Vault Buffer Consolidator 34 — 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 consolidator value under given operational constraints. The target value is assumed to be the sum of the first K elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Buffer Consolidator 34"
WHY DOES IT MATTER?
Prefix‑aggregation patterns turn repeated range queries into sub‑linear operations.
OPTIMIZATION CHALLENGE
Reduce per‑query time from O(K) to O(log N) by pre‑computing aggregates in a tree structure.
REAL-WORLD CONNECTION
Similar to caching cumulative sales in a hierarchical product catalog for fast reporting.
Keep node payload minimal—store only count and sum—to fit more nodes in cache and avoid pointer‑chasing overhead.
COMPLEXITY AT A GLANCE
O(N·L + Q·L)O(N·L)Core Theory — Why This Approach?
A Trie (prefix tree) excels at aggregating information over ordered prefixes, allowing O(L) updates and queries where L is the length of the key representation. By inserting each data element as a string of its index (or binary representation) and storing the cumulative sum of values in every node, we can retrieve the sum of the first K elements by traversing the path that represents K, accumulating the stored sums without scanning the entire array. Naïve approaches—such as iterating over the first K entries for each query—degenerate to O(N) per query and become infeasible when N and the number of queries reach 10^5 or higher. The optimal paradigm leverages the Trie’s hierarchical aggregation to transform repeated prefix‑sum queries into logarithmic or constant‑time operations, dramatically reducing total runtime.
The construction phase inserts each element once, updating O(L) nodes per insertion, yielding overall O(N·L) time, where L = ⌈log_B N⌉ for a base‑B digit decomposition (commonly base‑10 or base‑2). Queries then walk the same L‑depth path, reading the pre‑computed sum in O(L) time. This approach also supports dynamic updates (insertions, deletions, or value changes) with the same complexity, making it robust for real‑time consolidator systems where metrics continuously evolve.
Interview Questions on This Problem
Q1How does a Trie enable O(log N) prefix‑sum queries compared to a linear scan?
Each node stores the aggregate of its subtree, so traversing the key for K visits only O(log N) nodes. The sum is retrieved directly from the node representing the prefix K.
Q2What modifications are needed to support element updates in the Trie‑based consolidator?
When a value changes, walk the key path and adjust the stored sums at each visited node by the delta. This maintains correct aggregates without rebuilding the structure.
Q3Why might a binary‑indexed tree be preferred over a Trie for this problem?
A BIT offers O(log N) updates and queries with lower constant factors and simpler implementation for numeric indices. However, a Trie shines when keys are non‑numeric or have variable length.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100] K = 3
Output
180
Explanation: Step-by-step: Given an array of vault and buffer metrics, we need to find the maximum K elements that sum up to the target value. In this case, the target value is not specified, so we assume it as the sum of the first K elements. We sort the array in descending order and then use a sliding window approach to find the maximum sum of K elements.
Input
[1, 2, 3, 4, 5] K = 2
Output
7
Explanation: Step-by-step: Given an array of vault and buffer metrics, we need to find the maximum K elements that sum up to the target value. In this case, the target value is not specified, so we assume it as the sum of the first K elements. We sort the array in descending order and then use a sliding window approach to find the maximum sum of K elements.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Insert each element into a Trie that stores subtree sums; answer a query by traversing the K‑key path and reading the pre‑aggregated sum.
Brute Force Approach
Iterate over the first K elements and accumulate their values for each query.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0 || K === 0) return 0;
nums.sort((a, b) => b - a);
let target = nums.slice(0, K).reduce((a, b) => a + b, 0);
let left = 0, right = K - 1;
while (left < nums.length && right < nums.length) {
if (nums.slice(left, right + 1).reduce((a, b) => a + b, 0) === target) return target;
else if (nums.slice(left, right + 1).reduce((a, b) => a + b, 0) < target) left++;
else right++;
}
return 0;
}class Solution {
public:
int solution(vector<int> nums, int K) {
if (nums.size() === 0 || K === 0) return 0;
sort(nums.begin(), nums.end());
int target = 0;
for (int i = 0; i < K; i++) target += nums[i];
int left = 0, right = K - 1;
while (left < nums.size() && right < nums.size()) {
if (target === nums[left] + nums[left + 1] + ... + nums[right]) return target;
else if (target < nums[left] + nums[left + 1] + ... + nums[right]) left++;
else right++;
}
return 0;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length === 0 || K === 0) return 0;
Arrays.sort(nums);
int target = 0;
for (int i = 0; i < K; i++) target += nums[i];
int left = 0, right = K - 1;
while (left < nums.length && right < nums.length) {
if (target === nums[left] + nums[left + 1] + ... + nums[right]) return target;
else if (target < nums[left] + nums[left + 1] + ... + nums[right]) left++;
else right++;
}
return 0;
}
}def solution(nums, K):
if len(nums) === 0 or K === 0: return 0
nums.sort(reverse=True)
target = sum(nums[:K])
left = 0; right = K - 1
while left < len(nums) and right < len(nums):
if sum(nums[left:right + 1]) === target: return target
elif sum(nums[left:right + 1]) < target: left += 1
else: right += 1
return 0function solution(nums, K) {
if (nums.length === 0 || K === 0) return 0;
nums.sort((a, b) => b - a);
let target = nums.slice(0, K).reduce((a, b) => a + b, 0);
let left = 0, right = K - 1;
while (left < nums.length && right < nums.length) {
if (nums.slice(left, right + 1).reduce((a, b) => a + b, 0) === target) return target;
else if (nums.slice(left, right + 1).reduce((a, b) => a + b, 0) < target) left++;
else right++;
}
return 0;
}Asked in Top Tech Interviews
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