Tome Signal Partition 26 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and signal metrics, construct an optimal algorithm to evaluate and compute the target partition value under given operational constraints. The algorithm should sum up all elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Signal Partition 26"
WHY DOES IT MATTER?
The filter‑and‑aggregate pattern is essential because it guarantees linear time and constant space, which is critical for high‑throughput systems processing massive streams of data. It also maps directly to many real‑world scenarios where only a subset of data meets a quality or threshold criterion.
OPTIMIZATION CHALLENGE
The key insight is that the comparison operation is independent for each element; thus, we can discard the need for any auxiliary data structure and perform the aggregation in a single pass, reducing both time and space overhead.
REAL-WORLD CONNECTION
Think of a real‑time monitoring system that flags sensor readings above a safety threshold. The system must quickly sum or count these alerts to trigger alarms, and it cannot afford sorting or complex data structures due to latency constraints.
When explaining this to an interviewer, emphasize the O(n) time and O(1) space guarantees, and mention that the algorithm is trivially parallelizable across multiple cores or machines.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a simple linear scan: iterate through the array once, compare each element to the threshold K, and accumulate the sum of those that exceed K. This is a classic example of the "filter and aggregate" pattern, which can be solved in O(n) time and O(1) auxiliary space. Naive approaches that attempt to sort the array or use nested loops would introduce unnecessary O(n log n) or O(n^2) complexity, making them infeasible for large inputs (e.g., 10^7 elements). By recognizing that the operation is associative and commutative, we can avoid any expensive data structures and directly compute the result in a single pass, which is the optimal paradigm for this class of problems.
Interview Questions on This Problem
Q1How would you modify this algorithm if the requirement changed to find the sum of the top K largest elements instead of all elements greater than a threshold?
You would use a min-heap of size K to keep track of the largest K elements while scanning the array. Each element is compared to the heap's root; if larger, replace the root and heapify. After the scan, sum the heap contents. This runs in O(n log K) time and O(K) space.
Q2In a distributed system where the array is partitioned across multiple nodes, how would you compute the global sum of elements greater than K efficiently?
Each node independently computes its local sum of elements > K. The results are then reduced (e.g., via a tree‑reduce or MapReduce pattern) to a single global sum. This approach preserves linear time complexity overall and requires only O(1) extra space per node.
Q3What potential pitfalls should you watch for when implementing this algorithm in a language with 32-bit integer overflow, and how can you mitigate them?
If the sum can exceed the 32-bit integer range, use a 64-bit integer type (e.g., long in Java, long long in C++) or a BigInteger type. Additionally, validate input constraints and consider using modular arithmetic if the problem specifies a modulus.
Examples
Input
[1, 2, 3, 4, 5], 3
Output
12
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K = 3, we first filter the array to get elements greater than 3, which is [4, 5]. Then we sum up these elements, giving output 12.
Input
[1, 1, 1, 1, 1], 1
Output
0
Explanation: Step-by-step: with input [1, 1, 1, 1, 1] and K = 1, we first filter the array to get elements greater than 1, which is an empty array. Then we sum up these elements, giving output 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
The optimal approach scans the array once, compares each element to K, and accumulates the sum of those that exceed K. This runs in linear time and constant extra space.
Brute Force Approach
A naive approach might sort the array first and then iterate through the sorted list to find the cutoff point, or use nested loops to compare each element against every other element to find those greater than K. Both methods add unnecessary O(n log n) or O(n^2) overhead.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0) return 0;
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (nums.size() == 0) return 0;
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > K) sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0) return 0;
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > K) sum += nums[i];
}
return sum;
}
}def solution(nums, K):
if not nums:
return 0
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction solution(nums, K) {
if (nums.length === 0) return 0;
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) sum += nums[i];
}
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.