Payload Sequence Extractor 50 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a high-throughput data stream represented as an array of integers, where each integer denotes a specific payload metric. The system requires the extraction of a cumulative value based on a strict threshold constraint. Specifically, you must compute the sum of all elements in the sequence that are strictly greater than a given integer K. This operation simulates filtering out low-priority or invalid data points to isolate the significant payload contributions.
The input consists of an array of integers payloads and an integer K. Your objective is to iterate through the sequence and accumulate the sum of every element p such that p > K. If no elements satisfy this condition, the result should be 0. The solution must be efficient enough to handle large datasets within linear time complexity, ensuring optimal performance for real-time processing scenarios.
Return the computed sum as a 64-bit integer to accommodate potential overflow from large values. The algorithm should be deterministic and handle edge cases such as empty arrays or arrays where all elements are less than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Sequence Extractor 50"
WHY DOES IT MATTER?
Maintaining a bounded heap lets you answer “top‑K” style queries in real time without full re‑scans.
OPTIMIZATION CHALLENGE
The key is reducing per‑element work from O(n) to O(log K) while keeping the sum up‑to‑date.
REAL-WORLD CONNECTION
Streaming analytics platforms use similar heap windows to keep the most recent high‑value events for alerting.
Cache the current sum and adjust it only when the heap root changes to avoid recomputing from scratch.
COMPLEXITY AT A GLANCE
O(n log K)O(K)Core Theory — Why This Approach?
Heap data structures provide O(log n) insertion and removal while keeping a global ordering property, which is ideal for maintaining a dynamic threshold such as the K‑th largest element in a stream. A naive scan for each query would be O(n) per query and impossible for high‑throughput streams, whereas a min‑heap of size K lets us keep the K largest values at any moment, allowing us to compute the sum of all elements strictly greater than the K‑th largest in O(log n) per update and O(1) per query.
Interview Questions on This Problem
Q1How does a min‑heap of size K help you find the sum of elements greater than the K‑th largest value?
The heap always stores the K largest elements; the smallest in the heap is the K‑th largest overall. Any element larger than this root can be added to a running sum, and when the heap updates the root changes, we adjust the sum accordingly.
Q2What is the time complexity of inserting a new element and updating the sum in this approach?
Insertion is O(log K) because we may push into the heap and possibly pop the smallest; sum updates are O(1) with a simple addition or subtraction.
Q3Why is a single pass with a heap preferable to sorting the entire array for this problem?
Sorting costs O(n log n) once, but a heap maintains only K elements, giving O(n log K) total and using far less memory, which scales better for massive streams.
Examples
Input
payloads = [12, 5, 23, 8, 34], K = 10
Output
57
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 23 > 10 (add 23), 8 <= 10 (skip), 34 > 10 (add 34). Sum = 12 + 23 + 34 = 57.
Input
payloads = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: Iterate through the array: 1 <= 10, 2 <= 10, 3 <= 10, 4 <= 10, 5 <= 10. No elements are greater than 10. Sum = 0.
Input
payloads = [100, 200, 300], K = 150
Output
500
Explanation: Iterate through the array: 100 <= 150 (skip), 200 > 150 (add 200), 300 > 150 (add 300). Sum = 200 + 300 = 500.
Input
payloads = [-5, -10, 0, 5, 10], K = -3
Output
15
Explanation: Iterate through the array: -5 <= -3 (skip), -10 <= -3 (skip), 0 > -3 (add 0), 5 > -3 (add 5), 10 > -3 (add 10). Sum = 0 + 5 + 10 = 15.
Constraints
- 1 <= payloads.length <= 10^5
- -10^9 <= payloads[i] <= 10^9
- -10^9 <= K <= 10^9
- The sum of all elements greater than K will fit within a 64-bit signed integer.
Optimal Approach & Strategy
Maintain a min‑heap of size K and a running sum; update both in O(log K) per incoming element.
Brute Force Approach
Sort the entire array each time or scan all elements for every query, leading to O(n) per query.
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
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