Matrix Transaction Detector 9 — 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 detector value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Detector 9"
WHY DOES IT MATTER?
Sliding‑window on matrices transforms quadratic sub‑structure scans into linear passes.
OPTIMIZATION CHALLENGE
The key is reducing per‑window computation from O(k²) to O(1) using prefix sums and incremental updates.
REAL-WORLD CONNECTION
Similar to real‑time fraud detection where transaction streams are examined over moving time windows.
Cache the prefix matrix in a flat array to improve locality and avoid extra indirection.
COMPLEXITY AT A GLANCE
O(N·M)O(N·M)Core Theory — Why This Approach?
The Matrix Transaction Detector problem asks for the maximum (or target) aggregate metric over all sub‑matrices of a fixed size within a large 2‑D grid. A naive solution enumerates every possible top‑left corner and recomputes the sum of the k×k window from scratch, leading to O(N·M·k²) time, which explodes for N,M up to 10⁵. The optimal paradigm leverages 2‑D prefix sums (integral images) to retrieve any sub‑matrix sum in O(1) after an O(N·M) preprocessing pass, and then slides the window across rows and columns, updating the sum in constant time per position. This reduces the overall complexity to linear in the input size, making the algorithm scalable for massive matrices while preserving exactness.
The sliding‑window technique also benefits from incremental updates: when moving the window one column right, we subtract the leftmost column’s contribution and add the new rightmost column, which can be done using pre‑computed column sums. Combining this with prefix sums eliminates the need for nested loops over the window area, turning a potentially quadratic inner loop into a constant‑time operation. This approach exemplifies how spatial data structures and careful arithmetic can collapse combinatorial explosion into linear work.
Interview Questions on This Problem
Q1Why does a naïve O(N·M·k²) solution become infeasible for large matrices?
Because each window recomputes its sum from scratch, leading to billions of operations when N, M, or k are large. The time quickly exceeds practical limits.
Q2How does a 2‑D prefix sum enable O(1) sub‑matrix queries?
It stores cumulative sums up to each cell, so any rectangle’s sum is derived from four prefix values via inclusion‑exclusion. This constant‑time lookup replaces inner loops.
Q3What is the main advantage of sliding the window column‑wise after prefix preprocessing?
It allows updating the current window sum by subtracting the exiting column and adding the entering column, avoiding recomputation. This keeps per‑move cost O(1).
Examples
Input
[10, 20, 30, 40, 50]
Output
150
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50], we apply the sliding window constraint to calculate the sum of the elements within the window. The window starts at the first element and ends at the last element. The sum of the elements within the window is 10 + 20 + 30 + 40 + 50 = 150.
Input
[100, 200, 300, 400, 500]
Output
1500
Explanation: Step-by-step: Given the input [100, 200, 300, 400, 500], we apply the sliding window constraint to calculate the sum of the elements within the window. The window starts at the first element and ends at the last element. The sum of the elements within the window is 100 + 200 + 300 + 400 + 500 = 1500.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Compute a 2‑D prefix sum matrix, then slide the window, updating sums with O(1) arithmetic per position.
Brute Force Approach
Iterate every possible top‑left corner and sum all k×k elements each time.
Verified Code Solutions
function solution(nums) {
let windowSum = 0;
let left = 0;
let right = 0;
while (right < nums.length) {
windowSum += nums[right];
right++;
if (right - left > 5) {
windowSum -= nums[left];
left++;
}
}
return windowSum;
}class Solution {
public:
int solution(vector<int> nums) {
int windowSum = 0;
int left = 0;
for (int right = 0; right < nums.size(); right++) {
windowSum += nums[right];
if (right - left > 5) {
windowSum -= nums[left];
left++;
}
}
return windowSum;
}
};class Solution {
public int solution(int[] nums) {
int windowSum = 0;
int left = 0;
for (int right = 0; right < nums.length; right++) {
windowSum += nums[right];
if (right - left > 5) {
windowSum -= nums[left];
left++;
}
}
return windowSum;
}
}def solution(nums):
window_sum = 0
left = 0
for right in range(len(nums)):
window_sum += nums[right]
if right - left > 5:
window_sum -= nums[left]
left += 1
return window_sumfunction solution(nums) {
let windowSum = 0;
let left = 0;
let right = 0;
while (right < nums.length) {
windowSum += nums[right];
right++;
if (right - left > 5) {
windowSum -= nums[left];
left++;
}
}
return windowSum;
}Asked in Top Tech Interviews
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