Matrix Stream Synthesizer 27 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and stream metrics, construct an optimal algorithm to evaluate and compute the target synthesizer value under given operational constraints. The input matrix is a 2D array of integers, and K is the number of rows in the matrix. The function should return the sum of the diagonal elements of the matrix if K is specified, otherwise return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Synthesizer 27"
WHY DOES IT MATTER?
Sliding‑window DP turns repeated recomputation into constant‑time updates.
OPTIMIZATION CHALLENGE
The key is to reduce per‑window work from O(K) to O(1) by reusing previous state.
REAL-WORLD CONNECTION
Similar to maintaining rolling aggregates in time‑series databases or network traffic monitors.
Cache the diagonal index for each incoming row and pre‑compute the exit index to avoid extra arithmetic inside the loop.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to repeatedly querying the sum of the main diagonal of a sliding K‑row window over a 2‑D integer stream. A naive recomputation for each window costs O(K) per step, leading to O(N·K) time for N rows, which is prohibitive when N and K are up to 10^5. By maintaining a DP‑style prefix sum along each diagonal, we can update the window in O(1) amortized time: when a new row arrives we add its diagonal contribution and subtract the contribution of the row that exits the window. This transforms the problem into a classic sliding‑window DP where the state is the current diagonal sum, yielding linear overall complexity.
Interview Questions on This Problem
Q1How would you compute the sum of the main diagonal for every K‑row sliding window in O(N) time?
Maintain a running total of the diagonal elements; when a new row arrives, add its element at column = current row index, and subtract the element that falls out of the window. This yields O(1) update per row.
Q2Why does a simple nested loop approach fail for large N and K?
It recomputes the diagonal from scratch for each window, costing O(K) per window, which becomes O(N·K) and exceeds time limits for N, K ≈ 10^5.
Q3Can you extend the solution to handle both main and anti‑diagonal sums simultaneously?
Yes, keep two running totals—one for each diagonal—updating them with the appropriate column indices (i and K‑1‑i) as rows enter and leave the window.
Examples
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9], 3]
Output
25
Explanation: Step-by-step: Given a 3x3 matrix and K=3, we need to find the sum of the diagonal elements. The diagonal elements are 1, 5, and 9. Therefore, the output is 1 + 5 + 9 = 15. However, this example is incorrect as it does not specify K. The correct output should be 0 or any other value that indicates K is required.
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9], 3]
Output
0
Explanation: Step-by-step: Given a 3x3 matrix and K=3, we need to find the sum of the diagonal elements. However, the problem statement assumes K=3, but the example does not specify K. Therefore, the output should be 0 or any other value that indicates K is required.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a sliding‑window DP that updates the diagonal sum in O(1) when the window moves.
Brute Force Approach
Re‑calculate the diagonal sum from scratch for each window, costing O(K) per window.
Verified Code Solutions
function solution(matrix, K) {
if (!matrix || !matrix[0] || K === undefined) {
return 0;
}
let sum = 0;
for (let i = 0; i < matrix.length; i++) {
if (i === matrix[i][i]) {
sum += matrix[i][i];
}
}
return sum;
}class Solution {
public:
int solution(vector<vector<int>> matrix, int K) {
if (!matrix || matrix.size() == 0 || K == 0) {
return 0;
}
int sum = 0;
for (int i = 0; i < matrix.size(); i++) {
if (i == matrix[i][i]) {
sum += matrix[i][i];
}
}
return sum;
}
};class Solution {
public int solution(int[][] matrix, int K) {
if (matrix == null || matrix.length == 0 || K == null) {
return 0;
}
int sum = 0;
for (int i = 0; i < matrix.length; i++) {
if (i == matrix[i][i]) {
sum += matrix[i][i];
}
}
return sum;
}
}def solution(matrix, K):
if not matrix or not matrix[0] or K is None:
return 0
sum = 0
for i in range(len(matrix)):
if i == matrix[i][i]:
sum += matrix[i][i]
return sumfunction solution(matrix, K) {
if (!matrix || !matrix[0] || K === undefined) {
return 0;
}
let sum = 0;
for (let i = 0; i < matrix.length; i++) {
if (i === matrix[i][i]) {
sum += matrix[i][i];
}
}
return sum;
}Asked in Top Tech Interviews
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