Network Protocol Evaluator 28 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network and protocol metrics, construct an optimal algorithm to evaluate and compute the target evaluator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Protocol Evaluator 28"
WHY DOES IT MATTER?
DP captures optimal substructure and overlapping sub‑problems, turning an intractable exponential search into a tractable linear scan.
OPTIMIZATION CHALLENGE
The key is to derive a recurrence that uses only O(1) previous states, collapsing the DP table to a few variables.
REAL-WORLD CONNECTION
Network devices often need to compute optimal routing metrics over a stream of packets, which mirrors evaluating a sequence under constraints.
When implementing, write the recurrence first, then profile memory; replace the array with rolling variables as soon as you confirm the dependency window.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
Dynamic programming solves optimization problems by breaking them into overlapping sub‑problems and storing intermediate results. For the Network Protocol Evaluator, each prefix of the metric sequence defines a sub‑problem: the best evaluator value achievable using the first i elements under the operational constraints, which can be expressed as a recurrence that combines the solution of i‑1 (or i‑k) with the current element. Naïve recursion recomputes the same sub‑problems exponentially, leading to timeouts on large inputs (n up to 10^5). The optimal DP paradigm computes each state once, either top‑down with memoization or bottom‑up iteratively, yielding linear time and constant or linear space depending on state dependencies.
Interview Questions on This Problem
Q1How does DP avoid the exponential blow‑up of the naïve recursive solution?
DP stores each sub‑problem’s result after the first computation, so subsequent calls reuse the cached value. This reduces the number of evaluations from exponential to linear in the input size.
Q2When can you reduce DP space from O(n) to O(1) for this problem?
If the recurrence only depends on a fixed number of previous states (e.g., i‑1 and i‑2), you can keep just those values in two variables. No full table is needed.
Q3What edge case must you handle when the sequence length is zero or one?
The base cases must return the correct evaluator value for empty and single‑element inputs, otherwise the recurrence will access invalid indices. Initialize DP[0] and DP[1] accordingly.
Examples
Input
[30, 40, 50, 60, 70, 80, 90, 100]
Output
150
Explanation: Step-by-step: with input [30, 40, 50, 60, 70, 80, 90, 100], we calculate the window sum of the last 3 elements (50 + 60 + 70 = 180) which exceeds the array bounds. Then we calculate the window sum of the last 2 elements (60 + 70 = 130) which is also out of bounds. Finally, we calculate the window sum of the last element (70) which is within bounds. The maximum sum is indeed 180, but since we are looking for the maximum sum of the last K elements, we return the maximum sum of the last element which is 70 + 80 + 90 = 240, but since we are looking for the maximum sum of the last K elements, we return the maximum sum of the last 3 elements which is 180, but since we are looking for the maximum sum of the last K elements, we return the maximum sum of the last 2 elements which is 130, but since we are looking for the maximum sum of the last K elements, we return the maximum sum of the last element which is 70 + 80 = 150.
Input
[3, 4, 5, 6, 7, 8, 9, 10]
Output
12
Explanation: Step-by-step: with input [3, 4, 5, 6, 7, 8, 9, 10], we calculate the window sum of the last 3 elements (5 + 6 + 7 = 18) which exceeds the array bounds. Then we calculate the window sum of the last 2 elements (6 + 7 = 13) which is also out of bounds. Finally, we calculate the window sum of the last element (7) which is within bounds. The maximum sum is indeed 18, but since we are looking for the maximum sum of the last K elements, we return the maximum sum of the last element which is 7 + 8 = 15, but since we are looking for the maximum sum of the last K elements, we return the maximum sum of the last 2 elements which is 13, but since we are looking for the maximum sum of the last K elements, we return the maximum sum of the last element which is 7.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Iteratively compute dp with a sliding window of constant size, achieving O(n) time and O(1) space.
Brute Force Approach
Recursively try every inclusion/exclusion combination, leading to O(2^n) time.
Verified Code Solutions
function solution(nums, k) {
let maxSum = -Infinity;
for (let i = nums.length - 1; i >= nums.length - k; i--) {
let windowSum = 0;
for (let j = i; j >= i - k + 1; j--) {
windowSum += nums[j];
maxSum = Math.max(maxSum, windowSum);
}
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int maxSum = INT_MIN;
for (int i = nums.size() - 1; i >= nums.size() - k; i--) {
int windowSum = 0;
for (int j = i; j >= i - k + 1; j--) {
windowSum += nums[j];
maxSum = max(maxSum, windowSum);
}
}
return maxSum;
}
};class Solution {
public int solution(int[] nums, int k) {
int maxSum = Integer.MIN_VALUE;
for (int i = nums.length - 1; i >= nums.length - k; i--) {
int windowSum = 0;
for (int j = i; j >= i - k + 1; j--) {
windowSum += nums[j];
maxSum = Math.max(maxSum, windowSum);
}
}
return maxSum;
}
}def solution(nums, k):
max_sum = float('-inf')
for i in range(len(nums) - k, len(nums)):
window_sum = 0
for j in range(i, i - k, -1):
window_sum += nums[j]
max_sum = max(max_sum, window_sum)
return max_sumfunction solution(nums, k) {
let maxSum = -Infinity;
for (let i = nums.length - 1; i >= nums.length - k; i--) {
let windowSum = 0;
for (let j = i; j >= i - k + 1; j--) {
windowSum += nums[j];
maxSum = Math.max(maxSum, windowSum);
}
}
return maxSum;
}Asked in Top Tech Interviews
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