Maximized Network Stream Validator 6 — Problem Statement & Solution Guide

Sliding WindowHardSliding Window Maximum Deque

TimeO(n)

SpaceO(1)

Problem Description

In a high-throughput data pipeline, a stream of integer packets is processed in fixed-size batches to detect peak congestion levels. You are given an array stream representing the throughput values of consecutive packets and an integer windowSize denoting the batch capacity. Your task is to determine the maximum possible sum of any contiguous subarray of length windowSize. This metric is critical for validating network stability under peak load conditions.

To solve this efficiently, you must employ a Sliding Window technique augmented with a monotonic deque to maintain the maximum element within the current window in O(1) amortized time per step. The algorithm should track the sum of the current window and update it by subtracting the outgoing element and adding the incoming element as the window slides across the array. The deque must store indices of elements in decreasing order of their values, ensuring the front of the deque always holds the index of the maximum value in the current window.

Return the maximum sum observed across all valid windows of size windowSize. If the array length is less than windowSize, return -1 to indicate an invalid configuration.

DSA Pattern Breakdown

"Maximized Network Stream Validator 6"

hard

WHY DOES IT MATTER?

The sliding‑window pattern transforms problems that require repeated aggregation over overlapping intervals into constant‑time updates, dramatically reducing time complexity from quadratic to linear. It is a cornerstone for real‑time analytics, rate‑limiting, and any scenario where a moving aggregate is needed.

OPTIMIZATION CHALLENGE

The key insight is recognizing that consecutive windows share k‑1 elements; therefore, you can reuse the previous window's sum by a simple O(1) adjustment rather than recomputing from scratch.

REAL-WORLD CONNECTION

Think of a network router that monitors the total bytes transmitted over the last 5 seconds. Instead of recounting every packet each second, the router subtracts the bytes that fell out of the 5‑second window and adds the newest packet count—exactly the sliding‑window principle.

During an interview, compute the first window sum explicitly, then write a tight loop that slides the window while updating maxSum. Keep variables minimal (currentSum, maxSum, leftIdx) to avoid off‑by‑one errors.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem asks for the maximum sum of any contiguous subarray of a fixed length k (windowSize). A naïve solution would recompute the sum for each possible window by iterating over k elements, leading to O(n·k) time, which is prohibitive when both n (the length of the stream) and k are large (e.g., n up to 10^6). The optimal paradigm leverages the sliding‑window technique: once the sum of the first window is known, each subsequent window can be obtained in O(1) by subtracting the element that slides out of the window and adding the new element that slides in. This incremental update eliminates redundant work and yields a linear‑time algorithm. The sliding‑window pattern is a special case of prefix‑sum optimization, but it is more memory‑efficient because it only needs the current sum and does not store the entire prefix array.

Interview Questions on This Problem

Q1How would you modify the sliding‑window solution to also return the starting index of the window with the maximum sum?

Maintain a variable bestStart that records the start index whenever a new maximum sum is found. Initialize it to 0, update it when currentSum > maxSum, and finally return both maxSum and bestStart.

Q2If the array can contain negative numbers, does the sliding‑window approach still work for finding the maximum sum of a fixed‑size subarray?

Yes. The sliding‑window update (subtract left, add right) is independent of element sign, so the algorithm correctly handles negative values and still runs in O(n) time.

Q3Explain how you would adapt the algorithm to find the maximum average of any subarray of length at least k, not exactly k.

Compute prefix sums once, then for each end index i ≥ k, keep track of the minimum prefix sum among indices ≤ i‑k. The maximum average is (prefix[i]‑minPrefix)/windowLength, which can be found in O(n) using a deque or two‑pointer technique.

Examples

Example 1

Input

stream = [12, 4, 5, 6, 7, 8, 9, 10], windowSize = 4

Output

Explanation: Window 1: [12, 4, 5, 6] -> Sum = 27, Max = 12. Window 2: [4, 5, 6, 7] -> Sum = 22, Max = 7. Window 3: [5, 6, 7, 8] -> Sum = 26, Max = 8. Window 4: [6, 7, 8, 9] -> Sum = 30, Max = 9. Window 5: [7, 8, 9, 10] -> Sum = 34, Max = 10. The maximum sum is 34.

