Matrix Transaction Evaluator 12 — 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 evaluator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Evaluator 12"
WHY DOES IT MATTER?
Backtracking transforms an exponential brute force into a guided search that can solve realistic instances.
OPTIMIZATION CHALLENGE
The key is pruning invalid partial assignments to shrink the search tree dramatically.
REAL-WORLD CONNECTION
It mirrors transaction validation engines that must explore combinatorial rule sets efficiently.
Always sort inputs by constraint tightness and use immutable bitmask snapshots to keep recursion cheap.
COMPLEXITY AT A GLANCE
O(k·2^m)O(m)Core Theory — Why This Approach?
Backtracking systematically explores all feasible assignments of transaction metrics to matrix cells, pruning branches that violate constraints early. By representing the problem as a depth‑first search tree, each level fixes one element, and constraint checks eliminate impossible sub‑trees, turning an otherwise factorial search into a tractable exponential bound.
Naïve enumeration treats each element independently, leading to O(n!) or O(2^n) blow‑up even for modest matrix sizes, because it revisits identical sub‑states without memory. The optimal paradigm couples backtracking with state‑compression (e.g., bitmasking) and ordering heuristics such as most‑constrained‑first, which reduces the effective branching factor and yields a worst‑case O(k·2^m) where m is the number of mutable cells and k is a small constant from pruning.
Interview Questions on This Problem
Q1How does ordering the next cell by the most constrained heuristic improve backtracking performance?
It reduces the branching factor early, causing failures to be detected sooner. Fewer recursive calls mean lower overall runtime.
Q2Why is bitmasking a suitable state representation for this matrix transaction problem?
Bitmasking encodes used cells and transaction flags in O(1) operations. It enables fast copy‑on‑write and constant‑time constraint checks.
Q3What is the trade‑off between pure backtracking and adding memoization for sub‑states?
Memoization avoids recomputing identical sub‑problems, cutting exponential repeats. However, it increases memory usage and hash overhead.
Examples
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]]
Output
0
Explanation: Step-by-step: Given a 4x3 matrix, we iterate through each element. Since all elements are less than or equal to K (let's assume K = 10), the output is 0.
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 13]]
Output
1
Explanation: Step-by-step: Given a 4x3 matrix, we iterate through each element. The element 13 is greater than K (let's assume K = 10), so the output is 1.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use recursive backtracking with constraint checks, bitmask state, and most‑constrained‑first ordering to prune branches early.
Brute Force Approach
Generate all permutations of the transaction list and test each full matrix against constraints, which is factorial in the number of cells.
Verified Code Solutions
function solution(matrix, k) {
let count = 0;
for (let i = 0; i < matrix.length; i++) {
for (let j = 0; j < matrix[i].length; j++) {
if (matrix[i][j] > k) {
count++;
}
}
}
return count;
}class Solution {
public:
int solution(vector<vector<int>>& matrix, int k) {
int count = 0;
for (int i = 0; i < matrix.size(); i++) {
for (int j = 0; j < matrix[i].size(); j++) {
if (matrix[i][j] > k) {
count++;
}
}
}
return count;
}
};class Solution {
public int solution(int[][] matrix, int k) {
int count = 0;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
if (matrix[i][j] > k) {
count++;
}
}
}
return count;
}
}def solution(matrix, k):
count = 0
for i in range(len(matrix)):
for j in range(len(matrix[i])):
if matrix[i][j] > k:
count += 1
return countfunction solution(matrix, k) {
let count = 0;
for (let i = 0; i < matrix.length; i++) {
for (let j = 0; j < matrix[i].length; j++) {
if (matrix[i][j] > k) {
count++;
}
}
}
return count;
}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.