Vault Registry Consolidator 10 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the consolidation of a distributed vault registry system. The system is represented by an array metrics of length n, where each element metrics[i] denotes the operational load at index i. Your goal is to partition the array into contiguous segments such that the sum of the maximum values in each segment is minimized. This metric represents the total consolidation cost. However, you are constrained by a maximum segment length k. You must use a recursive backtracking approach with memoization to explore all valid partitions and determine the minimum possible consolidation cost. The solution must efficiently handle large inputs by leveraging the two-pointer technique to prune invalid states and optimize the search space during the recursive exploration.
Input: An array metrics of integers and an integer k representing the maximum allowed length of any segment. Output: A single integer representing the minimum sum of maximums across all valid partitions. If no valid partition exists (which is impossible given constraints), return -1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Registry Consolidator 10"
WHY DOES IT MATTER?
Two‑pointer with a monotonic structure turns a quadratic DP into linear time, crucial for hard‑level constraints.
OPTIMIZATION CHALLENGE
The key is reducing the recomputation of segment maxima from O(length) to O(1) per step.
REAL-WORLD CONNECTION
It mirrors load‑balancing where you merge jobs until a larger job would dominate the current batch's peak load.
Keep the deque clean – always discard stale indices and maintain decreasing order to avoid hidden O(n) spikes.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The naive solution enumerates every possible cut position, leading to exponential blow‑up because each of the n‑1 gaps can be either a cut or not. A DP that computes the best sum for the first i elements by trying every previous cut runs in O(n²) time, which fails for n up to 2·10⁵. The optimal paradigm treats the problem as a greedy two‑pointer scan: maintain a sliding window that expands until adding the next element would increase the current segment's maximum beyond a threshold that makes merging beneficial, then cut and start a new segment. A monotonic deque supplies the window maximum in O(1) amortized, allowing the whole array to be processed in linear time.
Interview Questions on This Problem
Q1Why does a simple DP with O(n²) time become infeasible for large n in this problem?
Because it examines every possible previous cut for each index, leading to ~10¹⁰ operations when n≈10⁵, which exceeds time limits.
Q2How does a monotonic deque help maintain the maximum of a sliding segment in O(1) amortized time?
It stores indices in decreasing order of value; when the window moves, elements out of range are popped from the front and smaller values are discarded from the back.
Q3What condition determines when to close the current segment and start a new one in the greedy two‑pointer solution?
Close the segment when the next element is larger than the current segment's maximum and merging would increase the total sum of maxima compared to starting a fresh segment.
Examples
Input
metrics = [3, 1, 2, 4, 5], k = 2
Output
10
Explanation: Valid partitions with max segment length 2: [3,1] [2,4] [5] -> max(3,1)=3, max(2,4)=4, max(5)=5 -> sum=12. [3] [1,2] [4,5] -> 3 + 2 + 5 = 10. [3] [1] [2,4] [5] -> 3+1+4+5=13. [3,1] [2] [4,5] -> 3+2+5=10. The minimum is 10.
Input
metrics = [1, 2, 3, 4, 5], k = 3
Output
12
Explanation: Partition [1,2,3] [4,5] -> max(1,2,3)=3, max(4,5)=5 -> sum=8. Wait, let's re-evaluate. [1,2,3] [4,5] is valid. Sum=8. Is there lower? [1,2] [3,4,5] -> 2+5=7. [1] [2,3,4] [5] -> 1+4+5=10. [1,2,3,4] is invalid (len 4 > 3). [1,2] [3,4] [5] -> 2+4+5=11. [1] [2,3] [4,5] -> 1+3+5=9. [1,2,3] [4] [5] -> 3+4+5=12. The minimum is 7 from [1,2] [3,4,5].
Input
metrics = [5, 5, 5, 5], k = 2
Output
10
Explanation: Any partition into segments of length <= 2 will have max 5 for each segment. To minimize the sum, we want the fewest segments. Max segment length is 2, so we can have [5,5] [5,5]. Sum = 5 + 5 = 10. Other partitions like [5] [5] [5] [5] sum to 20. Minimum is 10.
Constraints
- 1 <= metrics.length <= 10^5
- 1 <= metrics[i] <= 10^9
- 1 <= k <= metrics.length
- The sum of metrics.length over all test cases does not exceed 10^6
Optimal Approach & Strategy
Use a sliding window with two pointers and a monotonic deque to decide cuts greedily in O(n) time.
Brute Force Approach
Try every possible set of cut positions, compute the sum of segment maxima for each, and keep the minimum.
Verified Code Solutions
function solution(nums, k) {
if (k > nums.length) {
return 'Error: k is larger than the array length';
}
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k > nums.size()) {
return 'Error: k is larger than the array length';
}
sort(nums.rbegin(), nums.rend());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k > nums.length) {
return 'Error: k is larger than the array length';
}
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
if k > len(nums):
return 'Error: k is larger than the array length'
nums.sort(reverse=True)
sum = 0
for i in range(k):
sum += nums[i]
return sumfunction solution(nums, k) {
if (k > nums.length) {
return 'Error: k is larger than the array length';
}
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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