Tome Signal Architect 18 — Problem Statement & Solution Guide

Two PointersMediumRecursive Backtracking

TimeO(n)

SpaceO(1)

Problem Description

Given an array of tome and signal metrics, construct an optimal algorithm to evaluate and compute the target architect value under given operational constraints.

DSA Pattern Breakdown

"Tome Signal Architect 18"

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WHY DOES IT MATTER?

The two‑pointer pattern reduces time complexity from quadratic to linear, which is critical for large datasets and real‑time systems. It also simplifies code by avoiding nested loops and explicit subarray construction.

OPTIMIZATION CHALLENGE

The core insight is that the window’s property is monotonic with respect to its size, allowing us to adjust only one pointer at a time and maintain a running aggregate without recomputation.

REAL-WORLD CONNECTION

In network traffic monitoring, a sliding window tracks packet counts over the last N seconds to detect anomalies. The window expands with new packets and contracts as old packets expire, mirroring the two‑pointer technique.

When explaining this pattern in an interview, emphasize the invariant that the window always satisfies the constraint after each pointer move, and illustrate with a small example to show the pointer progression.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

Two pointers, also known as the sliding window technique, is a powerful paradigm for solving problems that involve contiguous subarrays or substrings. The key idea is to maintain two indices that define a window over the array and to move these indices in a coordinated fashion so that the window always satisfies a certain property (e.g., the sum of its elements is below a threshold). In contrast to the naive O(n^2) approach that checks every possible subarray, the two‑pointer method processes each element at most twice—once when it enters the window and once when it leaves—yielding an O(n) time complexity.

The naive solution typically involves nested loops: for each starting index, iterate over all possible ending indices, recompute the sum, and update the answer. This leads to quadratic time and is infeasible for large inputs (e.g., arrays of size 10^5). Moreover, recomputing the sum from scratch for each subarray incurs additional overhead. By contrast, the sliding window keeps a running sum and adjusts it incrementally as the window expands or contracts, eliminating redundant work.

Two pointers are especially effective when the array contains only non‑negative numbers, because expanding the window can only increase the sum, and contracting it can only decrease it. This monotonicity guarantees that once a window violates the constraint, moving the left pointer forward will eventually restore validity. The technique generalizes to many problems—minimum subarray length, maximum subarray sum under a bound, longest substring without repeating characters, and more—making it a staple in a senior engineer’s toolkit.

Interview Questions on This Problem

Q1What is the time complexity of the two‑pointer sliding window algorithm for finding the minimum length subarray with sum greater than or equal to a target value, and why does it achieve this complexity?

The algorithm runs in O(n) time because each element is added to the window once (when the right pointer moves) and removed at most once (when the left pointer moves). The pointers only move forward, never backward, ensuring linear traversal of the array.

Q2How would you modify the two‑pointer approach if the array contained negative numbers?

With negative numbers, the monotonicity of the sum no longer holds, so a simple sliding window may fail. One approach is to use a prefix sum array and a balanced BST or hash map to track the earliest index with a given prefix sum, enabling O(n log n) or O(n) solutions depending on constraints.

Q3In a distributed system, how can the two‑pointer pattern be applied to process a stream of metrics in real time?

Treat the stream as a sliding window over time: maintain a running sum and window bounds in a stateful microservice. As new metrics arrive, update the sum and adjust the window to satisfy constraints, allowing constant‑time updates per event and efficient real‑time analytics.

Examples

Example 1

Input

[10, 20, 30, 40, 50, 60, 70, 80, 90, 100], 5

Output

Explanation: Step-by-step: Given an array of tome and signal metrics, we first sort the array in descending order. Then, we initialize a variable to store the sum of the K largest values greater than K. We iterate through the sorted array, and for each element, we check if it is greater than K. If it is, we add it to the sum. Finally, we return the sum.

Example 2

Input

[10, 20, 30, 40, 50, 60, 70, 80, 90, 100], 3

Output

Constraints

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

Optimal Approach & Strategy

Maintain a sliding window with two pointers and a running sum. Expand the right pointer to increase the sum, and when the sum exceeds the limit, move the left pointer to shrink the window until the sum is valid again. This runs in O(n) time and O(1) space.

Brute Force Approach

Check every possible subarray by using two nested loops, recompute the sum for each subarray, and keep track of the best answer. This takes O(n^2) time and is impractical for large arrays.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums, k) {
   nums.sort((a, b) => b - a);
   let sum = 0;
   for (let i = 0; i < nums.length; i++) {
       if (nums[i] > k) {
           sum += nums[i];
       }
       if (i === k - 1) break;
   }
   return sum;
}

class Solution {
   public:
       int solution(vector<int>& nums, int k) {
           sort(nums.begin(), nums.end(), greater<int>());
           int sum = 0;
           for (int i = 0; i < nums.size(); i++) {
               if (nums[i] > k) {
                   sum += nums[i];
               }
               if (i == k - 1) 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];
           }
           if (i == k - 1) break;
       }
       return sum;
   }
}

def solution(nums, k):
   nums.sort(reverse=True)
   sum = 0
   for i in range(len(nums)):
       if nums[i] > k:
           sum += nums[i]
       if i == k - 1:
           break
   return sum

function solution(nums, k) {
   nums.sort((a, b) => b - a);
   let sum = 0;
   for (let i = 0; i < nums.length; i++) {
       if (nums[i] > k) {
           sum += nums[i];
       }
       if (i === k - 1) break;
   }
   return sum;
}

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Tome Signal Architect 18 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Tome Signal Architect 18 — 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