Tome Signal Aligner 35 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and signal metrics, and an integer K, construct an optimal algorithm to compute the sum of all elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Signal Aligner 35"
WHY DOES IT MATTER?
Single‑pass aggregation eliminates unnecessary passes and extra storage, which is critical for high‑throughput data pipelines.
OPTIMIZATION CHALLENGE
The key is reducing the problem from O(n log n) or O(n²) to a strict O(n) linear scan with constant extra memory.
REAL-WORLD CONNECTION
It mirrors real‑time analytics where metrics are summed on the fly as events stream in.
Keep the loop tight: compare, conditionally add, and move to next—avoid any auxiliary data structures or recursive calls.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a classic single‑pass aggregation over a singly linked list: as you traverse each node, you compare its value to K and, if larger, add it to a running total. This leverages the linear nature of linked lists, allowing O(n) time with O(1) auxiliary space because no random access or auxiliary containers are required. Naïve alternatives—such as copying nodes into an array, sorting, then summing—inflate the runtime to O(n log n) and consume extra memory, which becomes prohibitive for massive streams. The optimal paradigm embraces in‑place streaming: a deterministic linear scan that maintains only a scalar accumulator, guaranteeing the lowest possible asymptotic complexity for this task.
Interview Questions on This Problem
Q1What is the time and space complexity of summing all nodes greater than K in a singly linked list?
The algorithm runs in O(n) time because each node is visited once, and it uses O(1) extra space since only a few scalar variables are needed.
Q2How would you adapt the solution to also return the count of elements greater than K?
Introduce a second accumulator that increments each time a node's value exceeds K, updating it alongside the sum during the same traversal. This keeps the complexity unchanged.
Q3If the list is extremely large and stored on disk, what technique ensures the algorithm remains efficient?
Process the list in a streaming fashion, reading one node at a time and discarding it after updating the sum, thus avoiding loading the entire structure into memory. This preserves O(n) time and O(1) memory usage.
Examples
Input
[1, 2, 3, 4, 5], K = 3
Output
12
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K = 3, we iterate through the list and sum all elements greater than K. So, 4 + 5 = 9.
Input
[10, 20, 30, 40, 50], K = 25
Output
120
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and K = 25, we iterate through the list and sum all elements greater than K. So, 30 + 40 + 50 = 120.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Traverse the linked list once, maintaining a running sum and adding each node's value only if it exceeds K, achieving O(n) time and O(1) space.
Brute Force Approach
Copy all node values into an array, sort the array, then sum the tail segment greater than K, which costs 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): 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
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.