Vault Buffer Extractor 7 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the data extraction process for a high-security vault system. The system provides a sorted array of integer metrics, arranged in strictly descending order, representing the capacity levels of various buffer sectors. Your objective is to compute the 'Extractor Value', defined as the sum of all metric values that strictly exceed a given threshold K.
Given the sorted nature of the input, a linear scan is inefficient for large datasets. You must design an algorithm that leverages the sorted property to identify the boundary where values drop below or equal to K, and then efficiently compute the sum of the valid prefix. The solution should minimize the number of comparisons required to locate this boundary.
Input: A sorted array of integers metrics in descending order and an integer K representing the threshold.
Output: An integer representing the sum of all elements in metrics that are strictly greater than K. If no elements exceed K, return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Buffer Extractor 7"
WHY DOES IT MATTER?
This pattern, often referred to as 'Binary Search on Sorted Arrays' or 'Boundary Finding,' is essential for optimizing queries on large, ordered datasets. It demonstrates the ability to leverage data structure properties (sortedness) to reduce computational complexity from linear to logarithmic, a key skill for handling big data and real-time systems.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the array is sorted in descending order. Most candidates default to ascending order logic. The challenge is to correctly adjust the binary search conditions to find the *last* element that is greater than K, or equivalently, the *first* element that is less than or equal to K, and then sum the prefix up to that point.
REAL-WORLD CONNECTION
Consider a database index on a sorted column. When querying for all records where value > K, the database engine uses a B-tree index to jump directly to the first record satisfying the condition, rather than scanning the entire table. This problem simulates that index lookup and range aggregation.
In interviews, explicitly state the assumption about the array's order. If it's descending, clarify that you are looking for the boundary where the condition arr[i] > K flips to arr[i] <= K. This shows you are not just memorizing code but understanding the logic behind the search direction.
COMPLEXITY AT A GLANCE
O(log n + k)O(1)Core Theory — Why This Approach?
The problem presents a sorted array in strictly descending order, which is a critical structural constraint that invalidates standard linear scanning as the optimal solution. While a naive approach would iterate through the entire array to sum elements greater than K, this results in O(n) time complexity. In high-frequency trading or real-time data extraction systems, where n can be in the millions, this linear scan is computationally expensive and fails to leverage the sorted nature of the data. The underlying theory relies on the monotonicity of the array: once an element is found that is less than or equal to K, all subsequent elements (due to descending order) will also be less than or equal to K. This property allows us to truncate the search space significantly.
Interview Questions on This Problem
Q1How would you modify your solution if the array were sorted in ascending order instead of descending?
If the array is ascending, the elements greater than K will be at the end of the array. You would perform a binary search to find the first index where the value is greater than K. Then, you would sum the elements from that index to the end of the array. To optimize the summation, you could precompute a prefix sum array to answer the range sum query in O(1) after the O(log n) binary search, or simply iterate from the found index to the end if the range is small.
Q2What is the time complexity of your solution, and how does it scale if the array is not sorted?
For a sorted array, the time complexity is O(log n) to find the boundary using binary search, plus O(k) to sum the k elements greater than K, where k is the number of such elements. In the worst case where all elements are greater than K, it is O(n). If the array is not sorted, we must first sort it in O(n log n) time, making the total complexity O(n log n). However, if the array is unsorted and we cannot modify it, we are forced into an O(n) linear scan, as no sub-linear search is possible without prior sorting.
Q3Can you optimize the summation step to be O(1) regardless of the number of elements greater than K?
Yes, by precomputing a prefix sum array. If the array is static, we can build a prefix sum array in O(n) time and O(n) space. Then, once we find the index i via binary search where arr[i] > K and arr[i+1] <= K, the sum of elements greater than K is simply prefixSum[n] - prefixSum[i]. This reduces the query time to O(log n) for the binary search plus O(1) for the sum, assuming the prefix sums are precomputed.
Examples
Input
metrics = [100, 90, 80, 70, 60], K = 75
Output
270
Explanation: The array is [100, 90, 80, 70, 60]. We need elements > 75. 100 > 75 (include), 90 > 75 (include), 80 > 75 (include), 70 <= 75 (stop). Sum = 100 + 90 + 80 = 270.
Input
metrics = [50, 40, 30, 20, 10], K = 55
Output
0
Explanation: The array is [50, 40, 30, 20, 10]. We need elements > 55. The first element 50 is not greater than 55. Since the array is descending, no subsequent element can be greater. Sum = 0.
Input
metrics = [10, 9, 8, 7, 6], K = 5
Output
40
Explanation: The array is [10, 9, 8, 7, 6]. We need elements > 5. All elements are greater than 5. Sum = 10 + 9 + 8 + 7 + 6 = 40.
Input
metrics = [100, 50, 50, 50, 10], K = 50
Output
100
Explanation: The array is [100, 50, 50, 50, 10]. We need elements strictly > 50. 100 > 50 (include). 50 is not > 50 (stop). Sum = 100.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- -10^9 <= K <= 10^9
- metrics is sorted in strictly descending order
Optimal Approach & Strategy
Perform a binary search to find the first index where the element is less than or equal to K. Sum all elements from index 0 up to that index minus one. This reduces the search time to O(log n), and the summation time is O(k) where k is the number of elements greater than K, or O(1) with precomputed prefix sums.
Brute Force Approach
Iterate through the entire array from start to finish, adding each element to a running sum if it is strictly greater than K. This approach has a time complexity of O(n) and does not utilize the sorted nature of the array.
Verified Code Solutions
function solution(nums, k) {
if (nums.length === 0) return 0;
nums.sort((a, b) => b - a);
let extractorValue = 0;
for (let num of nums) {
if (num > k) extractorValue += num;
}
return extractorValue;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (nums.empty()) return 0;
sort(nums.rbegin(), nums.rend());
int extractorValue = 0;
for (int num : nums) {
if (num > k) extractorValue += num;
}
return extractorValue;
}
};class Solution {
public int solution(int[] nums, int k) {
if (nums.length == 0) return 0;
Arrays.sort(nums);
int extractorValue = 0;
for (int num : nums) {
if (num > k) extractorValue += num;
}
return extractorValue;
}
}def solution(nums, k):
if not nums:
return 0
nums.sort(reverse=True)
extractor_value = 0
for num in nums:
if num > k:
extractor_value += num
return extractor_valuefunction solution(nums, k) {
if (nums.length === 0) return 0;
nums.sort((a, b) => b - a);
let extractorValue = 0;
for (let num of nums) {
if (num > k) extractorValue += num;
}
return extractorValue;
}Asked in Top Tech Interviews
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