Matrix Stream Architect 8 — Problem Statement & Solution Guide

QueueHardBFS / Union Find

TimeO(N)

SpaceO(K)

Problem Description

You are tasked with processing a continuous stream of data packets that represent a dynamic grid of sensors. Each packet contains a timestamp, a row index, a column index, and a signal intensity value. The system must maintain a 2D matrix where each cell accumulates the maximum signal intensity received up to the current timestamp. However, the network imposes a strict bandwidth constraint: you can only process packets in batches of size K. If the total number of packets is not divisible by K, the remaining packets in the final partial batch must be discarded to maintain synchronization.

Your goal is to compute the 'Architect Value', defined as the sum of the maximum signal intensities in each row of the final processed matrix. If K is not provided (i.e., K is null or undefined), the system defaults to processing all packets individually (K=1). You must design an efficient algorithm to handle this stream, ensuring that the matrix state is updated correctly for each valid batch before computing the final row-wise maxima sum.

DSA Pattern Breakdown

"Matrix Stream Architect 8"

hard

WHY DOES IT MATTER?

The sliding window maximum pattern is essential for efficiently handling real-time data streams where only the most recent K elements are relevant. It is a cornerstone in designing high-throughput systems that require constant-time access to aggregate statistics.

OPTIMIZATION CHALLENGE

The key insight is using a monotonic deque to maintain a sorted subset of the window, allowing O(1) amortized time for both insertion and maximum retrieval, instead of O(K) for a brute-force scan.

REAL-WORLD CONNECTION

This pattern is analogous to monitoring network latency in a CDN, where you need to track the maximum latency over the last 100 requests to detect anomalies without storing all historical data.

In interviews, emphasize the amortized O(1) complexity of the monotonic deque and explain why a simple max-heap is less efficient due to O(log K) insertion and deletion costs.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(K)

Core Theory — Why This Approach?

The problem requires maintaining a dynamic 2D matrix where each cell holds the maximum signal intensity observed up to the current timestamp, under a strict bandwidth constraint that limits the number of updates or queries per time unit. A naive approach would involve scanning the entire matrix for every incoming packet to update the maximum value, resulting in O(R*C) time per operation, which is infeasible for large grids or high-frequency streams. The optimal paradigm leverages a sliding window maximum technique combined with a monotonic deque to efficiently track the maximum value in a fixed-size window of recent updates, reducing the per-operation cost to amortized O(1).

Interview Questions on This Problem

Q1How would you design a system to track the maximum value in a sliding window of size K over a continuous data stream?

Use a monotonic deque to maintain elements in decreasing order. For each new element, remove smaller elements from the back, add the new element, and remove elements from the front if they are outside the window. The front of the deque always holds the maximum value.

Q2In a distributed sensor network, how can you ensure that the maximum signal intensity is correctly maintained across multiple nodes with limited bandwidth?

Implement a local sliding window maximum tracker on each node and periodically synchronize the maximum values using a consensus algorithm like Raft or Paxos. This reduces the bandwidth required for full matrix synchronization.

Q3What data structure would you use to efficiently handle range maximum queries on a dynamic 2D matrix with point updates?

A 2D segment tree or a Fenwick tree (Binary Indexed Tree) can be used. For each row, maintain a 1D segment tree, and for the entire matrix, use a segment tree of segment trees to support O(log^2(R*C)) query and update times.

Examples

Example 1

Input

packets = [[1, 0, 0, 5], [2, 0, 1, 3], [3, 1, 0, 8], [4, 1, 1, 2]], K = 2

Output

Explanation: Batch 1: Packets [1,0,0,5] and [2,0,1,3]. Matrix becomes [[5,3],[0,0]]. Batch 2: Packets [3,1,0,8] and [4,1,1,2]. Matrix becomes [[5,3],[8,2]]. Row maxima: max(5,3)=5, max(8,2)=8. Sum = 5+8=13.

Example 2

Input

packets = [[1, 0, 0, 10], [2, 0, 1, 20], [3, 1, 0, 30]], K = 2

Output

Explanation: Batch 1: Packets [1,0,0,10] and [2,0,1,20]. Matrix becomes [[10,20],[0,0]]. Remaining packet [3,1,0,30] is discarded because 3 % 2 != 0. Row maxima: max(10,20)=20, max(0,0)=0. Sum = 20+0=20. Wait, let's re-evaluate. The problem says 'sum of maximum signal intensities in each row'. If row 1 is all zeros, max is 0. So 20+0=20. Let me correct the example to be clearer. Let's use a different example.

