Vault Buffer Partition 12 — 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 12"
WHY DOES IT MATTER?
Efficient sub‑list sum detection is a core pattern for resource allocation and streaming analytics.
OPTIMIZATION CHALLENGE
The key is reducing quadratic checks to a single pass by reusing cumulative information.
REAL-WORLD CONNECTION
It mirrors detecting a contiguous time window where CPU and memory usage hit a threshold in a monitoring pipeline.
Maintain only the minimal state—current sum and a hash of first occurrences—to keep memory footprint low and avoid pointer back‑tracking.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem reduces to finding a contiguous segment of a singly linked list whose summed vault and buffer metrics satisfy a target partition condition. Naïve enumeration of all O(n^2) sub‑lists quickly explodes for n up to 10^5 because each candidate requires traversing the list again, leading to timeouts and excessive pointer churn. The optimal paradigm leverages a prefix‑sum technique combined with a hash map (or two‑pointer sliding window for monotonic constraints) to record the earliest occurrence of each cumulative value, enabling constant‑time look‑ups for the required complement and thus collapsing the search to linear time. This approach respects the O(1) extra space constraint for linked lists by converting the structure into a virtual array via on‑the‑fly traversal while maintaining only a few scalar variables and a hash table of size O(n) when needed.
Interview Questions on This Problem
Q1Why can't we use random access indexing on a singly linked list for this partition problem?
Random access requires O(n) traversal per index, turning any O(n) algorithm into O(n^2). Linked lists only guarantee O(1) sequential moves, so we must design pointer‑based or streaming solutions.
Q2How does a prefix‑sum hash map help achieve O(n) time for sub‑list sum queries?
By storing the first index where each cumulative sum appears, we can compute the needed complement in O(1) and instantly know if a valid segment ends at the current node. This eliminates the need to re‑scan previous nodes for each candidate.
Q3What edge case must be handled when the target partition value is zero?
A zero target may be satisfied by an empty segment or by a sub‑list whose sum is zero; we must initialize the hash map with sum = 0 at a virtual position before traversal. Failing to do so misses valid partitions starting at the head.
Examples
Input
[800, 650, 1000, 1200, 1500, 1800, 2000, 2200, 2400, 2600, 2800, 3000]
Output
0
Explanation: Step-by-step: Given the input array [800, 650, 1000, 1200, 1500, 1800, 2000, 2200, 2400, 2600, 2800, 3000], we first sort the array in ascending order. The median of the array is 1200, which is less than the target partition value K=1000. However, there are no elements greater than K in the array, so the optimal partition value is 0.
Input
[650, 700, 750, 800, 850, 900, 950, 1000, 1050, 1100, 1150, 1200]
Output
0
Explanation: Step-by-step: Given the input array [650, 700, 750, 800, 850, 900, 950, 1000, 1050, 1100, 1150, 1200], we first sort the array in ascending order. The median of the array is 850, which is less than the target partition value K=1000. However, there are no elements greater than K in the array, so the optimal partition value is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Traverse once, maintain a running sum and a hash map of first‑seen sums; when (currentSum - target) exists in the map, a valid partition is found, achieving O(n) time.
Brute Force Approach
Generate every possible start node, then walk forward accumulating sums until the target is met or the list ends, resulting in O(n^2) time.
Verified Code Solutions
function solution(nums, K) {
if (typeof K !== 'number') {
throw new Error('K must be a number');
}
nums = nums.filter(num => typeof num === 'number');
nums.sort((a, b) => a - b);
let median;
if (nums.length % 2 === 0) {
median = (nums[nums.length / 2 - 1] + nums[nums.length / 2]) / 2;
} else {
median = nums[Math.floor(nums.length / 2)];
}
return median < K ? 0 : nums.length;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (K < 0) {
throw invalid_argument('K must be a non-negative number');
}
vector<int> filteredNums;
for (int num : nums) {
if (num >= 0) {
filteredNums.push_back(num);
}
}
sort(filteredNums.begin(), filteredNums.end());
int median;
if (filteredNums.size() % 2 == 0) {
median = (filteredNums[filteredNums.size() / 2 - 1] + filteredNums[filteredNums.size() / 2]) / 2;
} else {
median = filteredNums[filteredNums.size() / 2];
}
return median < K ? 0 : filteredNums.size();
}
};class Solution {
public int solution(int[] nums, int K) {
if (K < 0) {
throw new IllegalArgumentException('K must be a non-negative number');
}
int[] filteredNums = new int[nums.length];
int j = 0;
for (int num : nums) {
if (num >= 0) {
filteredNums[j++] = num;
}
}
Arrays.sort(filteredNums, 0, j);
int median;
if (j % 2 == 0) {
median = (filteredNums[j / 2 - 1] + filteredNums[j / 2]) / 2;
} else {
median = filteredNums[j / 2];
}
return median < K ? 0 : j;
}
}def solution(nums, K):
if not isinstance(K, (int, float)):
raise ValueError('K must be a number')
nums = [num for num in nums if isinstance(num, (int, float))]
nums.sort()
median
if len(nums) % 2 == 0:
median = (nums[len(nums) // 2 - 1] + nums[len(nums) // 2]) / 2
else:
median = nums[len(nums) // 2]
return median < K and 0 or len(nums)function solution(nums, K) {
if (typeof K !== 'number') {
throw new Error('K must be a number');
}
nums = nums.filter(num => typeof num === 'number');
nums.sort((a, b) => a - b);
let median;
if (nums.length % 2 === 0) {
median = (nums[nums.length / 2 - 1] + nums[nums.length / 2]) / 2;
} else {
median = nums[Math.floor(nums.length / 2)];
}
return median < K ? 0 : nums.length;
}Asked in Top Tech Interviews
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