Protocol Sensor Architect 40 — Problem Statement & Solution Guide

Binary SearchEasyMonotonic Stack

TimeO(n)

SpaceO(1)

Problem Description

You are tasked with analyzing a sequence of sensor readings to determine the 'architect value' for a specific protocol configuration. Given an array of integers representing sensor metrics, the goal is to identify the number of elements that are strictly greater than all elements to their right. This metric is critical for evaluating the stability of the protocol under monotonic constraints.

Formally, for an array metrics of length n, an element at index i is considered a 'valid architect node' if metrics[i] > metrics[j] for all j such that i < j < n. Note that the last element in the array is always considered a valid architect node because there are no elements to its right.

Your task is to compute the total count of such valid architect nodes in the given sequence. The solution must efficiently process the sequence to determine this count without resorting to brute-force pairwise comparisons for each element.

DSA Pattern Breakdown

"Protocol Sensor Architect 40"

easy

WHY DOES IT MATTER?

Identifying suffix‑dominant elements is a common sub‑problem in many ranking and filtering tasks.

OPTIMIZATION CHALLENGE

The key is reducing the quadratic suffix comparisons to a single pass by storing only the current maximum.

REAL-WORLD CONNECTION

It mirrors scenarios like finding peak stock prices after a given day or detecting dominant sensor thresholds in real‑time streams.

Cache the running max in a primitive variable and avoid extra data structures to keep memory footprint minimal.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The task is to find all "leaders" in an array—elements that are strictly greater than every element to their right. A naive O(n^2) scan compares each element with all following ones, which quickly becomes infeasible for large n due to quadratic blow‑up. The optimal paradigm leverages a single right‑to‑left pass, maintaining the maximum seen so far; each element is compared only once against this running maximum, yielding linear time. This approach exemplifies the greedy strategy of preserving only the necessary state (the current max) to make optimal local decisions that lead to a globally correct solution.

Interview Questions on This Problem

Q1What is the definition of a leader element in an array?

A leader is an element that is strictly greater than all elements to its right. The rightmost element is always a leader by definition.

Q2Why does scanning the array from right to left enable an O(n) solution?

Scanning right to left lets us keep the maximum of the suffix seen so far, so each element is compared only once. This eliminates the need for nested loops.

Q3How would you modify the algorithm to count leaders that are greater than or equal to elements on the right?

Change the strict comparison to a non‑strict one (>=) when updating the leader count. The rest of the reverse traversal logic remains identical.

Examples

Example 1

Input

metrics = [7, 4, 9, 2, 5]

Output

Explanation: We scan the array from right to left to track the maximum value seen so far. 1. Start at index 4 (value 5). It is the last element, so it is valid. Current max = 5. Count = 1. 2. Move to index 3 (value 2). 2 is not greater than current max (5). Not valid. Max remains 5. 3. Move to index 2 (value 9). 9 is greater than current max (5). Valid. Update max to 9. Count = 2. 4. Move to index 1 (value 4). 4 is not greater than current max (9). Not valid. Max remains 9. 5. Move to index 0 (value 7). 7 is not greater than current max (9). Not valid. Max remains 9. Total count is 2.

Example 2

Input

metrics = [1, 2, 3, 4, 5]

Output

Explanation: The array is strictly increasing. 1. Index 4 (value 5) is the last element. Valid. Max = 5. Count = 1. 2. Index 3 (value 4). 4 < 5. Not valid. 3. Index 2 (value 3). 3 < 5. Not valid. 4. Index 1 (value 2). 2 < 5. Not valid. 5. Index 0 (value 1). 1 < 5. Not valid. Only the last element is greater than all elements to its right (vacuously true for the last, and false for others). Total count is 1.

Example 3

Input

metrics = [5, 4, 3, 2, 1]

Output

Explanation: The array is strictly decreasing. 1. Index 4 (value 1). Last element. Valid. Max = 1. Count = 1. 2. Index 3 (value 2). 2 > 1. Valid. Max = 2. Count = 2. 3. Index 2 (value 3). 3 > 2. Valid. Max = 3. Count = 3. 4. Index 1 (value 4). 4 > 3. Valid. Max = 4. Count = 4. 5. Index 0 (value 5). 5 > 4. Valid. Max = 5. Count = 5. Every element is greater than all elements to its right. Total count is 5.

Example 4

Input

metrics = [10, 10, 10, 10]

Output

Explanation: The array contains equal elements. 1. Index 3 (value 10). Last element. Valid. Max = 10. Count = 1. 2. Index 2 (value 10). 10 is not strictly greater than 10. Not valid. 3. Index 1 (value 10). 10 is not strictly greater than 10. Not valid. 4. Index 0 (value 10). 10 is not strictly greater than 10. Not valid. Only the last element qualifies. Total count is 1.

Constraints

1 <= metrics.length <= 10^5
-10^9 <= metrics[i] <= 10^9
The input array will not be empty.

Optimal Approach & Strategy

Traverse the array from right to left, keep a running maximum, and increment the count whenever the current element exceeds this maximum.

Brute Force Approach

For each index, compare its value with every element to its right and count it if it’s larger.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums, K) {
      nums.sort((a, b) => a - b);
      let sum = 0;
      for (let i = 0; i < nums.length; i++) {
         if (nums[i] <= K) {
            sum += nums[i];
         } else {
            break;
         }
      }
      return sum;
   }

class Solution {
   public:
      int solution(vector<int>& nums, int K) {
         sort(nums.begin(), nums.end());
         int sum = 0;
         for (int i = 0; i < nums.size(); i++) {
            if (nums[i] <= K) {
               sum += nums[i];
            } else {
               break;
            }
         }
         return sum;
      }
   };

class Solution {
      public int solution(int[] nums, int K) {
         Arrays.sort(nums);
         int sum = 0;
         for (int i = 0; i < nums.length; i++) {
            if (nums[i] <= K) {
               sum += nums[i];
            } else {
               break;
            }
         }
         return sum;
      }
   }

def solution(nums, K):
      nums.sort()
      sum = 0
      for num in nums:
         if num <= K:
            sum += num
         else:
            break
      return sum

function solution(nums, K) {
      nums.sort((a, b) => a - b);
      let sum = 0;
      for (let i = 0; i < nums.length; i++) {
         if (nums[i] <= K) {
            sum += nums[i];
         } else {
            break;
         }
      }
      return sum;
   }

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Protocol Sensor Architect 40 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Binary Search

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Protocol Sensor Architect 40 — 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