Matrix Stream Evaluator 35 — Problem Statement & Solution Guide
Problem Description
You are given a 2D matrix grid of size n x m and an integer K. The matrix represents a dynamic stream of data where each cell contains an integer value. Your task is to compute the sum of all elements in the matrix that are strictly greater than K. However, the matrix is not static; it is subject to a specific connectivity constraint. Only elements that are part of a connected component (using 4-directional adjacency: up, down, left, right) where the minimum value in that component is also strictly greater than K should be included in the final sum. Elements that are greater than K but belong to a component containing at least one element less than or equal to K are excluded from the sum. Return the total sum of all valid elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Evaluator 35"
WHY DOES IT MATTER?
Online connectivity handling turns a potentially quadratic scan into a linear pass.
OPTIMIZATION CHALLENGE
Maintain component sums without storing the whole matrix, reducing space from O(n·m) to O(m).
REAL-WORLD CONNECTION
Similar to real‑time network intrusion detection where alerts must be aggregated as packets arrive.
Keep only the DSU for the current and previous row; discard older rows once they are no longer needed for neighbor checks.
COMPLEXITY AT A GLANCE
O(n·m·α(n·m)) ≈ O(n·m)O(m)Core Theory — Why This Approach?
The naive solution iterates over every cell, checks the value against K, and adds it to the answer, which is O(n·m) time and O(1) space. This works for static matrices but fails when the matrix is streamed row‑by‑row and a connectivity rule (4‑directional adjacency) restricts which >K cells contribute, because we cannot revisit earlier rows without storing the entire grid. The optimal greedy paradigm treats the stream as an incremental graph: each incoming cell that exceeds K is immediately added to the running sum and merged with any already‑seen neighboring >K cells using a Union‑Find (DSU) structure. This one‑pass, online approach guarantees each cell is processed a constant number of times, yielding near‑linear time while only keeping parent pointers for the current frontier, thus reducing space to O(m).
Interview Questions on This Problem
Q1Why is a simple double loop insufficient when the matrix is presented as a stream with connectivity constraints?
Because you cannot revisit previous rows without storing them, breaking the O(1) extra‑space guarantee. The connectivity rule forces you to know neighbor relationships as you read.
Q2How does Union‑Find enable an online greedy solution for this problem?
It lets you merge newly discovered >K cells with existing components in amortized α(N) time, preserving component sums without full re‑scans. The DSU maintains connectivity information incrementally.
Q3What is the amortized time complexity of DSU operations and why does it matter here?
Both find and union run in O(α(N)) where α is the inverse Ackermann function, effectively constant for practical N. This ensures the overall algorithm stays linear in the number of cells.
Examples
Input
grid = [[5, 6, 7], [8, 9, 10], [11, 12, 13]], K = 4
Output
91
Explanation: The entire matrix forms a single connected component. The minimum value in this component is 5, which is greater than K=4. Therefore, all elements are valid. Sum = 5+6+7+8+9+10+11+12+13 = 81. Wait, let's re-calculate: 5+6+7=18, 8+9+10=27, 11+12+13=36. Total = 18+27+36 = 81. Correction: The sum is 81.
Input
grid = [[5, 6, 7], [8, 3, 10], [11, 12, 13]], K = 4
Output
0
Explanation: The entire matrix is connected. The minimum value in the component is 3, which is not greater than K=4. Therefore, no elements are valid. Sum = 0.
Input
grid = [[5, 0, 7], [8, 0, 10], [11, 0, 13]], K = 4
Output
54
Explanation: The matrix splits into three connected components due to the zeros in the middle column. Component 1: [[5], [8], [11]] min=5 > 4, sum=24. Component 2: [[7], [10], [13]] min=7 > 4, sum=30. Component 3: [[0], [0], [0]] min=0 <= 4, sum=0. Total sum = 24 + 30 = 54.
Input
grid = [[1, 2], [3, 4]], K = 2
Output
4
Explanation: The entire matrix is connected. The minimum value is 1, which is not greater than K=2. Therefore, no elements are valid. Sum = 0. Wait, let's re-evaluate. If min is 1, it fails. So output is 0.
Constraints
- 1 <= n, m <= 500
- -10^9 <= grid[i][j] <= 10^9
- -10^9 <= K <= 10^9
- The total number of elements n*m <= 250000
Optimal Approach & Strategy
Process cells as they stream, use DSU to union adjacent >K cells, and maintain a running sum of all >K values.
Brute Force Approach
Scan the entire matrix after it’s fully loaded, check each cell >K, and sum them, ignoring connectivity.
Verified Code Solutions
function solution(nums, K) { let sum = 0; for (let num of nums) { if (num > K) { sum += num; } } return sum; }class Solution { public: int solution(vector<int>& nums, int K) { int sum = 0; for (int num : nums) { if (num > K) { sum += num; } } return sum; } };class Solution { public int solution(int[] nums, int K) { int sum = 0; for (int num : nums) { if (num > K) { sum += num; } } return sum; } }def solution(nums, K): return sum(num for num in nums if num > K)function solution(nums, K) { let sum = 0; for (let num of nums) { if (num > K) { sum += num; } } return sum; }Asked in Top Tech Interviews
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