Payload Token Extractor 26 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and token metrics, construct an optimal algorithm to evaluate and compute the target extractor value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Extractor 26"
WHY DOES IT MATTER?
Efficient token extraction is essential for real‑time parsing of network payloads and log streams.
OPTIMIZATION CHALLENGE
The key is collapsing a quadratic search space into a single linear scan by reusing previous computation.
REAL-WORLD CONNECTION
Think of deep packet inspection where firewalls must quickly locate signatures within high‑throughput traffic.
Pre‑compute token indices or use a frequency map, then slide pointers while updating counts in place for maximum cache friendliness.
COMPLEXITY AT A GLANCE
O(n)O(m)Core Theory — Why This Approach?
The Payload Token Extractor problem is a classic sliding‑window string processing challenge where we must locate the minimal segment of a payload that satisfies a set of token frequency constraints. A naive double‑loop enumerates every possible substring, leading to O(n²) time and quickly exhausting time limits on large inputs. The optimal paradigm leverages a two‑pointer (left‑right) window combined with a hash map (or frequency array) to track token counts in real time, allowing the window to expand until constraints are met and then contract to find the smallest valid segment. This linear‑time approach exploits the monotonic nature of the constraints: once a window satisfies the token requirements, any further expansion cannot improve the answer, so we can safely shrink from the left. By maintaining only the necessary state—current counts and the number of satisfied token types—we achieve O(n) time and O(m) auxiliary space, where m is the number of distinct tokens, which scales gracefully for massive payloads.
Interview Questions on This Problem
Q1How does the sliding‑window technique reduce the time complexity from O(n²) to O(n) for this problem?
It maintains a dynamic window that only moves forward, expanding and contracting in linear passes. Each character is visited at most twice, eliminating redundant recomputation of token counts.
Q2What data structure is ideal for tracking token frequencies within the window and why?
A hash map or fixed‑size array keyed by token identifiers provides O(1) updates and lookups. This constant‑time access is crucial for preserving the overall linear runtime.
Q3Why must we track the number of token types that have met their required count rather than just total token count?
Because the constraints are per‑token, not aggregate; a window may have enough total tokens but still miss a specific required token. Monitoring satisfied token types lets us know exactly when the window is valid.
Examples
Input
nums = [1, 2, 3, 4, 5], k = 5
Output
15
Explanation: Step-by-step: with input nums = [1, 2, 3, 4, 5] and k = 5, we sort the array in descending order, then sum the first k elements, which are 5, 4, 3, 2, and 1, giving output 15.
Input
nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 10
Output
55
Explanation: Step-by-step: with input nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and k = 10, we sort the array in descending order, then sum all elements, which are 10, 9, 8, 7, 6, 5, 4, 3, 2, and 1, giving output 55.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Apply a sliding‑window with a frequency map, expanding right until constraints hold, then shrinking left to tighten the window, achieving O(n) time.
Brute Force Approach
Enumerate every possible substring, count token frequencies for each, and keep the smallest valid one, resulting in O(n²) time.
Verified Code Solutions
function solution(nums, k) {
if (k >= nums.length) {
return nums.reduce((a, b) => a + b, 0);
} else {
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}
}class Solution {
public:
int solution(vector<int> nums, int k) {
if (k >= nums.size()) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
} else {
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}
};class Solution {
public int solution(int[] nums, int k) {
if (k >= nums.length) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
} else {
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}
}def solution(nums, k):
if k >= len(nums):
return sum(nums)
else:
return sum(nums[:k])function solution(nums, k) {
if (k >= nums.length) {
return nums.reduce((a, b) => a + b, 0);
} else {
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}
}Asked in Top Tech Interviews
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