Protocol Sensor Analyzer 19 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing protocol and sensor metrics, construct an optimal algorithm to evaluate and compute the target analyzer value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Analyzer 19"
WHY DOES IT MATTER?
DP captures optimal substructure and overlapping sub‑problems, which are common in protocol and sensor metric aggregations.
OPTIMIZATION CHALLENGE
The key is to reduce the exponential combination of metric sequences to a single pass over the data.
REAL-WORLD CONNECTION
Think of a network router aggregating packet statistics where each step builds on the previous cumulative state.
Initialize the DP array with sentinel values and update in place to avoid extra memory churn.
COMPLEXITY AT A GLANCE
O(n^2)O(n)Core Theory — Why This Approach?
Dynamic programming solves optimization problems by breaking them into overlapping sub‑problems and storing intermediate results. For the Protocol Sensor Analyzer, the optimal value can be expressed as a recurrence that depends on previously computed analyzer values, turning an exponential search space into a linear or quadratic one. Naïve enumeration tries every possible subsequence or state transition, leading to O(2^n) or O(n!) time, which explodes even for moderate n. The DP paradigm replaces repeated work with a memo table, guaranteeing each sub‑problem is solved once and enabling the algorithm to run in polynomial time while preserving optimality.
Interview Questions on This Problem
Q1How does memoization convert an exponential brute‑force solution into a polynomial DP solution?
Memoization caches the result of each sub‑problem the first time it is computed. Subsequent calls return the cached value in O(1), eliminating redundant recursive branches.
Q2When can the DP state be reduced from O(n^2) to O(n) for this problem?
If the recurrence only depends on the best value of the immediate predecessor, a single rolling variable suffices. This collapses the two‑dimensional table into a constant‑size buffer.
Q3What is the significance of choosing the correct base case in the DP table?
The base case anchors the recurrence and prevents undefined accesses. An incorrect base propagates wrong values throughout the table, breaking optimality.
Examples
Input
[10, 5, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100]
Output
100
Explanation: Step-by-step: Given the input array [10, 5, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100] and K = 5, we iterate through the array and find the maximum element greater than or equal to K, which is 100. Therefore, the output is 100.
Input
[5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5]
Output
5
Explanation: Step-by-step: Given the input array [5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5] and K = 5, we iterate through the array and find the maximum element greater than or equal to K, which is 5. Therefore, the output is 5.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Build a DP table where dp[i] holds the optimal value ending at position i, updating it using the recurrence derived from problem constraints.
Brute Force Approach
Enumerate every possible subsequence or state transition and compute its analyzer value, which is exponential in the input size.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0) return 0;
let max = -Infinity;
for (let num of nums) {
if (num >= K) {
max = Math.max(max, num);
}
}
return max;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (nums.empty()) return 0;
int max = INT_MIN;
for (int num : nums) {
if (num >= K) {
max = std::max(max, num);
}
}
return max;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0) return 0;
int max = Integer.MIN_VALUE;
for (int num : nums) {
if (num >= K) {
max = Math.max(max, num);
}
}
return max;
}
}def solution(nums, K):
if not nums:
return 0
max_val = float('-inf')
for num in nums:
if num >= K:
max_val = max(max_val, num)
return max_valfunction solution(nums, K) {
if (nums.length === 0) return 0;
let max = -Infinity;
for (let num of nums) {
if (num >= K) {
max = Math.max(max, num);
}
}
return max;
}Asked in Top Tech Interviews
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