Network Network Partition 33 — Problem Statement & Solution Guide

Linked ListEasyBFS / Union Find

TimeO(n)

SpaceO(1)

Problem Description

Consider a linear sequence of nodes representing a singly linked list, where each node contains an integer value. You are provided with the head of this list and a threshold integer K. Your task is to traverse the entire structure and compute the aggregate sum of all node values that strictly exceed the threshold K. If no node value satisfies this condition, the result must be zero. The traversal must be performed in a single pass to ensure optimal time complexity, leveraging the sequential nature of the linked list pointers.

DSA Pattern Breakdown

"Network Network Partition 33"

easy

WHY DOES IT MATTER?

This pattern is essential because it reinforces the understanding of linked list traversal, which is a prerequisite for more complex operations like merging, reversing, or detecting cycles. It also highlights the importance of conditional logic within a loop, a skill that is frequently tested in coding interviews.

OPTIMIZATION CHALLENGE

The key insight is that no additional data structures are needed. By using a single accumulator variable and a pointer to traverse the list, you achieve O(1) space complexity. This is optimal because any solution that stores intermediate results or creates new lists would increase space complexity unnecessarily.

REAL-WORLD CONNECTION

In distributed systems, this pattern is analogous to processing a stream of events where only certain events (those exceeding a threshold) contribute to a global metric. For example, in a monitoring system, you might sum up all CPU usage spikes that exceed a baseline threshold to identify potential performance bottlenecks.

During the interview, explicitly state that you are using a single-pass approach to achieve O(n) time and O(1) space. Mention that you are using a 64-bit integer for the sum to handle potential overflow, which demonstrates attention to detail and robustness.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem of summing node values in a singly linked list that exceed a threshold K is a fundamental application of linear traversal. Unlike arrays, linked lists do not support random access, meaning you cannot jump to the i-th element directly. Therefore, the only way to inspect every node is to iterate sequentially from the head to the tail. This makes the problem a canonical example of O(n) time complexity where n is the number of nodes. The algorithm relies on maintaining a running accumulator (sum) that is updated only when the current node's value satisfies the condition value > K. This conditional accumulation ensures that we process each node exactly once, making the solution both time and space efficient.

Interview Questions on This Problem

Q1At a fintech platform, you need to calculate the total value of transactions in a stream that exceed a certain fraud detection threshold. How would you model this if the data arrives as a linked list of transaction nodes?

Model the transaction stream as a singly linked list where each node holds the transaction amount. Traverse the list once, maintaining a running sum of amounts strictly greater than the threshold. This approach ensures O(n) time complexity and O(1) space complexity, which is critical for real-time processing where memory overhead must be minimized.

Q2In a high-growth startup, you are building a feature to filter user engagement metrics. If the metrics are stored in a linked list, how would you efficiently compute the sum of metrics above a dynamic threshold K?

Implement a single-pass traversal of the linked list. For each node, compare its value to K. If it exceeds K, add it to a cumulative sum. Return the final sum. This method is optimal because it avoids storing intermediate results or creating new lists, thus preserving O(1) auxiliary space while maintaining linear time performance.

Q3At a global product company, you are optimizing a data pipeline that processes sensor readings stored in linked lists. How would you handle the case where K is negative, and what edge cases should you consider?

If K is negative, all positive sensor readings will exceed the threshold, so the sum will include all positive values. Edge cases to consider include an empty list (return 0), a list with all values less than or equal to K (return 0), and integer overflow if the sum exceeds the maximum integer limit. Use a 64-bit integer for the sum to prevent overflow.

Examples

Example 1

Input

head = [12, 5, 18, 3, 22], K = 10

Output

Explanation: Traverse the list: Node 1 (12) > 10, add 12. Node 2 (5) <= 10, skip. Node 3 (18) > 10, add 18. Node 4 (3) <= 10, skip. Node 5 (22) > 10, add 22. Total sum = 12 + 18 + 22 = 52.

Example 2

Input

head = [1, 2, 3, 4, 5], K = 100

Output

Explanation: Traverse the list: All node values (1, 2, 3, 4, 5) are less than or equal to 100. No values are added to the sum. Total sum = 0.

Example 3

Input

head = [10, 10, 10], K = 9

Output

Explanation: Traverse the list: Node 1 (10) > 9, add 10. Node 2 (10) > 9, add 10. Node 3 (10) > 9, add 10. Total sum = 10 + 10 + 10 = 30.

Example 4

Input

head = [-5, -1, 0, 2, 7], K = 0

Output

Explanation: Traverse the list: Node 1 (-5) <= 0, skip. Node 2 (-1) <= 0, skip. Node 3 (0) <= 0, skip. Node 4 (2) > 0, add 2. Node 5 (7) > 0, add 7. Total sum = 2 + 7 = 9.

Constraints

1 <= length of linked list <= 10^5
-10^9 <= node.val <= 10^9
-10^9 <= K <= 10^9
The linked list is guaranteed to be valid with no cycles.

Optimal Approach & Strategy

The optimized approach is to traverse the linked list directly using a pointer, maintaining a running sum of values greater than K. This approach uses O(1) additional space and O(n) time, which is optimal for this problem.

Brute Force Approach

The brute force approach would involve converting the linked list into an array, then iterating through the array to sum the values greater than K. This approach is inefficient because it requires O(n) additional space to store the array and O(n) time to convert the list, making it O(n) in both time and 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 sumAboveThreshold = function(head, K) {
    let totalSum = 0;
    let current = head;
    while (current !== null) {
        if (current.val > K) {
            totalSum += current.val;
        }
        current = current.next;
    }
    return totalSum;
};

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

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

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

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

class Solution:
    def sumAboveThreshold(self, head: Optional[ListNode], K: int) -> int:
        total_sum = 0
        current = head
        while current:
            if current.val > K:
                total_sum += current.val
            current = current.next
        return total_sum

/**
 * 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 sumAboveThreshold = function(head, K) {
    let totalSum = 0;
    let current = head;
    while (current !== null) {
        if (current.val > K) {
            totalSum += current.val;
        }
        current = current.next;
    }
    return totalSum;
};

Asked in Top Tech Interviews

GoogleAmazonMicrosoft

Solve in Interative Editor

Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.

Network Network Partition 33 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 Network Partition 33 — 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