Example 2

Input

stream = [3, 1, 4, 1, 5, 9, 2, 6], windowSize = 3

Output

Explanation: Window 1: [3, 1, 4] -> Sum = 8. Window 2: [1, 4, 1] -> Sum = 6. Window 3: [4, 1, 5] -> Sum = 10. Window 4: [1, 5, 9] -> Sum = 15. Window 5: [5, 9, 2] -> Sum = 16. Window 6: [9, 2, 6] -> Sum = 17. Wait, let's re-calculate. 9+2+6=17. Let's check previous. 1+5+9=15. 5+9+2=16. 9+2+6=17. The max is 17. Let me adjust the example to be distinct. Let's use stream = [3, 1, 4, 1, 5, 9, 2, 6], windowSize = 3. Max sum is 17. I will update the output to 17.

Example 3

Input

stream = [100, 200, 300, 400, 500], windowSize = 2

Output

Explanation: Window 1: [100, 200] -> Sum = 300. Window 2: [200, 300] -> Sum = 500. Window 3: [300, 400] -> Sum = 700. Window 4: [400, 500] -> Sum = 900. The maximum sum is 900.

Constraints

1 <= stream.length <= 10^5
-10^9 <= stream[i] <= 10^9
1 <= windowSize <= stream.length
The sum of all elements in any window may exceed 32-bit integer limits, so use 64-bit integers for accumulation.

Optimal Approach & Strategy

Compute the sum of the first window, then slide the window across the array, updating the sum by removing the leftmost element and adding the new rightmost element, achieving O(n) time.

Brute Force Approach

Iterate over every possible start index, sum the next k elements for each window, and track the maximum; this costs O(n·k) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums, k) {
    let maxSum = -Infinity;
    let windowSum = 0;
    let deque = [];

    for (let i = 0; i < nums.length; i++) {
        while (deque.length > 0 && nums[deque[deque.length - 1]] < nums[i]) {
            deque.pop();
        }
        deque.push(i);

        if (deque[0] <= i - k) {
            deque.shift();
        }

        windowSum = 0;
        for (let j = deque[0]; j <= deque[deque.length - 1]; j++) {
            windowSum += nums[j];
        }

        maxSum = Math.max(maxSum, windowSum);
    }

    return maxSum;
}

class Solution {
public:
    int solution(vector<int>& nums, int k) {
        int maxSum = INT_MIN;
        int windowSum = 0;
        deque<int> deque;

        for (int i = 0; i < nums.size(); i++) {
            while (!deque.empty() && nums[deque.back()] < nums[i]) {
                deque.pop_back();
            }
            deque.push_back(i);

            if (deque.front() <= i - k) {
                deque.pop_front();
            }

            windowSum = 0;
            for (int j = deque.front(); j <= deque.back(); j++) {
                windowSum += nums[j];
            }

            maxSum = max(maxSum, windowSum);
        }

        return maxSum;
    }
};

class Solution {
    public int solution(int[] nums, int k) {
        int maxSum = Integer.MIN_VALUE;
        int windowSum = 0;
        Deque<Integer> deque = new ArrayDeque<>();

        for (int i = 0; i < nums.length; i++) {
            while (deque.size() > 0 && nums[deque.peekLast()] < nums[i]) {
                deque.removeLast();
            }
            deque.addLast(i);

            if (deque.peekFirst() <= i - k) {
                deque.removeFirst();
            }

            windowSum = 0;
            for (int j = deque.peekFirst(); j <= deque.peekLast(); j++) {
                windowSum += nums[j];
            }

            maxSum = Math.max(maxSum, windowSum);
        }

        return maxSum;
    }
}

def solution(nums, k):
    max_sum = float('-inf')
    window_sum = 0
    deque = []

    for i in range(len(nums)):
        while deque and nums[deque[-1]] < nums[i]:
            deque.pop()
        deque.append(i)

        if deque[0] <= i - k:
            deque.pop(0)

        window_sum = 0
        for j in range(deque[0], deque[-1] + 1):
            window_sum += nums[j]

        max_sum = max(max_sum, window_sum)

    return max_sum

function solution(nums, k) {
    let maxSum = -Infinity;
    let windowSum = 0;
    let deque = [];

    for (let i = 0; i < nums.length; i++) {
        while (deque.length > 0 && nums[deque[deque.length - 1]] < nums[i]) {
            deque.pop();
        }
        deque.push(i);

        if (deque[0] <= i - k) {
            deque.shift();
        }

        windowSum = 0;
        for (let j = deque[0]; j <= deque[deque.length - 1]; j++) {
            windowSum += nums[j];
        }

        maxSum = Math.max(maxSum, windowSum);
    }

    return maxSum;
}

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Morgan StanleyUber

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Maximized Network Stream Validator 6 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Sliding Window

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Maximized Network Stream Validator 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