Protocol Tome Detector 6 — Problem Statement & Solution Guide

StackEasyBitmasking

TimeO(n)

SpaceO(n)

Problem Description

Given a sequence of data elements representing protocol and tome metrics, construct an optimal algorithm to evaluate and compute the target detector value under given operational constraints.

DSA Pattern Breakdown

"Protocol Tome Detector 6"

easy

WHY DOES IT MATTER?

Monotonic stack patterns turn quadratic pairwise comparisons into linear scans, a critical optimization for any real‑time analytics or monitoring service where latency and throughput are paramount.

OPTIMIZATION CHALLENGE

The breakthrough is recognizing that once an element is popped, it can never influence any future computation, allowing us to discard it permanently and avoid revisiting it.

REAL-WORLD CONNECTION

Think of a conveyor belt with packages of varying weight. As each new package arrives, you need to know how many lighter packages are directly behind it before a heavier one appears. The belt itself acts like a stack: you keep lighter packages in order until a heavier one forces you to discard them, mirroring how a monotonic stack resolves dependencies in streaming data.

When coding, always store indices (not just values) on the stack; this lets you compute distances or spans directly without extra passes.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(n)

Core Theory — Why This Approach?

The "Protocol Tome Detector" problem maps directly to the classic monotonic stack pattern. The goal is to process a linear sequence of metrics and, for each element, efficiently determine a related value (e.g., the next greater metric, the nearest smaller metric, or a cumulative span). A naive double‑loop scans every possible pair, leading to O(n²) time, which quickly becomes infeasible for large n (10⁵‑10⁶) common in production telemetry pipelines. By maintaining a stack that stores indices or values in a monotonic order (either increasing or decreasing), we can guarantee that each element is pushed and popped at most once, collapsing the total work to linear time. The optimal paradigm therefore hinges on two insights: (1) the stack preserves a “candidate” set that respects the problem’s ordering constraint, and (2) the stack’s LIFO nature lets us discard dominated candidates as soon as a new element invalidates them, ensuring O(1) amortized work per element.

Interview Questions on This Problem

Q1How would you compute the span of each day's stock price (the number of consecutive days before it with price ≤ current) in O(n) time?

Use a decreasing monotonic stack that stores pairs of (price, span). While the stack top price ≤ current price, pop and accumulate its span. Push the current price with its total span onto the stack. This yields O(n) time and O(n) space.

Q2Explain why a simple nested loop to find the next greater element for each array entry is unacceptable for n = 10⁶, and how a stack resolves the issue.

A nested loop performs up to n·(n‑1)/2 comparisons, i.e., O(n²), which exceeds time limits for n = 10⁶ (≈10¹² operations). A monotonic stack processes each element once, pushing it onto the stack and popping elements that are smaller, guaranteeing each element is handled at most twice, thus O(n) total work.

Q3In a distributed log‑processing system, you need to compute for each event the distance to the next event of higher priority. Which data structure would you choose and why?

A monotonic decreasing stack is ideal because it maintains a list of pending events ordered by priority. As new events arrive, you pop all lower‑priority events, instantly resolving their distance to the current higher‑priority event, achieving linear processing time and minimal memory overhead.

Examples

Example 1

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Output

Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we do the following: 1. Initialize sum = 0, left pointer = 0, right pointer = 0. 2. In the first iteration, sum = 1, right pointer = 1, left pointer = 0. 3. In the second iteration, sum = 3, right pointer = 2, left pointer = 0. 4. In the third iteration, sum = 6, right pointer = 3, left pointer = 0. 5. In the fourth iteration, sum = 10, right pointer = 4, left pointer = 0. 6. In the fifth iteration, sum = 15, right pointer = 5, left pointer = 0. 7. In the sixth iteration, sum = 21, right pointer = 6, left pointer = 0. 8. In the seventh iteration, sum = 28, right pointer = 7, left pointer = 0. 9. In the eighth iteration, sum = 36, right pointer = 8, left pointer = 0. 10. In the ninth iteration, sum = 45, right pointer = 9, left pointer = 0. 11. In the tenth iteration, sum = 55, right pointer = 10, left pointer = 0. Therefore, the output is 55.

Example 2

Input

[1, 2, 3, 4, 5, 6]

Output

Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6], we do the following: 1. Initialize sum = 0, left pointer = 0, right pointer = 0. 2. In the first iteration, sum = 1, right pointer = 1, left pointer = 0. 3. In the second iteration, sum = 3, right pointer = 2, left pointer = 0. 4. In the third iteration, sum = 6, right pointer = 3, left pointer = 0. 5. In the fourth iteration, sum = 10, right pointer = 4, left pointer = 0. 6. In the fifth iteration, sum = 15, right pointer = 5, left pointer = 0. 7. In the sixth iteration, sum = 21, right pointer = 6, left pointer = 0. Therefore, the output is 21.

Constraints

1 <= N <= 10^5
-10^4 <= metrics[i] <= 10^4
1 <= K <= N

Optimal Approach & Strategy

Traverse the array once while maintaining a monotonic stack; pop elements that no longer satisfy the condition and compute their results on the fly, achieving linear time.

Brute Force Approach

For each element, scan forward until you find the first element that satisfies the required condition (e.g., greater value), recording the distance; repeat for all elements.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums) {
   if (nums.length === 0) return 0;
   let sum = 0;
   let left = 0;
   let right = 0;
   while (right < nums.length) {
       sum += nums[right];
       right++;
   }
   let result = 0;
   while (left < nums.length) {
       result += sum;
       sum -= nums[left];
       left++;
   }
   return result;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           if (nums.size() == 0) return 0;
           int sum = 0;
           int left = 0;
           int right = 0;
           while (right < nums.size()) {
               sum += nums[right];
               right++;
           }
           int result = 0;
           while (left < nums.size()) {
               result += sum;
               sum -= nums[left];
               left++;
           }
           return result;
       }
   };

class Solution {
   public int solution(int[] nums) {
       if (nums.length == 0) return 0;
       int sum = 0;
       int left = 0;
       int right = 0;
       while (right < nums.length) {
           sum += nums[right];
           right++;
       }
       int result = 0;
       while (left < nums.length) {
           result += sum;
           sum -= nums[left];
           left++;
       }
       return result;
   }
}

def solution(nums):
   if not nums:
       return 0
   sum_val = 0
   left = 0
   right = 0
   while right < len(nums):
       sum_val += nums[right]
       right += 1
   result = 0
   while left < len(nums):
       result += sum_val
       sum_val -= nums[left]
       left += 1
   return result

function solution(nums) {
   if (nums.length === 0) return 0;
   let sum = 0;
   let left = 0;
   let right = 0;
   while (right < nums.length) {
       sum += nums[right];
       right++;
   }
   let result = 0;
   while (left < nums.length) {
       result += sum;
       sum -= nums[left];
       left++;
   }
   return result;
}

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Protocol Tome Detector 6 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Stack

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Protocol Tome Detector 6 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor