Matrix Transaction Extractor 22 — Problem Statement & Solution Guide
Problem Description
Matrix Transaction Extractor 22
You are given an integer array metrics of length N and an integer K (1 ≤ K ≤ N). Each element of metrics represents a transaction metric extracted from a matrix. Your task is to compute the maximum possible sum obtainable by selecting exactly K distinct metrics. The selection must be performed using a recursive backtracking approach that explores combinations of indices, but you may incorporate pruning techniques to achieve acceptable performance within the given constraints.
**Input**: The first line contains two integers N and K. The second line contains N space‑separated integers representing metrics.
**Output**: A single integer – the sum of the K largest metrics (i.e., the maximum sum achievable by any selection of K distinct elements).
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Extractor 22"
WHY DOES IT MATTER?
Top‑K selection is a fundamental pattern for summarizing large data streams efficiently.
OPTIMIZATION CHALLENGE
Reducing the naive O(N log N) sort to O(N log K) or O(N) while preserving correctness under distinctness constraints.
REAL-WORLD CONNECTION
Search engines rank the most relevant pages, and recommendation systems surface the highest‑scoring items in real time.
Always pair the selection structure with a hash set to enforce uniqueness without sacrificing performance.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The problem reduces to selecting the K largest distinct values from an unsorted array, which is a classic selection‑top‑K scenario. A naive sort (O(N log N)) or exhaustive combination (O(N^K)) quickly becomes infeasible for large N, especially under hard constraints, because the combinatorial explosion dwarfs any realistic time budget. The optimal paradigm leverages a priority queue (max‑heap) or QuickSelect to isolate the K highest metrics in linearithmic or linear expected time, respectively, while guaranteeing distinctness by tracking visited values. This approach transforms the exponential search space into a manageable O(N log K) or O(N) operation, aligning with the constraints of high‑performance interview problems.
Interview Questions on This Problem
Q1How would you retrieve the K largest distinct elements from an unsorted array in O(N log K) time?
Maintain a min‑heap of size K; iterate the array, pushing each new distinct value and popping when the heap exceeds K. This keeps the K biggest distinct values at the end.
Q2Why is QuickSelect not always safe for finding the K largest distinct elements?
QuickSelect works on values, not uniqueness; duplicate values can cause the partition to return fewer than K distinct elements, requiring extra handling.
Q3What is the space trade‑off between using a heap versus an array‑based QuickSelect for this problem?
A heap needs O(K) extra space, while QuickSelect can operate in‑place with O(1) auxiliary space but may need additional structures to deduplicate.
Examples
Input
5 3 4 -1 7 3 2
Output
14
Explanation: The three largest metrics are 7, 4 and 3. Their sum is 7 + 4 + 3 = 14.
Input
6 2 -5 -2 -9 -1 -3 -4
Output
-3
Explanation: Even though all numbers are negative, we must pick two. The two greatest values are -1 and -2. Their sum is -1 + (-2) = -3.
Input
8 5 10 20 5 15 25 30 2 8
Output
100
Explanation: The five largest metrics are 30, 25, 20, 15 and 10. Adding them yields 30 + 25 + 20 + 15 + 10 = 100.
Constraints
- 1 <= N <= 30
- 1 <= K <= N
- -10^9 <= metrics[i] <= 10^9
- The algorithm should run in O(2^N) worst‑case time but must employ pruning to handle the maximum N within the time limit.
Optimal Approach & Strategy
Iterate once, insert each unseen metric into a min‑heap of capacity K, evict the smallest when overflow, then sum the heap contents.
Brute Force Approach
Generate all combinations of K indices, compute each sum, and keep the maximum – exponential time and impossible for large N.
Verified Code Solutions
function solution(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 solution(vector<int>& nums, int k) {
sort(nums.rbegin(), nums.rend());
int sum = 0;
for (int i = nums.size() - k; i < nums.size(); i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = nums.length - k; i < nums.length; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
return sum(nums[:k])function solution(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
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.