigm leverages the linear nature of linked lists: by maintaining a running sum while iterating once, we achieve O(N) time with O(1) extra space, because each node is visited exactly once and no additional containers are required. This approach aligns with the classic "stream processing" model where data is consumed in a single pass, making it ideal for large inputs where memory constraints and cache locality are critical.

When dealing with linked lists, pointer manipulation is the only way to move forward; random access is impossible. Therefore, any algorithm that attempts to revisit nodes or backtrack incurs extra traversals, which quickly becomes prohibitive as N grows. The optimal solution embraces the forward-only constraint, using a simple accumulator variable and a conditional check at each step. This not only guarantees linear time but also preserves the original list structure, a requirement in many interview settings where mutating the input is disallowed.

Interview Questions on This Problem

Q1How would you modify the solution if the list were doubly linked and you needed the sum of nodes greater than K while also removing those nodes from the list?

Traverse the list once, keeping a running sum. For each node with value > K, adjust its previous and next pointers to bypass it, effectively deleting it in O(1) time per removal. The overall complexity remains O(N) time and O(1) extra space.

Q2Can you compute the sum of nodes greater than K without using an explicit accumulator variable, perhaps using recursion?

Yes, define a recursive function that returns the sum for the sublist starting at the current node: sum(node) = (node.val > K ? node.val : 0) + sum(node.next). The recursion depth equals the list length, so time is O(N) but space becomes O(N) due to call stack.

Q3If the list is extremely large and stored on disk (e.g., a memory‑mapped file), what considerations change for this algorithm?

The algorithm remains a single sequential scan, which is optimal for disk‑based data because it minimizes random I/O. However, you must ensure you read the list in blocks that fit in memory and maintain the accumulator across block boundaries, still achieving O(N) time with minimal additional memory.

Examples

Example 1

Input

nums = [12, 5, 8, 15, 3], K = 10

Output

Explanation: Iterate through the list: 12 > 10 (add 12), 5 <= 10 (skip), 8 <= 10 (skip), 15 > 10 (add 15), 3 <= 10 (skip). Sum = 12 + 15 = 27.

Example 2

Input

nums = [1, 2, 3, 4, 5], K = 10

Output

Explanation: Iterate through the list: All elements (1, 2, 3, 4, 5) are less than or equal to 10. No elements are added to the sum. Final sum = 0.

Example 3

Input

nums = [10, 10, 10, 11], K = 10

Output

Explanation: Iterate through the list: 10 <= 10 (skip), 10 <= 10 (skip), 10 <= 10 (skip), 11 > 10 (add 11). Sum = 11.

Example 4

Input

nums = [-5, -2, 0, 2, 5], K = -1

Output

Explanation: Iterate through the list: -5 <= -1 (skip), -2 <= -1 (skip), 0 > -1 (add 0), 2 > -1 (add 2), 5 > -1 (add 5). Sum = 0 + 2 + 5 = 7.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
-10^9 <= K <= 10^9

Optimal Approach & Strategy

Perform a single linear traversal, checking each node's value against K and updating a running total when the condition holds. This yields O(N) time with O(1) auxiliary space.

Brute Force Approach

Repeatedly scan the list for each node to check if it exceeds K, leading to O(N^2) time. This also often uses extra storage to keep track of visited nodes.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

function solution(nums, K) {
   let sum = 0;
   for (let i = 0; i < nums.length; i++) {
       if (nums[i] > K) {
           sum += nums[i];
       }
   }
   return sum;
}

class Solution {
public:
   int solution(vector<int> nums, int K) {
       int sum = 0;
       for (int i = 0; i < nums.size(); i++) {
           if (nums[i] > K) {
               sum += nums[i];
           }
       }
       return sum;
   }
};

class Solution {
   public int solution(int[] nums, int K) {
       int sum = 0;
       for (int i = 0; i < nums.length; i++) {
           if (nums[i] > K) {
               sum += nums[i];
           }
       }
       return sum;
   }
}

def solution(nums, K):
   sum = 0
   for i in range(len(nums)):
       if nums[i] > K:
           sum += nums[i]
   return sum

function solution(nums, K) {
   let sum = 0;
   for (let i = 0; i < nums.length; i++) {
       if (nums[i] > K) {
           sum += nums[i];
       }
   }
   return sum;
}

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Network Node Validator 19 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Linked List

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Network Node Validator 19 — 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