Matrix Transaction Optimizer 30 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and transaction metrics, construct an optimal algorithm to evaluate and compute the target optimizer value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Optimizer 30"
WHY DOES IT MATTER?
Heap selection patterns turn exponential‑size searches into linear‑ith‑log‑k scans.
OPTIMIZATION CHALLENGE
The key is reducing the combinatorial explosion of matrix sub‑structures to a bounded heap size.
REAL-WORLD CONNECTION
Similar to priority‑queue scheduling in transaction processing systems where only the most critical jobs are kept in memory.
Initialize the heap with the first k elements, then stream‑process the rest to keep the heap size constant.
COMPLEXITY AT A GLANCE
O(N·M log k)O(k)Core Theory — Why This Approach?
The naive solution treats the matrix as a flat list, sorting all N × M elements or evaluating every possible sub‑matrix, which incurs O(N·M·log(N·M)) time and blows up memory for large inputs. By recognizing that the optimizer value depends on the k‑largest (or k‑smallest) transaction metrics, we can maintain a fixed‑size heap while streaming through the matrix, guaranteeing O(N·M log k) time and O(k) extra space. The heap‑based paradigm leverages the fact that insertion and extraction are logarithmic, allowing us to prune irrelevant elements early and avoid the full sort. This shift from global ordering to incremental selection is the cornerstone of the optimal algorithm for the problem.
Interview Questions on This Problem
Q1Why does a heap reduce the time complexity compared to sorting the entire matrix?
A heap maintains only the top‑k elements, so each insertion is O(log k) instead of O(log (N·M)). This avoids the O(N·M log(N·M)) cost of a full sort.
Q2How would you handle duplicate transaction values when using a heap?
Store each element with its coordinates or a unique identifier to differentiate duplicates. The heap ordering remains based on the metric value alone.
Q3What is the impact of using a max‑heap versus a min‑heap for this problem?
A max‑heap lets you quickly discard values larger than the current k‑th smallest, while a min‑heap is suited for extracting the k‑largest. Choose the orientation that matches the optimizer’s definition.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], K = 5
Output
45
Explanation: Step 1: Sort the array in ascending order. The sorted array is [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]. Step 2: Initialize a variable sum to 0. Step 3: Iterate through the array from the last element to the first element. Step 4: For each element, if it is greater than or equal to K, add it to the sum. Step 5: Return the sum.
Input
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1], K = 5
Output
45
Explanation: Step 1: Sort the array in ascending order. The sorted array is [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]. Step 2: Initialize a variable sum to 0. Step 3: Iterate through the array from the last element to the first element. Step 4: For each element, if it is greater than or equal to K, add it to the sum. Step 5: Return the sum.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Traverse the matrix once, maintaining a size‑k min‑heap (or max‑heap) to keep only the most relevant metrics, updating the answer on the fly.
Brute Force Approach
Flatten the matrix, sort all elements, then compute the optimizer value from the sorted list.
Verified Code Solutions
function solution(nums, K) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = nums.length - 1; i >= 0; i--) {
if (nums[i] >= K) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = nums.size() - 1; i >= 0; i--) {
if (nums[i] >= K) {
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 - 1; i >= 0; i--) {
if (nums[i] >= K) {
sum += nums[i];
}
}
return sum;
}
}def solution(nums, K):
nums.sort(reverse=True)
sum = 0
for i in range(len(nums)):
if nums[i] >= K:
sum += nums[i]
return sumfunction solution(nums, K) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = nums.length - 1; i >= 0; i--) {
if (nums[i] >= K) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
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