Sensor Cluster Consolidator 44 — Problem Statement & Solution Guide

Linked ListMediumFrequency Hash Map

TimeO(n)

SpaceO(1)

Problem Description

In a distributed IoT network, sensor nodes transmit raw metric values that must be aggregated for cluster-level analysis. You are provided with a singly linked list where each node contains an integer representing a specific sensor reading. Your task is to process this linked list and compute the total sum of all node values that strictly exceed a given threshold K.

The input consists of a reference to the head of the linked list and an integer K. You must traverse the list exactly once to identify qualifying nodes. The output should be a single integer representing the cumulative sum of these values. If no node value exceeds K, the result must be 0. This operation is critical for real-time anomaly detection where only high-magnitude readings contribute to the alert score.

DSA Pattern Breakdown

"Sensor Cluster Consolidator 44"

medium

WHY DOES IT MATTER?

Streaming aggregates on linked structures enable real‑time analytics with minimal memory footprint.

OPTIMIZATION CHALLENGE

Eliminating auxiliary containers reduces both time and space from O(n) to O(1) extra memory, crucial for edge devices.

REAL-WORLD CONNECTION

IoT gateways often process sensor streams on the fly, summing or filtering readings without storing the entire dataset.

Always traverse once, update a running total, and avoid mutable state that requires a second pass or extra data structures.

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 > threshold). Since each node can be visited only once, a linear-time algorithm is optimal, leveraging the inherent sequential nature of linked lists without extra indexing structures. Naïve approaches might attempt to convert the list to an array, sort it, or use nested loops, which inflate time to O(n log n) or O(n^2) and waste memory, making them unsuitable for large IoT streams where node counts can reach millions. The optimal paradigm is a streaming aggregate: maintain a running sum while iterating, achieving O(n) time and O(1) auxiliary space, which aligns with real‑time processing constraints of sensor networks.

Interview Questions on This Problem

Q1Why is a single traversal sufficient to compute the sum of values greater than a threshold in a singly linked list?

Each node's value is independent of others for the sum, so we can evaluate the predicate and update the accumulator on the fly. No back‑tracking or random access is required, guaranteeing linear time.

Q2What would be the impact of converting the linked list to an array before processing?

Conversion adds O(n) time and O(n) extra space, which is unnecessary for a simple aggregate. It also introduces cache‑miss overhead and defeats the memory‑efficiency of the linked list.

Q3How would you handle potential integer overflow when summing large sensor readings?

Use a wider numeric type (e.g., 64‑bit long) or check for overflow before each addition. In languages with built‑in big integers, leverage those to guarantee correctness.

Examples

Example 1

Input

head = [12, 5, 20, 8, 30], K = 10

Output

Explanation: Traverse the linked list: 12 > 10 (add 12), 5 <= 10 (skip), 20 > 10 (add 20), 8 <= 10 (skip), 30 > 10 (add 30). Total sum = 12 + 20 + 30 = 62.

Example 2

Input

head = [3, 7, 2, 9, 4], K = 5

Output

Explanation: Traverse the linked list: 3 <= 5 (skip), 7 > 5 (add 7), 2 <= 5 (skip), 9 > 5 (add 9), 4 <= 5 (skip). Total sum = 7 + 9 = 16.

Example 3

Input

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

Output

Explanation: Traverse the linked list: All values (1, 2, 3, 4, 5) are less than or equal to 10. No values are added. Total sum = 0.

Example 4

Input

head = [100, 200, 300], K = 150

Output

Explanation: Traverse the linked list: 100 <= 150 (skip), 200 > 150 (add 200), 300 > 150 (add 300). Total sum = 200 + 300 = 500.

Constraints

1 <= number of nodes in linked list <= 10^5
-10^9 <= node.val <= 10^9
-10^9 <= K <= 10^9
The linked list is guaranteed to be non-circular and properly terminated with null.

Optimal Approach & Strategy

Traverse the list once, conditionally accumulate values in a single variable, achieving O(n) time and O(1) extra space.

Brute Force Approach

Convert the list to an array, sort it, then sum elements greater than the threshold—adds unnecessary O(n log n) time and O(n) space.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */

/**
 * @param {ListNode} head
 * @param {number} K
 * @return {number}
 */
var consolidateSensors = function(head, K) {
    let sum = 0;
    let current = head;
    while (current !== null) {
        if (current.val > K) {
            sum += current.val;
        }
        current = current.next;
    }
    return sum;
};

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

class Solution {
public:
    int consolidateSensors(ListNode* head, int K) {
        int sum = 0;
        ListNode* current = head;
        while (current != nullptr) {
            if (current->val > K) {
                sum += current->val;
            }
            current = current->next;
        }
        return sum;
    }
};

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int consolidateSensors(ListNode head, int K) {
        int sum = 0;
        ListNode current = head;
        while (current != null) {
            if (current.val > K) {
                sum += current.val;
            }
            current = current.next;
        }
        return sum;
    }
}

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def consolidateSensors(self, head: 'ListNode', K: int) -> int:
        total = 0
        current = head
        while current:
            if current.val > K:
                total += current.val
            current = current.next
        return total

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */

/**
 * @param {ListNode} head
 * @param {number} K
 * @return {number}
 */
var consolidateSensors = function(head, K) {
    let sum = 0;
    let current = head;
    while (current !== null) {
        if (current.val > K) {
            sum += current.val;
        }
        current = current.next;
    }
    return sum;
};

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Sensor Cluster Consolidator 44 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

Sensor Cluster Consolidator 44 — 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