Vault Buffer Partition 47 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and buffer metrics, construct an optimal algorithm to evaluate and compute the target partition value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Buffer Partition 47"
WHY DOES IT MATTER?
Sliding windows turn quadratic sub‑array problems into linear scans, essential for real‑time analytics.
OPTIMIZATION CHALLENGE
The key is reducing repeated recomputation of the window metric to O(1) per move.
REAL-WORLD CONNECTION
Think of monitoring a vault's temperature buffer where you continuously evaluate the last k readings to trigger alerts.
Keep the window state minimal—use counters or deques and update them incrementally rather than rebuilding.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
Sliding‑window techniques convert a naïve quadratic scan into a linear pass by maintaining a dynamic interval over the input and updating aggregate metrics (sum, max, counts) as the interval expands or contracts. In the Vault Buffer Partition problem the window represents a candidate partition; each step we adjust the left or right bound to satisfy the operational constraints while tracking the optimal value, guaranteeing O(n) traversal for any n‑length sequence. Naïve approaches recompute the partition metric from scratch for every possible sub‑segment, leading to O(n²) time and quickly exhausting time limits on large datasets. The optimal paradigm leverages two pointers that move monotonically, ensuring each element is added and removed at most once, which yields linear time and constant auxiliary space.
The sliding window is a special case of the more general two‑pointer method, but its power lies in the ability to maintain complex state (e.g., frequency maps, min‑max deques) in O(1) amortized updates. By abstracting the constraint checks into incremental updates, the algorithm avoids repeated full scans and can be extended to variable‑size windows, making it the go‑to strategy for any problem that asks for optimal sub‑array/segment values under additive or monotonic constraints.
Interview Questions on This Problem
Q1How does a sliding window achieve O(n) time where a double loop yields O(n²)?
Each element enters and leaves the window at most once, so total work is proportional to n. The inner loop’s work is amortized across the outer loop, eliminating nested full scans.
Q2When would you need a deque instead of a simple sum in a sliding window?
A deque efficiently tracks the minimum or maximum of the current window in O(1) amortized time. Simple sums cannot answer order‑statistic queries without extra data structures.
Q3What edge case must you guard against when the window can shrink to zero length?
If constraints force the left pointer past the right, you must reset the window or handle empty‑window logic to avoid invalid accesses. Properly checking bounds prevents infinite loops or runtime errors.
Examples
Input
[120, 120, 120, 120, 120], 3
Output
360
Explanation: To find the target partition value, we apply the sliding window technique with a size of 3. We start by summing the first three elements: 120 + 120 + 120 = 360. Then, we slide the window to the right by subtracting the first element (120) and adding the second element (120), resulting in a sum of 240. Next, we slide the window to the right by subtracting the second element (120) and adding the third element (120), resulting in a sum of 360. We repeat this process until we reach the end of the array, maintaining a sum of 360.
Input
[10, 20, 30, 40, 50], 3
Output
120
Explanation: To find the target partition value, we apply the sliding window technique with a size of 3. We start by summing the first three elements: 10 + 20 + 30 = 60. Then, we slide the window to the right by subtracting the first element (10) and adding the second element (20), resulting in a sum of 70. Next, we slide the window to the right by subtracting the second element (20) and adding the third element (30), resulting in a sum of 80. Finally, we slide the window to the right by subtracting the third element (30) and adding the fourth element (40), resulting in a sum of 100. Then, we slide the window to the right by subtracting the fourth element (40) and adding the fifth element (50), resulting in a sum of 120.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use two pointers to expand and contract a window while updating the metric incrementally, discarding the need for full recomputation.
Brute Force Approach
Enumerate every possible sub‑segment, recompute the partition metric from scratch, and keep the best result.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
let result = 0;
for (let i = 0; i < nums.length; i++) {
if (i >= k) {
sum -= nums[i - k];
}
sum += nums[i];
if (i >= k - 1) {
result = Math.max(result, sum);
}
}
return result;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
int result = 0;
for (int i = 0; i < nums.size(); i++) {
if (i >= k) {
sum -= nums[i - k];
}
sum += nums[i];
if (i >= k - 1) {
result = max(result, sum);
}
}
return result;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
int result = 0;
for (int i = 0; i < nums.length; i++) {
if (i >= k) {
sum -= nums[i - k];
}
sum += nums[i];
if (i >= k - 1) {
result = Math.max(result, sum);
}
}
return result;
}
}def solution(nums, k):
sum_val = 0
result = 0
for i in range(len(nums)):
if i >= k:
sum_val -= nums[i - k]
sum_val += nums[i]
if i >= k - 1:
result = max(result, sum_val)
return resultfunction solution(nums, k) {
let sum = 0;
let result = 0;
for (let i = 0; i < nums.length; i++) {
if (i >= k) {
sum -= nums[i - k];
}
sum += nums[i];
if (i >= k - 1) {
result = Math.max(result, sum);
}
}
return result;
}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.