Minimized Capacity Window — Problem Statement & Solution Guide
Problem Description
You are provided with a linear sequence of integers representing discrete capacity units and a fixed integer k denoting the window size. Your objective is to identify the maximum aggregate value achievable by summing exactly k consecutive elements within the sequence. The subarray must be contiguous, and indices are 0-based. If the sequence length is less than k, the operation is undefined, but per constraints, n is always greater than or equal to k.
The input consists of two lines. The first line contains two space-separated integers: n, the total number of elements in the sequence, and k, the fixed length of the window. The second line contains n space-separated integers representing the values in the sequence.
The output must be a single integer representing the highest sum found among all possible windows of length k. This problem requires an efficient approach, typically utilizing a sliding window technique, to avoid redundant calculations and ensure optimal performance for large inputs.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Minimized Capacity Window"
WHY DOES IT MATTER?
The sliding‑window pattern is essential because many real‑world metrics (throughput, latency, moving averages) require aggregating over a fixed-size recent segment of data. Mastering this pattern enables engineers to write code that processes streams in linear time without redundant recomputation.
OPTIMIZATION CHALLENGE
The key insight is that adjacent windows overlap by k‑1 elements, so the sum of the next window can be derived from the previous one with just two arithmetic operations, eliminating the need to traverse the entire window again.
REAL-WORLD CONNECTION
Think of a network router that monitors the total bytes transmitted over the last 5 seconds. As each new packet arrives, the router subtracts the bytes that fell out of the 5‑second window and adds the new packet size, keeping the metric up‑to‑date instantly—exactly what the sliding window does for array sums.
During an interview, compute the first window sum explicitly, then loop from index k to n‑1 updating the sum in‑place; always keep track of the maximum and, if required, its start index. This pattern is easy to code, hard to mess up, and signals to the interviewer that you think in terms of incremental state.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the maximum sum of any contiguous subarray of length exactly k. A naïve solution would recompute the sum for every possible window, leading to O(n·k) time, which quickly becomes prohibitive when n and k are large (e.g., n = 10^6). The optimal paradigm leverages the sliding‑window technique: after computing the sum of the first k elements, each subsequent window can be obtained by subtracting the element that slides out of the window and adding the new element that slides in. This constant‑time update transforms the overall runtime to linear O(n). The approach also uses only a few scalar variables, so the auxiliary space remains O(1). The sliding‑window pattern is a special case of prefix‑sum optimization, but it avoids the extra O(n) storage by maintaining a running total, making it ideal for real‑time streaming or memory‑constrained environments.
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 an additional variable bestStart that records the start index whenever a new maximum sum is found. Initialize it to 0, and each time currentSum > maxSum, set maxSum = currentSum and bestStart = currentWindowStart (which increments as the window slides).
Q2If the array can contain negative numbers, does the sliding‑window algorithm still work for finding the maximum sum of exactly k elements? Why or why not?
Yes, because the window size is fixed at k; the algorithm simply tracks the sum of each fixed‑length segment, regardless of sign. The presence of negatives does not affect the O(1) update rule (subtract left, add right), only the final comparison of sums.
Q3Explain how you could extend this problem to find the maximum average of any subarray with length at least k, and discuss the time complexity of your approach.
Use binary search on the answer (average) combined with a prefix‑sum check: for a guessed average m, transform each element to (a[i] - m) and check if any subarray of length ≥ k has non‑negative sum using prefix minima. This yields O(n·log range) time, where log range is the binary‑search precision, which is more complex than the fixed‑k case.
Examples
Input
5 3 1 2 3 4 5
Output
12
Explanation: The array is [1, 2, 3, 4, 5] and k is 3. The possible windows of length 3 are: [1, 2, 3] with sum 6, [2, 3, 4] with sum 9, and [3, 4, 5] with sum 12. The maximum sum is 12.
Input
4 2 -1 -2 -3 -4
Output
-3
Explanation: The array is [-1, -2, -3, -4] and k is 2. The possible windows are: [-1, -2] with sum -3, [-2, -3] with sum -5, and [-3, -4] with sum -7. The maximum sum is -3.
Input
6 4 10 20 30 40 50 60
Output
180
Explanation: The array is [10, 20, 30, 40, 50, 60] and k is 4. The windows are: [10, 20, 30, 40] sum 100, [20, 30, 40, 50] sum 140, and [30, 40, 50, 60] sum 180. The maximum sum is 180.
Input
3 3 5 5 5
Output
15
Explanation: The array is [5, 5, 5] and k is 3. There is only one window: [5, 5, 5] with sum 15. The maximum sum is 15.
Constraints
- 1 <= n <= 10^5
- 1 <= k <= n
- -10^9 <= nums[i] <= 10^9
- The sum of elements in any window will fit within a 64-bit integer.
Optimal Approach & Strategy
Compute the first window sum, then slide the window across the array updating the sum in O(1) per step, achieving O(n) total time.
Brute Force Approach
Iterate over every possible start index, sum the k elements for each window, and keep the maximum; this costs O(n·k) time.
Verified Code Solutions
function solution(nums, k) {
if (k > nums.length) return 0;
let windowSum = 0;
let maxSum = -Infinity;
for (let i = 0; i < k; i++) {
windowSum += nums[i];
}
maxSum = Math.max(maxSum, windowSum);
for (let i = k; i < nums.length; i++) {
windowSum = windowSum - nums[i - k] + nums[i];
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k > nums.size()) return 0;
int windowSum = 0;
int maxSum = INT_MIN;
for (int i = 0; i < k; i++) {
windowSum += nums[i];
}
maxSum = max(maxSum, windowSum);
for (int i = k; i < nums.size(); i++) {
windowSum = windowSum - nums[i - k] + nums[i];
maxSum = max(maxSum, windowSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k > nums.length) return 0;
int windowSum = 0;
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < k; i++) {
windowSum += nums[i];
}
maxSum = Math.max(maxSum, windowSum);
for (int i = k; i < nums.length; i++) {
windowSum = windowSum - nums[i - k] + nums[i];
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}
}def solution(nums, k):
if k > len(nums):
return 0
window_sum = 0
max_sum = float('-inf')
for i in range(k):
window_sum += nums[i]
max_sum = max(max_sum, window_sum)
for i in range(k, len(nums)):
window_sum = window_sum - nums[i - k] + nums[i]
max_sum = max(max_sum, window_sum)
return max_sumfunction solution(nums, k) {
if (k > nums.length) return 0;
let windowSum = 0;
let maxSum = -Infinity;
for (let i = 0; i < k; i++) {
windowSum += nums[i];
}
maxSum = Math.max(maxSum, windowSum);
for (let i = k; i < nums.length; i++) {
windowSum = windowSum - nums[i - k] + nums[i];
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}Asked in Top Tech Interviews
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