Verified Capacity Window — Problem Statement & Solution Guide
Problem Description
A distributed monitoring system logs integer-valued throughput metrics into a linear buffer. To evaluate peak load stability, the engineering team requires identifying the maximum aggregate throughput observed within any fixed-duration interval. Given an array metrics representing sequential throughput values and an integer windowSize denoting the interval length, compute the highest sum achievable by any contiguous subarray of exactly windowSize elements.
The solution must efficiently process the buffer by sliding the observation window across the array, updating the cumulative sum incrementally rather than recalculating from scratch for each position. This approach ensures optimal performance for large-scale data streams.
Input: An array metrics of integers and an integer windowSize.
Output: A single integer representing the maximum sum of any contiguous subarray of length windowSize.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Verified Capacity Window"
WHY DOES IT MATTER?
Fixed‑size sliding‑window problems appear in performance monitoring, financial time‑series analysis, and any scenario where recent history must be aggregated quickly. Mastering this pattern enables engineers to build low‑latency analytics pipelines that scale to massive data volumes.
OPTIMIZATION CHALLENGE
The key insight is recognizing that consecutive windows share windowSize‑1 elements, allowing the sum to be updated by a simple subtraction and addition instead of recomputing from scratch.
REAL-WORLD CONNECTION
Think of a rolling 5‑minute average of CPU usage displayed on a dashboard. As each new measurement arrives, the system drops the oldest reading and adds the newest, updating the average in constant time—exactly the sliding‑window principle.
During an interview, write the initial sum of the first window first, then loop from windowSize to n‑1, updating the sum in place. This reduces mental overhead and avoids off‑by‑one bugs.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the maximum sum of any contiguous sub‑array of length windowSize. This is a classic sliding‑window scenario where the window’s size is fixed, allowing us to update the aggregate in O(1) time as the window moves one position to the right. A naive solution would recompute the sum for each possible window by iterating over windowSize elements, leading to O(n·windowSize) time, which quickly becomes prohibitive when both n and windowSize are large (e.g., n ≈ 10⁶). The optimal paradigm leverages the fact that consecutive windows overlap by windowSize‑1 elements; by subtracting the element that leaves the window and adding the new element that enters, we maintain the current sum in constant time.
The sliding‑window technique is a special case of the more general two‑pointer method used for problems involving contiguous sub‑structures. It transforms a potentially quadratic scan into a linear pass, preserving the order of elements while avoiding extra storage. The algorithm therefore runs in O(n) time and O(1) auxiliary space, which is optimal because every element must be examined at least once to guarantee the correct maximum.
Understanding why the naive approach fails is crucial: recomputing sums duplicates work for overlapping portions of windows. By recognizing the overlap and reusing previously computed information, we eliminate this redundancy. This insight is the cornerstone of many high‑performance solutions in streaming analytics, real‑time monitoring, and any domain where fixed‑size windows over a data stream are required.
Interview Questions on This Problem
Q1How would you modify the sliding‑window solution to also return the starting index of the window that yields the maximum sum?
Maintain a variable maxStart that records the current window’s start index whenever a new maximum sum is found. As you slide the window, compare the updated sum to the global maximum; if it exceeds, set maxStart = i - windowSize + 1 (where i is the rightmost index). Return both the maximum sum and maxStart.
Q2If the array can contain negative numbers, does the sliding‑window approach still work for a fixed window size? Why or why not?
Yes, it still works because the window size is fixed; the algorithm does not depend on the sign of the numbers. It simply maintains the sum of the current window, and the maximum sum among all windows is still correctly identified, even if some windows have negative contributions.
Q3Explain how you would adapt the algorithm to handle a stream of metrics where the total length is unknown in advance.
Use a circular buffer (or queue) of size windowSize to store the last windowSize elements. As each new metric arrives, dequeue the oldest element, subtract its value from the running sum, enqueue the new metric, add its value, and update the maximum if needed. This keeps O(1) per‑element processing and O(windowSize) memory, suitable for unbounded streams.
Examples
Input
metrics = [4, 2, 1, 7, 3, 5], windowSize = 3
Output
15
Explanation: Step 1: Calculate the sum of the first window [4, 2, 1] = 7. Set maxSum = 7. Step 2: Slide the window to [2, 1, 7]. Subtract the leftmost element (4) and add the new rightmost element (7): 7 - 4 + 7 = 10. Update maxSum = 10. Step 3: Slide the window to [1, 7, 3]. Subtract 2 and add 3: 10 - 2 + 3 = 11. Update maxSum = 11. Step 4: Slide the window to [7, 3, 5]. Subtract 1 and add 5: 11 - 1 + 5 = 15. Update maxSum = 15. Final result: 15.
Input
metrics = [10, -2, 3, 8, -1, 4], windowSize = 4
Output
19
Explanation: Step 1: Sum of first window [10, -2, 3, 8] = 19. Set maxSum = 19. Step 2: Slide to [-2, 3, 8, -1]. Subtract 10 and add -1: 19 - 10 + (-1) = 8. maxSum remains 19. Step 3: Slide to [3, 8, -1, 4]. Subtract -2 and add 4: 8 - (-2) + 4 = 14. maxSum remains 19. Final result: 19.
Input
metrics = [5, 5, 5, 5], windowSize = 2
Output
10
Explanation: Step 1: Sum of first window [5, 5] = 10. Set maxSum = 10. Step 2: Slide to [5, 5]. Subtract 5 and add 5: 10 - 5 + 5 = 10. maxSum remains 10. Step 3: Slide to [5, 5]. Subtract 5 and add 5: 10 - 5 + 5 = 10. maxSum remains 10. Final result: 10.
Input
metrics = [-3, -1, -4, -2], windowSize = 3
Output
-6
Explanation: Step 1: Sum of first window [-3, -1, -4] = -8. Set maxSum = -8. Step 2: Slide to [-1, -4, -2]. Subtract -3 and add -2: -8 - (-3) + (-2) = -7. Update maxSum = -7. Final result: -7.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- 1 <= windowSize <= metrics.length
- The sum of any subarray fits within a 64-bit integer.
Optimal Approach & Strategy
Use a sliding window: keep a running sum, subtract the element exiting the window and add the new element entering, updating the maximum in O(1) per step for O(n) total time.
Brute Force Approach
Re‑calculate the sum for every possible window by iterating over windowSize elements each time, resulting in O(n·windowSize) time.
Verified Code Solutions
function solution(nums, k) {
let maxSum = -Infinity;
for (let i = 0; i <= nums.length - k; i++) {
let sum = 0;
for (let j = i; j < i + k; j++) {
sum += nums[j];
}
maxSum = Math.max(maxSum, sum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int maxSum = INT_MIN;
for (int i = 0; i <= nums.size() - k; i++) {
int sum = 0;
for (int j = i; j < i + k; j++) {
sum += nums[j];
}
maxSum = max(maxSum, sum);
}
return maxSum;
}
}class Solution {
public int solution(int[] nums, int k) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i <= nums.length - k; i++) {
int sum = 0;
for (int j = i; j < i + k; j++) {
sum += nums[j];
}
maxSum = Math.max(maxSum, sum);
}
return maxSum;
}
}def solution(nums, k):
max_sum = float('-inf')
for i in range(len(nums) - k + 1):
subarray_sum = sum(nums[i:i+k])
max_sum = max(max_sum, subarray_sum)
return max_sumfunction solution(nums, k) {
let maxSum = -Infinity;
for (let i = 0; i <= nums.length - k; i++) {
let sum = 0;
for (let j = i; j < i + k; j++) {
sum += nums[j];
}
maxSum = Math.max(maxSum, sum);
}
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.