Maximized Network Stream Validator 2 — Problem Statement & Solution Guide

Sliding WindowHardSliding Window Maximum Deque

TimeO(N)

SpaceO(k)

Problem Description

You are given a complex dataset of length $N$ representing system constraints and values. Your task is to calculate the maximized network stream using the **Sliding Window Maximum Deque** methodology.

DSA Pattern Breakdown

"Maximized Network Stream Validator 2"

hard

WHY DOES IT MATTER?

The sliding window maximum pattern exemplifies how to transform a seemingly quadratic problem into linear time by exploiting monotonicity, a technique that recurs in many streaming and real‑time analytics scenarios.

OPTIMIZATION CHALLENGE

The key insight is that any element smaller than a newer entrant can never become the maximum for any future window that includes the newer element, allowing us to prune it immediately and keep the deque size bounded by k.

REAL-WORLD CONNECTION

Think of a network router that continuously monitors the highest bandwidth usage over the last 5 seconds; the deque acts like a rolling leaderboard that discards stale or irrelevant measurements instantly.

During an interview, initialize the deque with the first k elements, then loop from k to N‑1, remembering to remove out‑of‑range indices before pushing the new element; this order prevents off‑by‑one bugs.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(k)

Core Theory — Why This Approach?

The sliding window maximum problem asks for the greatest element in every contiguous sub‑array of size k across a length‑N sequence. A naïve solution recomputes the maximum for each window, leading to O(N·k) time, which quickly becomes infeasible when N and k approach 10^6 or larger. The optimal paradigm leverages a double‑ended queue (deque) to maintain candidates for the maximum in a monotonic decreasing order. As the window slides, elements that fall out of the range are popped from the front, while newly entered elements purge any smaller values from the back, guaranteeing that the deque’s front always holds the current window’s maximum.

This monotonic deque technique achieves linear time because each array element is inserted and removed at most once. The approach also uses O(k) auxiliary space, bounded by the window size, which is dramatically smaller than the O(N) space of auxiliary structures like segment trees or heaps. The underlying theory rests on the observation that any element smaller than a newer element can never become a maximum for any future window that includes the newer element, allowing us to discard it safely.

By converting the problem into a series of constant‑time front‑peek operations after O(1) amortized updates, the algorithm scales gracefully for massive data streams, making it the de‑facto solution in real‑time analytics, network traffic monitoring, and financial tick‑data processing.

Interview Questions on This Problem

Q1How would you modify the sliding window maximum algorithm to also return the index of each maximum element?

Store pairs of (value, index) in the deque instead of just values. When popping from the front due to window slide, compare the stored index with the current window’s left bound and discard if it’s out of range. The front of the deque then gives both the maximum value and its original index.

Q2Can you solve the sliding window maximum problem in O(N log k) time using a different data structure? Explain the trade‑offs.

Yes, a balanced binary search tree or a max‑heap with lazy deletions can maintain the window’s elements, offering O(log k) insertion and removal, thus O(N log k) total time. However, the constant factors are higher, and handling duplicate values or lazy deletions adds complexity compared to the O(N) deque solution.

Q3In a distributed system where each node processes a segment of a massive stream, how would you compute the global sliding window maximum efficiently?

Each node computes local maxima for its segment using the deque method and also emits the prefix and suffix maxima of length k‑1. A coordinator then merges overlapping prefix/suffix windows to produce the correct global maxima, achieving near‑linear overall time with minimal inter‑node communication.

Examples

Example 1

Input

[15, 12, 12, 12, 12, 15, 12, 12, 12, 12, 15, 12, 12, 12, 12, 15, 12, 12, 12, 12, 19, 18, 17, 16, 15]

Output

Explanation: Step-by-step: 1. Initialize the deque and result variable. 2. Iterate through the array, maintaining the deque with the maximum values in the current window. 3. For each element, add the maximum value in the deque to the result variable. 4. Return the result variable.

Example 2

Input

[19, 18, 17, 16, 15, 19, 18, 17, 16, 15, 19, 18, 17, 16, 15, 19, 18, 17, 16, 15, 19, 18, 17, 16, 15]

Output

Constraints

1 <= N <= 2 * 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity: O(N) or O(N log N)
Space Complexity: O(N) or O(1)

Optimal Approach & Strategy

Maintain a monotonic decreasing deque that supports O(1) amortized insert and delete, yielding O(N) total time.

Brute Force Approach

Recompute the maximum for each window by scanning all k elements, resulting in O(N·k) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

function solution(nums) {
   const n = nums.length;
   const dq = [];
   let result = 0;
   for (let i = 0; i < n; i++) {
       while (dq.length && dq[0] < i - 3) dq.shift();
       while (dq.length && nums[dq[dq.length - 1]] < nums[i]) dq.pop();
       dq.push(i);
       result += nums[dq[0]];
   }
   return result;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           int n = nums.size();
           deque<int> dq;
           int result = 0;
           for (int i = 0; i < n; i++) {
               while (!dq.empty() && dq.front() < i - 3) dq.pop_front();
               while (!dq.empty() && nums[dq.back()] < nums[i]) dq.pop_back();
               dq.push_back(i);
               result += nums[dq.front()];
           }
           return result;
       }
   };

class Solution {
   public int solution(int[] nums) {
       int n = nums.length;
       Deque<Integer> dq = new ArrayDeque<>();
       int result = 0;
       for (int i = 0; i < n; i++) {
           while (dq.size() > 0 && dq.peekFirst() < i - 3) dq.pollFirst();
           while (dq.size() > 0 && nums[dq.peekLast()] < nums[i]) dq.pollLast();
           dq.addLast(i);
           result += nums[dq.peekFirst()];
       }
       return result;
   }
}

def solution(nums):
   n = len(nums)
   dq = []
   result = 0
   for i in range(n):
       while dq and dq[0] < i - 3: dq.pop(0)
       while dq and nums[dq[-1]] < nums[i]: dq.pop()
       dq.append(i)
       result += nums[dq[0]]
   return result

function solution(nums) {
   const n = nums.length;
   const dq = [];
   let result = 0;
   for (let i = 0; i < n; i++) {
       while (dq.length && dq[0] < i - 3) dq.shift();
       while (dq.length && nums[dq[dq.length - 1]] < nums[i]) dq.pop();
       dq.push(i);
       result += nums[dq[0]];
   }
   return result;
}

Asked in Top Tech Interviews

Morgan StanleyUber

Solve in Interative Editor

Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.

Maximized Network Stream Validator 2 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Sliding Window

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Maximized Network Stream Validator 2 — 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