Node Payload Optimizer 20 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing data throughput in a distributed network where each node transmits a payload with a specific weight. The system operates on a fixed-size buffer that can hold exactly K consecutive payloads at any given time. Your objective is to determine the maximum total weight of any contiguous subsequence of length K within the provided stream of payload weights.
Given an array weights representing the payload sizes and an integer K representing the buffer capacity, compute the maximum sum of any sliding window of size K. If the array length is less than K, return 0, as no valid window can be formed. The solution must efficiently process the stream without recalculating the entire sum for every position, leveraging the overlapping nature of consecutive windows.
Input: An integer array weights and an integer K.
Output: An integer representing the maximum sum of any contiguous subarray of length K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Optimizer 20"
WHY DOES IT MATTER?
Sliding windows turn quadratic scans into linear passes, a core performance lever for real‑time systems.
OPTIMIZATION CHALLENGE
The key is to update the window's aggregate in O(1) instead of recomputing it from scratch each step.
REAL-WORLD CONNECTION
Think of a network buffer that continuously drops the oldest packet while receiving a new one, always keeping the last K packets in view.
Initialize the first window fully, then loop once, adjusting the sum by subtracting arr[i‑K] and adding arr[i] for each i.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The sliding window technique transforms a problem that naively requires recomputing a function over every possible subarray into a linear‑time solution by reusing work from the previous step. For a fixed‑size window K, the sum of the next window can be obtained by subtracting the element that leaves the window and adding the new element that enters, eliminating the O(K) recomputation for each position.
A brute‑force O(N·K) approach quickly becomes infeasible when N reaches 10^5 or higher, as the inner loop dominates runtime. The optimal paradigm leverages a running aggregate (e.g., sum, max) and updates it in O(1) per shift, yielding an overall O(N) algorithm that scales gracefully for large inputs while using only constant extra space.
Interview Questions on This Problem
Q1How does the sliding window technique reduce the time complexity for fixed‑size subarray problems?
It maintains a running aggregate that can be updated in constant time as the window moves, avoiding recomputation of the entire window. This changes the complexity from O(N·K) to O(N).
Q2When is a sliding window not the appropriate strategy?
If the window size varies based on data conditions or the operation isn’t easily updatable (e.g., median without extra structures), a simple sliding window fails. In such cases, more advanced data structures or two‑pointer techniques are needed.
Q3What extra care is needed when K equals the array length or exceeds it?
When K equals N, the answer is simply the aggregate of the whole array; when K > N, the problem is undefined and should be handled with an early return or error. Edge‑case checks prevent out‑of‑bounds access.
Examples
Input
weights = [2, 1, 5, 1, 3, 2], K = 3
Output
9
Explanation: The possible windows of size 3 are: [2,1,5] sum=8, [1,5,1] sum=7, [5,1,3] sum=9, [1,3,2] sum=6. The maximum sum is 9.
Input
weights = [10, 20, 30, 40], K = 2
Output
70
Explanation: The possible windows of size 2 are: [10,20] sum=30, [20,30] sum=50, [30,40] sum=70. The maximum sum is 70.
Input
weights = [5], K = 2
Output
0
Explanation: The array length (1) is less than K (2), so no valid window exists. Return 0.
Input
weights = [1, 1, 1, 1, 1], K = 5
Output
5
Explanation: There is only one window of size 5: [1,1,1,1,1] sum=5. The maximum sum is 5.
Constraints
- 1 <= weights.length <= 10^5
- 1 <= K <= 10^5
- -10^4 <= weights[i] <= 10^4
- K <= weights.length for valid non-zero outputs
Optimal Approach & Strategy
Compute the sum of the first K elements, then slide the window across the array, updating the sum in O(1) per move and recording the max.
Brute Force Approach
Iterate over every possible start index, sum K elements each time, and keep the maximum; this is O(N·K) time.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num <= K) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num <= K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num <= K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for num in nums:
if num <= K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num <= K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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