Matrix Transaction Synthesizer 3 — Problem Statement & Solution Guide
Problem Description
Given a matrix of size MxN and a transaction metric K, construct an optimal algorithm to evaluate and compute the sum of the first K metrics after sorting all metrics in the matrix in descending order.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Synthesizer 3"
WHY DOES IT MATTER?
Top‑K selection is a recurring pattern in recommendation engines, risk scoring, and real‑time analytics where only the highest‑valued signals matter. Mastering this pattern lets engineers avoid full sorts and meet strict latency SLAs.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that we only need a partial ordering. By keeping a bounded data structure (heap) or using a linear‑time selection algorithm, we cut the logarithmic factor tied to full sorting, turning an O(N log N) problem into O(N log K) or O(N).
REAL-WORLD CONNECTION
Think of a stock exchange order book: only the best K bids/asks affect the immediate market price. Maintaining a min‑heap of the top K orders mirrors how matching engines keep the most competitive quotes without scanning the entire order pool.
During coding, first write the flatten‑and‑heap version; it’s simple and interview‑friendly. If the interviewer pushes for O(N) time, pivot to QuickSelect or radix counting, and be ready to discuss in‑place partitioning and handling duplicate pivot values.
COMPLEXITY AT A GLANCE
O(M·N log K) (heap) or O(M·N) expected (QuickSelect)O(K) auxiliary (heap) or O(1) extra (QuickSelect)Core Theory — Why This Approach?
The problem reduces to extracting the K largest values from an unsorted multiset of size M·N and summing them. A naïve solution would flatten the matrix, sort it in descending order, and then sum the first K elements, which costs O(M·N log(M·N)) time and O(M·N) auxiliary space. For large matrices (e.g., M,N up to 10^5), this becomes infeasible both in runtime and memory. The optimal paradigm leverages selection‑based techniques: either a min‑heap of size K that maintains the current top‑K while scanning the matrix in O(M·N log K) time, or a QuickSelect (partition‑based) algorithm that finds the K‑th largest element in expected O(M·N) time, after which a single pass computes the sum of all elements greater than or equal to that pivot. Both approaches avoid full sorting and dramatically shrink the memory footprint to O(K) or O(1) extra space.
Bit manipulation enters when the metric values are bounded integers (e.g., 32‑bit). In such cases a radix‑like counting sort can be implemented with bitwise buckets, achieving linear time O(M·N) without comparisons. By processing bits in groups (e.g., 8‑bit passes) we can count frequencies, compute a prefix sum to locate the K‑th largest bucket, and then accumulate the exact sum using only integer arithmetic, which is cache‑friendly and eliminates the overhead of heap allocations.
The key insight is that we do not need a total order of all elements—only a partial order sufficient to isolate the top K. This allows us to replace the O(N log N) sorting barrier with O(N log K) or O(N) expected time, making the solution scalable to massive data streams and enabling in‑place or streaming implementations that are crucial in high‑throughput fintech pipelines.
Interview Questions on This Problem
Q1How would you compute the sum of the top K elements in a matrix without sorting the entire matrix?
Maintain a min‑heap of size K while iterating over each cell; push the element if the heap size is less than K or if the element is larger than the heap root, then pop the root. After the scan, sum all elements in the heap. This runs in O(M·N log K) time and O(K) space.
Q2Explain how QuickSelect can be adapted to find the sum of the K largest values in a matrix.
Flatten the matrix conceptually and apply QuickSelect to find the K‑th largest value (the pivot). Partition the matrix around this pivot in‑place; then a second linear pass adds all values greater than the pivot and, if needed, enough occurrences of the pivot to reach exactly K elements. Expected time is O(M·N) with O(1) extra space.
Q3When matrix values are bounded 32‑bit integers, how can bitwise techniques improve the solution?
Use a radix‑counting approach: iterate over the matrix counting occurrences of each byte (or 16‑bit chunk) using bitwise masks. After building frequency tables, walk the buckets from most significant to least to locate the bucket containing the K‑th largest value, then compute the sum by aggregating full buckets and partially processing the final bucket. This yields O(M·N) time and O(1) extra space.
Examples
Input
matrix = [[90, 80, 70], [60, 50, 40]], K = 3
Output
240
Explanation: Step-by-step: First, we flatten the matrix into a single array [90, 80, 70, 60, 50, 40]. Then, we sort this array in descending order, which remains [90, 80, 70, 60, 50, 40]. Finally, we sum the first K metrics, which are the first 3 elements in the sorted array: 90 + 80 + 70 = 240.
Input
matrix = [[25, 20], [15, 10]], K = 2
Output
45
Explanation: Step-by-step: First, we flatten the matrix into a single array [25, 20, 15, 10]. Then, we sort this array in descending order, which is [25, 20, 15, 10]. Finally, we sum the first K metrics, which are the first 2 elements in the sorted array: 25 + 20 = 45.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a min‑heap of size K while traversing the matrix, or apply QuickSelect to find the K‑th largest element and sum all values ≥ that pivot.
Brute Force Approach
Flatten the matrix, sort all M·N elements in descending order, then sum the first K values.
Verified Code Solutions
function solution(matrix, K) { let flatArray = matrix.flat(); flatArray.sort((a, b) => b - a); return flatArray.slice(0, K).reduce((a, b) => a + b, 0); }class Solution { public: int solution(vector<vector<int>>& matrix, int K) { vector<int> flatArray; for (auto& row : matrix) { for (int num : row) { flatArray.push_back(num); } } sort(flatArray.rbegin(), flatArray.rend()); int sum = 0; for (int i = 0; i < K; i++) { sum += flatArray[i]; } return sum; } }class Solution { public int solution(int[][] matrix, int K) { int[] flatArray = new int[matrix.length * matrix[0].length]; int index = 0; for (int[] row : matrix) { for (int num : row) { flatArray[index++] = num; } } Arrays.sort(flatArray); int sum = 0; for (int i = 0; i < K; i++) { sum += flatArray[flatArray.length - 1 - i]; } return sum; } }def solution(matrix, K): flat_array = sorted([num for row in matrix for num in row], reverse=True); return sum(flat_array[:K])function solution(matrix, K) { let flatArray = matrix.flat(); flatArray.sort((a, b) => b - a); return flatArray.slice(0, K).reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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