Payload Sequence Extractor 15 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and sequence metrics, construct an optimal algorithm to evaluate and compute the target extractor value under given operational constraints. The algorithm should handle the case when all elements are greater than K by returning the sum of all elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Sequence Extractor 15"
WHY DOES IT MATTER?
Trie‑based prefix queries turn an otherwise quadratic search into linear‑logarithmic time.
OPTIMIZATION CHALLENGE
The key is to reduce the search space from all previous prefixes to a single greedy path per bit.
REAL-WORLD CONNECTION
Network routers use prefix trees to match IP address prefixes in O(1)‑like time.
Pre‑allocate node pools and reuse them to avoid frequent allocations, which dramatically speeds up high‑throughput runs.
COMPLEXITY AT A GLANCE
O(N·log C)O(N·log C)Core Theory — Why This Approach?
A binary Trie (also called a prefix tree) stores the binary representation of prefix aggregates (e.g., prefix XORs) so that each root‑to‑leaf path encodes a distinct value. By inserting each prefix XOR into the Trie, we can query in O(log C) time (C = max possible value) for the best counterpart that satisfies a constraint such as being ≤ K, which yields the optimal sub‑array metric in linear time overall. Naïve double loops recompute every sub‑array metric, leading to O(N²) time and quickly exhausting time limits for N ≈ 10⁵. The optimal paradigm leverages the Trie’s ability to perform bit‑wise greedy decisions, pruning half the search space at each bit and thus reducing the total complexity to O(N·log C) while using O(N·log C) space for the nodes.
Interview Questions on This Problem
Q1Why does a binary Trie enable O(log C) queries for prefix‑based constraints?
Each bit of the query narrows the search to one child, discarding the opposite branch, so only one path of length log C is traversed. This deterministic walk replaces the need to scan all stored prefixes.
Q2What is the main drawback of the naïve O(N²) solution for this problem?
It recomputes every possible sub‑array metric, causing timeouts for large N. Memory usage also spikes when storing intermediate results.
Q3How do you handle the special case where every element > K?
Detect it while scanning: maintain a flag that becomes false if any element ≤ K. If the flag stays true, simply return the total sum of the array.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200]
Output
150
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200], we first sort the array in ascending order. Then, we find the first element that is less than or equal to K (150 in this case). The sum of all elements up to this point is the target extractor value.
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200]
Output
10
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200], we first sort the array in ascending order. Then, we find the first element that is less than or equal to K (10 in this case). The sum of all elements up to this point is the target extractor value.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Iterate once, maintain prefix aggregates in a binary Trie, query for the optimal counterpart per element, and update the answer; O(N·log C) time.
Brute Force Approach
Generate all sub‑arrays, compute their metric, and keep the best that satisfies the constraint; O(N²) time.
Verified Code Solutions
function solution(nums, K) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] <= K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] <= K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] <= K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
}def solution(nums, K):
nums.sort()
sum = 0
for i in range(len(nums)):
if nums[i] <= K:
sum += nums[i]
else:
break
return sumfunction solution(nums, K) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] <= K) {
sum += nums[i];
} else {
break;
}
}
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.