Sensor Cluster Tracker 1 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a linear sequence of telemetry readings from a distributed sensor network. The data is structured as a singly linked list where each node contains an integer value representing a specific metric. Your objective is to traverse this linked list and compute the aggregate sum of all node values that strictly exceed a given threshold K.
The input consists of a head pointer to the linked list and an integer K. You must iterate through the list from head to tail, checking each node's value against K. If a node's value is greater than K, add it to a running total. The final result is this total sum. If no values exceed K, the result should be 0.
This problem tests your ability to perform linear traversal on a linked list structure while applying a simple filtering condition. Ensure your solution handles edge cases such as an empty list or a list where no elements meet the criteria efficiently.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Cluster Tracker 1"
WHY DOES IT MATTER?
Single-pass aggregation on linked lists is a fundamental pattern because it demonstrates mastery of pointer traversal, constant‑space computation, and the ability to reason about data streams without random access.
OPTIMIZATION CHALLENGE
The key insight is recognizing that you do not need auxiliary data structures or multiple passes; by maintaining a running total while iterating, you achieve the optimal O(n) time and O(1) space.
REAL-WORLD CONNECTION
Think of a telemetry pipeline where each sensor reading arrives as a node in a stream; engineers often need to compute metrics (e.g., total high‑temperature events) on‑the‑fly without storing the entire history, mirroring the linked‑list sum pattern.
During an interview, write the loop first, then immediately add the conditional check (value > K) inside it—this prevents the common mistake of a second pass or extra list traversal.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single-pass traversal of a singly linked list, accumulating the values of nodes that satisfy the predicate value > K. In a linked list, random access is O(n) because each node only knows its successor, so any algorithm that attempts to revisit nodes or perform nested traversals quickly escalates to O(n²) time, which is unacceptable for large telemetry streams that can contain millions of readings. The optimal paradigm leverages the inherent sequential nature of the structure: a linear scan that maintains a running sum while moving the pointer from head to tail. This approach respects the O(n) time bound and O(1) auxiliary space, because only a few scalar variables (current node reference and accumulator) are needed regardless of list length.
Naïve solutions often try to first convert the linked list into an array or use recursion to accumulate sums. Converting to an array incurs O(n) extra space and adds overhead for copying, while recursive depth can cause stack overflow for deep lists, violating the space constraints. The optimal iterative method sidesteps these pitfalls by directly processing each node as it is visited, ensuring constant extra memory and eliminating the risk of stack overflow. This pattern—single-pass aggregation—is a cornerstone of linked‑list interview questions, reinforcing the importance of understanding pointer manipulation and in‑place computation.
Interview Questions on This Problem
Q1How would you modify the solution if the list could contain negative values and you needed the sum of nodes whose absolute value exceeds K?
Traverse the list once, and for each node check if Math.abs(node.val) > K; if true, add node.val to the accumulator. The algorithmic complexity remains O(n) time and O(1) space because the absolute check is O(1) per node.
Q2What changes are required if the list is doubly linked and you must also return the count of qualifying nodes?
The same linear scan works for a doubly linked list; you simply maintain two counters: one for the sum and one for the count. No additional traversal is needed, preserving O(n) time and O(1) extra space.
Q3In a distributed sensor system, how could you parallelize the sum computation across multiple machines while preserving correctness?
Partition the linked list into contiguous sub‑lists (e.g., by breaking the list at known sentinel nodes), assign each sub‑list to a worker that computes a local sum of values > K, then aggregate the local sums in a reduction step. The overall complexity stays O(n) total work, with O(log p) time for the reduction across p workers, and the approach respects data locality.
Examples
Input
head = [12, 5, 20, 3, 15], K = 10
Output
47
Explanation: Traverse the list: 12 > 10 (sum=12), 5 <= 10 (skip), 20 > 10 (sum=32), 3 <= 10 (skip), 15 > 10 (sum=47). Final sum is 47.
Input
head = [1, 2, 3], K = 5
Output
0
Explanation: Traverse the list: 1 <= 5, 2 <= 5, 3 <= 5. No elements exceed K. Final sum is 0.
Input
head = [100, 200, 300], K = 50
Output
600
Explanation: Traverse the list: 100 > 50 (sum=100), 200 > 50 (sum=300), 300 > 50 (sum=600). Final sum is 600.
Input
head = [], K = 10
Output
0
Explanation: The list is empty. No nodes to process. Final sum is 0.
Constraints
- 0 <= number of nodes in the linked list <= 10^5
- -10^9 <= node.val <= 10^9
- -10^9 <= K <= 10^9
- The linked list is guaranteed to be acyclic.
Optimal Approach & Strategy
Iterate the linked list once, checking each node's value against K and updating a running sum in place.
Brute Force Approach
Convert the linked list to an array, then iterate the array twice—once to filter values > K and once to sum them.
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): return sum(num for num in nums if num > K)function 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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