Payload Sequence Consolidator 42 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and sequence metrics, and an integer K, construct an optimal algorithm to evaluate and compute the sum of all numbers greater than or equal to K in the sequence.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Sequence Consolidator 42"
WHY DOES IT MATTER?
Efficient aggregation of filtered data is a core building block for analytics pipelines.
OPTIMIZATION CHALLENGE
Eliminate unnecessary sorting or auxiliary structures to achieve linear time and constant space.
REAL-WORLD CONNECTION
Think of a network router tallying packets larger than a size threshold to enforce QoS policies.
When streaming, always keep the state minimal—just the accumulator and the threshold.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a linear scan where each element is compared against the threshold K and, if it meets the condition, added to a running total. This single‑pass approach leverages the principle of stream processing, allowing us to handle arbitrarily large sequences without storing them, which is crucial for memory‑constrained environments. Naïve alternatives—such as sorting the entire array (O(N log N)) or using nested loops to recompute partial sums—inflate both time and space complexity, making them infeasible for big‑data scenarios. The optimal paradigm embraces an accumulator pattern combined with a queue‑like consumption model, ensuring O(N) time and O(1) auxiliary space while preserving order‑agnostic correctness.
Interview Questions on This Problem
Q1How would you compute the sum of all numbers ≥ K in a single pass?
Initialize a sum variable to zero, iterate through the sequence, and add each element to the sum if it is greater than or equal to K. This yields O(N) time and O(1) extra space.
Q2Why is sorting the array before summing not optimal for this problem?
Sorting incurs O(N log N) time, which is unnecessary because the order of elements does not affect the sum condition. A linear scan achieves the same result faster.
Q3Can this algorithm be adapted for a streaming input where the total length is unknown?
Yes; maintain the accumulator and process each incoming element on the fly, discarding it after the check, which keeps memory usage constant. This is ideal for real‑time telemetry or log processing.
Examples
Input
[10, 20, 30, 40, 50, 5, 15, 25, 35, 45, 55]
Output
100
Explanation: Step-by-step: with input [10, 20, 30, 40, 50, 5, 15, 25, 35, 45, 55], we iterate through the array and add numbers greater than or equal to K (20) to the sum. We skip numbers less than K (5). The sum is 100.
Input
[10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 5, 10, 15]
Output
60
Explanation: Step-by-step: with input [10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 5, 10, 15], we iterate through the array and add numbers greater than or equal to K (10) to the sum. We skip numbers less than K (5). The sum is 60.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
The optimal solution iterates once, conditionally adds to an accumulator, achieving O(N) time and O(1) space.
Brute Force Approach
A brute‑force method might sort the array then sum from the first element ≥ K, costing O(N log N) time.
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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