Optimizing Warehouse Storage 2 — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums of length n and an integer k (1 ≤ k ≤ n). A subarray must begin at index 0 and must contain the first k elements. After those k elements, you may optionally extend the subarray by adding any number of consecutive elements to its right, including none. Your task is to compute the maximum possible sum of such a subarray.
Input format:
- The first line contains two integers n and k.
- The second line contains n space‑separated integers representing nums.
Output format:
- Output a single integer: the maximum achievable sum.
The problem requires an efficient solution that runs in linear time, as n can be large.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimizing Warehouse Storage 2"
WHY DOES IT MATTER?
This pattern—maintaining a running aggregate and a running maximum—avoids redundant work and is a cornerstone of linear‑time solutions for subarray problems. It ensures that each element is processed only once, which is critical for large datasets and real‑time systems.
OPTIMIZATION CHALLENGE
The key insight is that once you have the sum up to index i, you can decide whether to extend the subarray without recomputing sums for every possible end. This reduces the time from quadratic to linear and the space from O(n) to O(1).
REAL-WORLD CONNECTION
Think of a warehouse where you must pick at least k items from the front, then optionally continue picking more. The running sum represents the total weight carried so far, and the running maximum is the heaviest load you could have carried at any point. Just as a forklift operator keeps track of the current load to avoid overloading, the algorithm keeps a running sum to decide when to stop.
When explaining this in an interview, emphasize that the mandatory k elements guarantee a starting point, so you can safely start the loop at index k-1 and treat the running sum as the candidate subarray sum. Highlight that the algorithm is essentially a one‑pass scan with a simple comparison.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to finding the maximum sum of a prefix of the array that is at least length k. A naive approach would enumerate all possible extensions beyond the mandatory k elements, compute each subarray sum from scratch, and keep the maximum. This results in O(n^2) time, which is infeasible for large n.
The optimal solution leverages prefix sums. By iterating once through the array, we maintain a running total of the sum of elements seen so far. Starting from index k-1 (the end of the mandatory segment), we compare the current running total to a stored maximum and update it if larger. This single pass yields O(n) time.
Because we only need the current sum and the maximum seen so far, we can do this in constant additional space. The algorithm is essentially a special case of Kadane’s algorithm where the subarray must start at the beginning and have a minimum length, but the simplicity of the prefix sum approach makes it both fast and easy to implement.
Interview Questions on This Problem
Q1How would you modify the algorithm if the subarray could start at any index but still must contain at least k consecutive elements?
You would use a sliding window of size k to maintain the sum of the last k elements, then extend the window to the right while keeping track of the maximum sum. This is equivalent to computing prefix sums and for each right endpoint r, considering the sum of the subarray ending at r with length at least k, which can be done in O(n) time.
Q2In a distributed system, how might you compute the maximum prefix sum across multiple shards of data?
Each shard can compute its local prefix sums and the maximum prefix sum that starts at the shard’s first element. Then, a coordinator aggregates these results by maintaining the global maximum and the cumulative sum of all preceding shards to adjust the prefix sums of subsequent shards. This two‑phase reduction achieves O(n) total work with minimal communication.
Q3What is the time complexity if you precompute all prefix sums and then iterate over all possible extensions?
Precomputing prefix sums is O(n). However, iterating over all extensions still requires O(n^2) comparisons unless you use the running maximum trick. The optimal O(n) solution avoids the second loop by updating the maximum on the fly.
Examples
Input
5 2 1 2 -3 4 5
Output
9
Explanation: The subarray must start with the first two elements: 1 + 2 = 3. We can extend to the right: - Adding -3 gives 0. - Adding 4 gives 4. - Adding 5 gives 9. The largest sum is 9.
Input
4 3 -5 -2 -3 10
Output
0
Explanation: The first three elements sum to -10. Extending by the last element gives -10 + 10 = 0. Any other extension is impossible because the subarray already ends at the last element. The maximum sum is 0.
Input
6 1 -1 2 3 -4 5 -6
Output
5
Explanation: Start with the first element: -1. - Extend by 2: sum = 1. - Extend by 3: sum = 4. - Extend by -4: sum = 0. - Extend by 5: sum = 5. - Extend by -6: sum = -1. The maximum sum achieved is 5.
Input
3 3 0 0 0
Output
0
Explanation: The subarray must include all three elements: 0 + 0 + 0 = 0. No further extension is possible. The maximum sum is 0.
Constraints
- 1 <= n <= 100000
- 1 <= k <= n
- -1000000000 <= nums[i] <= 1000000000
- The answer fits within a 64‑bit signed integer
Optimal Approach & Strategy
Iterate once through the array, maintaining a running sum and a maximum value. After processing the first k elements, update the maximum whenever the running sum increases. This runs in O(n) time and O(1) extra space.
Brute Force Approach
Enumerate all subarrays that start at 0 and have length at least k, compute each sum by summing elements from scratch, and track the maximum. This takes O(n^2) time and is impractical for large arrays.
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var maxSubarraySum = function(nums, k) {
let prefixSum = 0;
for (let i = 0; i < k; i++) {
prefixSum += nums[i];
}
let maxSum = prefixSum;
for (let i = k; i < nums.length; i++) {
prefixSum += nums[i];
if (prefixSum > maxSum) {
maxSum = prefixSum;
}
}
return maxSum;
};class Solution {
public:
int maxSubarraySum(vector<int>& nums, int k) {
int n = nums.size();
int prefixSum = 0;
for (int i = 0; i < k; ++i) {
prefixSum += nums[i];
}
int maxSum = prefixSum;
for (int i = k; i < n; ++i) {
prefixSum += nums[i];
if (prefixSum > maxSum) {
maxSum = prefixSum;
}
}
return maxSum;
}
};class Solution {
public int maxSubarraySum(int[] nums, int k) {
int prefixSum = 0;
for (int i = 0; i < k; i++) {
prefixSum += nums[i];
}
int maxSum = prefixSum;
for (int i = k; i < nums.length; i++) {
prefixSum += nums[i];
if (prefixSum > maxSum) {
maxSum = prefixSum;
}
}
return maxSum;
}
}class Solution:
def maxSubarraySum(self, nums: List[int], k: int) -> int:
prefix_sum = sum(nums[:k])
max_sum = prefix_sum
for i in range(k, len(nums)):
prefix_sum += nums[i]
if prefix_sum > max_sum:
max_sum = prefix_sum
return max_sum/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var maxSubarraySum = function(nums, k) {
let prefixSum = 0;
for (let i = 0; i < k; i++) {
prefixSum += nums[i];
}
let maxSum = prefixSum;
for (let i = k; i < nums.length; i++) {
prefixSum += nums[i];
if (prefixSum > maxSum) {
maxSum = prefixSum;
}
}
return maxSum;
};Asked in Top Tech Interviews
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