Payload Sequence Resolver 44 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and sequence metrics, construct an optimal algorithm to evaluate and compute the target resolver value under given operational constraints. The algorithm should handle the case when the window size is greater than K and find the maximum sum of elements in the window.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Sequence Resolver 44"
WHY DOES IT MATTER?
Sliding‑window patterns turn quadratic scans into linear passes, crucial for real‑time analytics.
OPTIMIZATION CHALLENGE
The key is to update aggregates incrementally instead of recomputing from scratch for each window.
REAL-WORLD CONNECTION
Network routers compute moving averages of packet sizes to detect congestion using the same principle.
Always keep a running aggregate and remember to handle the element that leaves the window first.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The naive solution iterates over every possible window of size K, summing its elements in O(K) time per window, which leads to O(N·K) overall complexity for an input of length N. This quickly becomes infeasible for large N (e.g., N = 10⁶) because the repeated recomputation of overlapping sums wastes linear work.
The optimal paradigm leverages the sliding‑window technique: maintain a running sum (or a deque for variable‑size windows) and update it in O(1) when the window slides one position. By adding the incoming element and subtracting the element that exits, each array element is processed a constant number of times, yielding an O(N) time algorithm with O(1) auxiliary space (or O(K) if a deque is used for max‑value tracking).
Interview Questions on This Problem
Q1Why does a sliding‑window approach reduce the time complexity from O(N·K) to O(N)?
Because each element is added and removed from the window at most once, eliminating redundant summations.
Q2How would you modify the algorithm to return the maximum sum for any window size ≤ K?
Maintain the current sum while expanding the window up to K, and track the maximum after each addition; shrinking is optional if you only need ≤ K.
Q3What data structure helps retrieve the maximum element in a variable‑size window in O(1) amortized time?
A monotonic deque stores candidates in decreasing order, allowing constant‑time max extraction.
Examples
Input
[3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
40
Explanation: Step-by-step: with input [3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we slide the window to the right by subtracting the first element of the window from the current window sum and incrementing the window start index. The maximum sum is 40.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
Output
95
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20], we slide the window to the right by subtracting the first element of the window from the current window sum and incrementing the window start index. The maximum sum is 95.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Compute the sum of the first K elements, then slide the window, updating the sum in O(1) per step for O(N) total.
Brute Force Approach
Loop over every start index, sum K elements each time, resulting in O(N·K) time.
Verified Code Solutions
function solution(nums, k) {
let maxSum = -Infinity;
let windowSum = 0;
let windowStart = 0;
for (let windowEnd = 0; windowEnd < nums.length; windowEnd++) {
windowSum += nums[windowEnd];
if (windowEnd >= k - 1) {
maxSum = Math.max(maxSum, windowSum);
windowSum -= nums[windowStart];
windowStart++;
}
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int maxSum = INT_MIN;
int windowSum = 0;
int windowStart = 0;
for (int windowEnd = 0; windowEnd < nums.size(); windowEnd++) {
windowSum += nums[windowEnd];
if (windowEnd >= k - 1) {
maxSum = max(maxSum, windowSum);
windowSum -= nums[windowStart];
windowStart++;
}
}
return maxSum;
}
};class Solution {
public int solution(int[] nums, int k) {
int maxSum = Integer.MIN_VALUE;
int windowSum = 0;
int windowStart = 0;
for (int windowEnd = 0; windowEnd < nums.length; windowEnd++) {
windowSum += nums[windowEnd];
if (windowEnd >= k - 1) {
maxSum = Math.max(maxSum, windowSum);
windowSum -= nums[windowStart];
windowStart++;
}
}
return maxSum;
}
}def solution(nums, k):
max_sum = float('-inf')
window_sum = 0
window_start = 0
for window_end in range(len(nums)):
window_sum += nums[window_end]
if window_end >= k - 1:
max_sum = max(max_sum, window_sum)
window_sum -= nums[window_start]
window_start += 1
return max_sumfunction solution(nums, k) {
let maxSum = -Infinity;
let windowSum = 0;
let windowStart = 0;
for (let windowEnd = 0; windowEnd < nums.length; windowEnd++) {
windowSum += nums[windowEnd];
if (windowEnd >= k - 1) {
maxSum = Math.max(maxSum, windowSum);
windowSum -= nums[windowStart];
windowStart++;
}
}
return maxSum;
}Asked in Top Tech Interviews
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.