Example 3

Input

packets = [[1, 0, 0, 10], [2, 0, 1, 20], [3, 1, 0, 30], [4, 1, 1, 40]], K = 3

Output

Explanation: Batch 1: Packets [1,0,0,10], [2,0,1,20], [3,1,0,30]. Matrix becomes [[10,20],[30,0]]. Remaining packet [4,1,1,40] is discarded. Row maxima: max(10,20)=20, max(30,0)=30. Sum = 20+30=50.

Example 4

Input

packets = [[1, 0, 0, 5], [2, 0, 1, 3], [3, 1, 0, 8], [4, 1, 1, 2]], K = null

Output

Explanation: K is null, so K=1. Each packet is processed individually. Final matrix is [[5,3],[8,2]]. Row maxima: 5 and 8. Sum = 13.

Constraints

1 <= packets.length <= 10^5
1 <= packets[i][0] <= 10^9 (timestamp)
0 <= packets[i][1], packets[i][2] < 10^3 (row, col indices)
1 <= packets[i][3] <= 10^9 (signal intensity)
1 <= K <= 10^5 or K is null

Optimal Approach & Strategy

Use a monotonic deque to maintain the maximum value in a sliding window of recent updates for each cell. For each new packet, update the deque in O(1) amortized time and retrieve the maximum value from the front of the deque.

Brute Force Approach

For each incoming packet, scan the entire 2D matrix to find the current maximum value and update the cell if the new value is larger. This results in O(R*C) time per operation, which is too slow for large grids.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

/**
 * @param {number[][]} packets
 * @param {number} K
 * @return {number}
 */
var processPackets = function(packets, K) {
    packets.sort((a, b) => a[0] - b[0]);
    
    const matrix = new Map();
    let total = 0;
    
    for (const [t, r, c, val] of packets) {
        const key = `${r},${c}`;
        const current = matrix.get(key) || 0;
        
        if (val > current) {
            total += val - current;
            matrix.set(key, val);
        }
    }
    
    return total;
};

class Solution {
public:
    int processPackets(vector<vector<int>>& packets, int K) {
        sort(packets.begin(), packets.end(), [](const vector<int>& a, const vector<int>& b) {
            return a[0] < b[0];
        });
        
        unordered_map<int, unordered_map<int, int>> matrix;
        int total = 0;
        
        for (const auto& p : packets) {
            int t = p[0], r = p[1], c = p[2], val = p[3];
            
            if (matrix[r][c] < val) {
                total += val - matrix[r][c];
                matrix[r][c] = val;
            }
        }
        
        return total;
    }
};

class Solution {
    public int processPackets(int[][] packets, int K) {
        Arrays.sort(packets, (a, b) -> a[0] - b[0]);
        
        Map<Integer, Map<Integer, Integer>> matrix = new HashMap<>();
        int total = 0;
        
        for (int[] p : packets) {
            int t = p[0], r = p[1], c = p[2], val = p[3];
            
            Map<Integer, Integer> row = matrix.computeIfAbsent(r, k -> new HashMap<>());
            int current = row.getOrDefault(c, 0);
            
            if (val > current) {
                total += val - current;
                row.put(c, val);
            }
        }
        
        return total;
    }
}

class Solution:
    def processPackets(self, packets: List[List[int]], K: int) -> int:
        packets.sort(key=lambda x: x[0])
        
        matrix = {}
        total = 0
        
        for t, r, c, val in packets:
            key = (r, c)
            current = matrix.get(key, 0)
            
            if val > current:
                total += val - current
                matrix[key] = val
        
        return total

/**
 * @param {number[][]} packets
 * @param {number} K
 * @return {number}
 */
var processPackets = function(packets, K) {
    packets.sort((a, b) => a[0] - b[0]);
    
    const matrix = new Map();
    let total = 0;
    
    for (const [t, r, c, val] of packets) {
        const key = `${r},${c}`;
        const current = matrix.get(key) || 0;
        
        if (val > current) {
            total += val - current;
            matrix.set(key, val);
        }
    }
    
    return total;
};

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Matrix Stream Architect 8 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Queue

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Matrix Stream Architect 8 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor