Tome Voyage Analyzer 20 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a linked list of integer values representing telemetry data from a series of automated survey drones. Each node in the list contains a single integer metric. Your objective is to compute the 'Aggregate Excess Score', defined as the sum of all node values that are strictly greater than a specified threshold K.
The input consists of a singly linked list where each node holds an integer value, and an integer K representing the threshold. You must traverse the list exactly once to identify qualifying nodes and accumulate their values. The solution must handle empty lists and cases where no values exceed the threshold efficiently.
Return the computed sum as a 64-bit integer. If the list is empty or no values exceed K, return 0. The algorithm must operate in O(n) time complexity and O(1) auxiliary space, where n is the number of nodes in the linked list.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Analyzer 20"
WHY DOES IT MATTER?
Linear traversal with conditional accumulation is a fundamental pattern for streaming data structures.
OPTIMIZATION CHALLENGE
The key is to avoid extra passes or auxiliary containers, keeping both time and memory at their theoretical minima.
REAL-WORLD CONNECTION
It mirrors real‑time telemetry pipelines where each sensor reading is processed on arrival without storing the entire history.
Initialize the accumulator outside the loop and update it only when node->val > K to minimize branch mispredictions.
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 values that exceed a threshold K. Since each node is visited exactly once, the algorithm runs in linear time, which is optimal for unsorted linked structures where random access is unavailable.
Naïve alternatives, such as converting the list to an array and then sorting or using nested loops to compare each element against every other, inflate both time and space complexity to O(N log N) or O(N^2). The optimal paradigm leverages the inherent sequential nature of linked lists, applying a straightforward accumulator pattern while maintaining O(1) auxiliary space.
Interview Questions on This Problem
Q1How would you handle a list that may contain negative integers when computing the Aggregate Excess Score?
The algorithm treats negative values like any other integer; they are added only if they exceed K. Ensure K can also be negative, so the comparison remains value > K.
Q2What is the time and space complexity of traversing a singly linked list to compute a sum conditionally?
Time complexity is O(N) because each node is visited once. Space complexity is O(1) as only a few scalar variables are used.
Q3Why is it unnecessary to modify the list structure for this problem?
The required result depends solely on node values, not on list ordering or structure. Preserving the original list avoids side‑effects and keeps the solution safe for reuse.
Examples
Input
linked_list = [12, 5, 18, 7, 22], K = 10
Output
52
Explanation: Traverse the list: 12 > 10 (add 12), 5 <= 10 (skip), 18 > 10 (add 18), 7 <= 10 (skip), 22 > 10 (add 22). Sum = 12 + 18 + 22 = 52.
Input
linked_list = [3, 3, 3], K = 3
Output
0
Explanation: Traverse the list: 3 is not strictly greater than 3, so it is skipped. All elements are equal to K. Sum = 0.
Input
linked_list = [], K = 5
Output
0
Explanation: The linked list is empty. No nodes to traverse. Sum = 0.
Input
linked_list = [-5, -1, 0, 2, 10], K = -2
Output
12
Explanation: Traverse the list: -5 <= -2 (skip), -1 > -2 (add -1), 0 > -2 (add 0), 2 > -2 (add 2), 10 > -2 (add 10). Sum = -1 + 0 + 2 + 10 = 11. Wait, let me re-calculate: -1 + 0 + 2 + 10 = 11. Correction: The values greater than -2 are -1, 0, 2, 10. Sum = -1 + 0 + 2 + 10 = 11.
Constraints
- 0 <= number of nodes in linked list <= 10^5
- -10^9 <= node value <= 10^9
- -10^9 <= K <= 10^9
- The sum of all qualifying values will fit within a 64-bit signed integer.
Optimal Approach & Strategy
Traverse the list once, accumulating values that satisfy node->val > K, using only constant extra memory.
Brute Force Approach
Convert the list to an array, sort it, then sum elements greater than K; this adds unnecessary O(N log N) time and O(N) space.
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):
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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.