Payload Cipher Evaluator 27 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and cipher metrics, construct an optimal algorithm to evaluate and compute the target evaluator value under given operational constraints. K is the threshold value.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Cipher Evaluator 27"
WHY DOES IT MATTER?
Sliding‑window queue patterns turn quadratic scans into linear passes.
OPTIMIZATION CHALLENGE
The key is to ensure each element is processed a constant number of times.
REAL-WORLD CONNECTION
Network routers use similar queues to keep the most recent packets for QoS calculations.
Initialize the deque before the main loop and always purge stale indices before reading the front.
COMPLEXITY AT A GLANCE
O(N)O(K)Core Theory — Why This Approach?
The problem reduces to maintaining a dynamic view of the last K payload‑cipher metrics while efficiently extracting the required evaluator (e.g., maximum, minimum, or a custom aggregate). A monotonic queue (deque) provides O(1) amortized access to the extremal element by storing indices in decreasing (or increasing) order and discarding out‑of‑range entries, thus supporting both insertion and removal in constant time.
A naïve solution recomputes the evaluator for every window by scanning K elements, leading to O(N·K) time which explodes for N up to 10^6 and K up to 10^5. The optimal paradigm leverages the sliding‑window property: each element enters and leaves the window exactly once, and the deque guarantees that each element is pushed and popped at most once, yielding a linear O(N) overall runtime.
Interview Questions on This Problem
Q1Why does a monotonic queue guarantee O(1) amortized removal of obsolete elements?
Each element is inserted once and removed at most once when it falls out of the window. The deque operations are constant time, so the total work per element is O(1).
Q2How would you adapt the solution if the evaluator required the sum of the window instead of the max?
Maintain a running sum variable, adding the new element and subtracting the element that exits the window. This keeps the sum update in O(1) without a deque.
Q3What edge case must you handle when K equals 1 or N?
When K = 1, every element is its own window, so the answer is the original array; when K = N, the evaluator is computed once over the whole array. Both cases must avoid out‑of‑bounds deque accesses.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100], K = 3
Output
0
Explanation: Step-by-step: Given the input array and K = 3, we iterate through the array. Since all numbers are less than K, we return 0.
Input
[100, 200, 300, 400, 500, 600, 700, 800, 900, 1000], K = 1000
Output
0
Explanation: Step-by-step: Given the input array and K = 1000, we iterate through the array. Since all numbers are greater than or equal to K, we return the sum of all numbers in the array.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a monotonic deque to maintain candidates, updating it as the window slides for O(N) total time.
Brute Force Approach
For each position, scan the next K elements to compute the evaluator, resulting in O(N·K) 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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