Network Network Architect 25 — Problem Statement & Solution Guide

Sliding WindowEasyDFS Traversal

TimeO(N)

SpaceO(1)

Problem Description

You are tasked with optimizing the throughput of a distributed network architecture. The system processes a stream of integer metrics, where each value represents the load on a specific node. To ensure stability, the architecture must identify the most efficient contiguous segment of exactly K nodes that yields the maximum total load. This metric is critical for determining the peak capacity of the network segment.

Given an array of integers representing the node loads and an integer K representing the window size, compute the maximum sum of any contiguous subarray of length K. If the array length is less than K, return 0, as no valid window exists.

The solution must efficiently traverse the data stream, maintaining a sliding window of size K to calculate the sum of the current window and update the maximum value found. This approach ensures optimal performance for large-scale network data.

DSA Pattern Breakdown

"Network Network Architect 25"

easy

WHY DOES IT MATTER?

Sliding window eliminates redundant recomputation for fixed‑size subarrays.

OPTIMIZATION CHALLENGE

It reduces the brute‑force O(N·K) cost to linear O(N) by reusing the previous sum.

REAL-WORLD CONNECTION

Network monitors often need the highest traffic load over a rolling K‑second window.

Keep a running sum variable and update it in place; never recompute the whole window.

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 exactly K elements. A naïve solution enumerates every possible window, recomputing the sum each time, which leads to O(N·K) time and quickly becomes infeasible for large N.

The optimal paradigm uses the sliding window technique: compute the sum of the first K elements, then slide the window one position at a time, subtracting the element leaving the window and adding the new element. This reuses the previous computation, guaranteeing O(N) time while using only O(1) extra space.

Interview Questions on This Problem

Q1What is the time and space complexity of the optimal sliding‑window solution?

The time complexity is O(N) because each array element is added and removed at most once. The space complexity is O(1) as only a few scalar variables are needed.

Q2How does the algorithm behave when the array contains negative numbers?

The sliding window still works because the window size is fixed; the sum may decrease, but we still compare it against the current maximum. The final answer can be negative if all possible K‑length sums are negative.

Q3How would you modify the solution to also return the start and end indices of the optimal segment?

Maintain a variable that records the start index whenever a new maximum sum is found. After the loop, return that start index and start+K‑1 as the segment boundaries.

Examples

Example 1

Input

nums = [2, 1, 5, 1, 3, 2], K = 3

Output

Explanation: The array has length 6, which is >= K (3). We slide a window of size 3 across the array. Window 1: [2, 1, 5] -> sum = 8. Window 2: [1, 5, 1] -> sum = 7. Window 3: [5, 1, 3] -> sum = 9. Window 4: [1, 3, 2] -> sum = 6. The maximum sum encountered is 9.

Example 2

Input

nums = [10, -2, 3, 4, 5], K = 2

Output

Explanation: The array has length 5, which is >= K (2). We slide a window of size 2. Window 1: [10, -2] -> sum = 8. Window 2: [-2, 3] -> sum = 1. Window 3: [3, 4] -> sum = 7. Window 4: [4, 5] -> sum = 9. The maximum sum is 9.

Example 3

Input

nums = [1, 2], K = 3

Output

Explanation: The array length is 2, which is less than K (3). Since no valid window of size 3 can be formed, the function returns 0 as per the problem constraints.

Example 4

Input

nums = [5, 5, 5, 5], K = 4

Output

Explanation: The array length is 4, which equals K (4). There is only one possible window: [5, 5, 5, 5]. The sum is 5 + 5 + 5 + 5 = 20. This is the maximum sum.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
1 <= K <= 10^5
If nums.length < K, return 0

Optimal Approach & Strategy

Compute the sum of the first K elements, then slide the window, adjusting the sum by subtracting the leftmost element and adding the new rightmost one.

Brute Force Approach

Iterate over every possible start index, sum K elements each time, and track the maximum.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

/**
 * @param {number[]} nums
 * @param {number} K
 * @return {number}
 */
var maxSumSubarray = function(nums, K) {
    const n = nums.length;
    if (n < K) return 0;
    
    let currentSum = 0;
    let maxSum = 0;
    
    for (let i = 0; i < K; i++) {
        currentSum += nums[i];
    }
    maxSum = currentSum;
    
    for (let i = K; i < n; i++) {
        currentSum += nums[i] - nums[i - K];
        maxSum = Math.max(maxSum, currentSum);
    }
    
    return maxSum;
};

class Solution {
public:
    int maxSumSubarray(vector<int>& nums, int K) {
        int n = nums.size();
        if (n < K) return 0;
        
        int currentSum = 0;
        int maxSum = 0;
        
        for (int i = 0; i < K; i++) {
            currentSum += nums[i];
        }
        maxSum = currentSum;
        
        for (int i = K; i < n; i++) {
            currentSum += nums[i] - nums[i - K];
            maxSum = max(maxSum, currentSum);
        }
        
        return maxSum;
    }
};

class Solution {
    public int maxSumSubarray(int[] nums, int K) {
        int n = nums.length;
        if (n < K) return 0;
        
        int currentSum = 0;
        int maxSum = 0;
        
        for (int i = 0; i < K; i++) {
            currentSum += nums[i];
        }
        maxSum = currentSum;
        
        for (int i = K; i < n; i++) {
            currentSum += nums[i] - nums[i - K];
            maxSum = Math.max(maxSum, currentSum);
        }
        
        return maxSum;
    }
}

class Solution:
    def maxSumSubarray(self, nums: List[int], K: int) -> int:
        n = len(nums)
        if n < K:
            return 0
        
        current_sum = sum(nums[:K])
        max_sum = current_sum
        
        for i in range(K, n):
            current_sum += nums[i] - nums[i - K]
            max_sum = max(max_sum, current_sum)
        
        return max_sum

/**
 * @param {number[]} nums
 * @param {number} K
 * @return {number}
 */
var maxSumSubarray = function(nums, K) {
    const n = nums.length;
    if (n < K) return 0;
    
    let currentSum = 0;
    let maxSum = 0;
    
    for (let i = 0; i < K; i++) {
        currentSum += nums[i];
    }
    maxSum = currentSum;
    
    for (let i = K; i < n; i++) {
        currentSum += nums[i] - nums[i - K];
        maxSum = Math.max(maxSum, currentSum);
    }
    
    return maxSum;
};

Asked in Top Tech Interviews

GoogleAmazonMicrosoft

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.

Network Network Architect 25 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Sliding Window

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Network Network Architect 25 — 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