Matrix Transaction Analyzer 33 — 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 analyzer value under given operational constraints. The algorithm should handle edge cases such as empty or single-element input arrays, negative numbers, and special values.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Analyzer 33"
WHY DOES IT MATTER?
Transforming a 2‑D search into repeated 1‑D scans cuts exponential blow‑up to polynomial time.
OPTIMIZATION CHALLENGE
The key is reducing the rectangle enumeration from O(N^2 M^2) to O(N^2 M) by reusing column sums.
REAL-WORLD CONNECTION
It mirrors financial risk engines that aggregate daily transaction streams across multiple accounts to spot peak exposure.
Cache the running column totals for each top row and update incrementally as you slide the bottom row to avoid recomputation.
COMPLEXITY AT A GLANCE
O(N^2 M)O(M)Core Theory — Why This Approach?
The Matrix Transaction Analyzer reduces to finding a sub‑matrix whose weighted sum (row‑wise transaction metric multiplied by cell value) is maximal. By collapsing rows with prefix sums we transform the 2‑D problem into a series of 1‑D maximum subarray problems, solvable with Kadane’s algorithm. Naïve enumeration of all possible top, bottom, left, and right boundaries incurs O(N^2 M^2) time, which explodes for N,M≈10^3. The optimal paradigm leverages cumulative column sums for each pair of row boundaries, yielding O(N^2 M) time and O(M) extra space, making the solution tractable for hard‑level constraints.
Interview Questions on This Problem
Q1How does compressing rows with prefix sums convert a 2‑D maximum‑sum problem into a 1‑D problem?
Summing columns between two row indices gives a 1‑D array of column aggregates. Kadane’s algorithm then finds the best left‑right span in linear time.
Q2Why does the naïve O(N^2 M^2) enumeration become infeasible for large matrices?
It requires examining every possible rectangle, leading to ~10^12 operations for N,M≈10^3. Such quadratic‑quadratic growth exceeds typical time limits.
Q3What space‑optimisation does the algorithm achieve compared to storing full 2‑D prefix sums?
Only a single column‑wise accumulator of size M is needed for each row‑pair iteration. This reduces auxiliary space from O(N M) to O(M).
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
35
Explanation: Step-by-step: With input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we first calculate the sum of the first 5 numbers, which is 15. Then, we check if the numbers 6, 7, 8, 9, 10 are included in the sum of the first 5 numbers. Since they are not, we add them to the sum, giving us a final output of 35.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
Output
15
Explanation: Step-by-step: With input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], we first calculate the sum of the first 5 numbers, which is 15. Then, we check if the numbers 6, 7, 8, 9, 10 are included in the sum of the first 5 numbers. Since they are, we do not add them to the sum, giving us a final output of 15.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Fix top and bottom rows, maintain column‑wise sums, and apply Kadane’s algorithm to find the best left‑right span for each row pair.
Brute Force Approach
Enumerate every possible top, bottom, left, and right boundary and compute the weighted sum directly.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let i = 0; i < K; i++) {
sum += nums[i];
}
for (let i = K; i < nums.length; i++) {
if (!nums.slice(0, K).includes(nums[i])) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int i = 0; i < K; i++) {
sum += nums[i];
}
for (int i = K; i < nums.size(); i++) {
if (find(nums.begin(), nums.begin() + K, nums[i]) == nums.begin() + K) {
sum += nums[i];
}
}
return sum;
}
}class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int i = 0; i < K; i++) {
sum += nums[i];
}
for (int i = K; i < nums.length; i++) {
if (!contains(nums, 0, K, nums[i])) {
sum += nums[i];
}
}
return sum;
}
private boolean contains(int[] nums, int start, int end, int target) {
for (int i = start; i < end; i++) {
if (nums[i] == target) {
return true;
}
}
return false;
}
}def solution(nums, K):
sum = 0
for i in range(K):
sum += nums[i]
for i in range(K, len(nums)):
if nums[i] not in nums[:K]:
sum += nums[i]
return sumfunction solution(nums, K) {
let sum = 0;
for (let i = 0; i < K; i++) {
sum += nums[i];
}
for (let i = K; i < nums.length; i++) {
if (!nums.slice(0, K).includes(nums[i])) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
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