Matrix Transaction Detector 44 — Problem Statement & Solution Guide
Problem Description
You are tasked with analyzing a high-frequency trading log represented as a sequence of integer transaction values. The system requires identifying the top k highest-value transactions to calculate a specific risk metric. Given an array of integers representing transaction amounts and an integer k, determine the sum of the k largest elements in the array. If the array contains fewer than k elements, return the sum of all available elements. The solution must be efficient enough to handle large-scale data streams, implying an optimal time complexity better than O(n log n) for sorting, typically achieved via a min-heap or selection algorithm.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Detector 44"
WHY DOES IT MATTER?
Selecting top‑k elements is a fundamental building block for ranking, recommendation, and risk‑assessment systems.
OPTIMIZATION CHALLENGE
Reducing from O(n log n) to O(n) cuts latency dramatically, enabling near‑instant metric updates.
REAL-WORLD CONNECTION
High‑frequency trading platforms continuously compute the sum of the largest trades to gauge exposure in real time.
Prefer an in‑place partition with careful pivot choice (e.g., median‑of‑three) to avoid pathological O(n²) cases.
COMPLEXITY AT A GLANCE
O(n) average, O(n²) worst‑caseO(1) auxiliaryCore Theory — Why This Approach?
The naive solution—sorting the entire array and summing the last k elements—runs in O(n log n) time, which becomes a bottleneck for massive high‑frequency trading logs where n can reach millions. Moreover, sorting mutates the input and incurs additional memory overhead, making it unsuitable for real‑time risk calculations. The optimal paradigm leverages the selection problem: using a QuickSelect‑style partition (a two‑pointer Lomuto or Hoare scheme) we can isolate the k largest values in linear average time without fully sorting the data. By repeatedly partitioning around a pivot and narrowing the search space to the segment that contains the top‑k elements, we achieve O(n) expected time and O(1) auxiliary space, enabling fast, in‑place computation of the required sum.
Interview Questions on This Problem
Q1How does QuickSelect differ from QuickSort in terms of recursion depth and work done?
QuickSelect only recurses into the partition that contains the k‑th order statistic, reducing work to O(n) on average, whereas QuickSort recurses into both partitions, yielding O(n log n).
Q2When would a min‑heap of size k be preferable to QuickSelect for this problem?
A min‑heap guarantees O(n log k) worst‑case time and is easier to implement correctly when k is much smaller than n or when streaming data is involved.
Q3What edge case must you handle when all array elements are identical?
The partition algorithm must still correctly identify k elements without infinite loops; using <= or >= consistently in the two‑pointer swaps avoids stagnation.
Examples
Input
nums = [3, 1, 4, 1, 5, 9, 2, 6], k = 3
Output
20
Explanation: The three largest elements in the array are 9, 6, and 5. Their sum is 9 + 6 + 5 = 20.
Input
nums = [10, 20, 30, 40, 50], k = 2
Output
90
Explanation: The two largest elements are 50 and 40. Their sum is 50 + 40 = 90.
Input
nums = [5, 5, 5, 5], k = 10
Output
20
Explanation: Since k (10) is greater than the length of the array (4), we sum all elements. 5 + 5 + 5 + 5 = 20.
Input
nums = [-1, -2, -3, -4], k = 2
Output
-3
Explanation: The two largest elements (closest to positive infinity) are -1 and -2. Their sum is -1 + (-2) = -3.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- 1 <= k <= 10^5
Optimal Approach & Strategy
Apply QuickSelect to partition the array so that the k largest values are at the end, then sum them (average O(n)).
Brute Force Approach
Sort the array and sum the last k elements (O(n log n)).
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var detectTopKTransactions = function(nums, k) {
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 detectTopKTransactions(vector<int>& nums, int k) {
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int i = 0; i < k; ++i) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int detectTopKTransactions(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = nums.length - 1; i >= nums.length - k; i--) {
sum += nums[i];
}
return sum;
}
}class Solution:
def detectTopKTransactions(self, nums: List[int], k: int) -> int:
nums.sort(reverse=True)
return sum(nums[:k])/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var detectTopKTransactions = function(nums, k) {
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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