Protocol Tome Architect 17 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a 2D grid of integer values representing signal strengths in a distributed network. Given a 2D array grid of size m x n and an integer threshold K, your objective is to compute the total sum of all elements in the grid that are strictly greater than K. This operation simulates the aggregation of high-priority data packets that exceed a specific noise floor.
The input consists of a 2D list of integers grid where each element represents a signal strength, and an integer K representing the threshold. The output should be a single integer representing the sum of all elements grid[i][j] such that grid[i][j] > K. If no elements exceed the threshold, the result is 0.
This problem requires iterating through the entire grid and applying a conditional summation. Although the topic is tagged as Binary Trees and the pattern as 2D Grid DP, the core logic here is a straightforward linear scan with a filter condition, suitable for an easy difficulty level. Ensure your solution handles edge cases such as empty grids or thresholds that are higher than all values in the grid.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Architect 17"
WHY DOES IT MATTER?
Scanning and aggregating is a foundational pattern for extracting metrics from large data streams.
OPTIMIZATION CHALLENGE
The key is to avoid extra passes or auxiliary containers, reducing complexity to linear time and constant space.
REAL-WORLD CONNECTION
It mirrors real‑time telemetry aggregation where only high‑priority signals are summed.
Use a simple nested for‑loop and update a running total; early exit isn’t applicable, but keep the loop tight for cache efficiency.
COMPLEXITY AT A GLANCE
O(m*n)O(1)Core Theory — Why This Approach?
The task reduces to a linear scan of a matrix, where each cell is examined exactly once to decide if it contributes to the aggregate sum. A naive approach might attempt nested loops with additional data structures or repeated passes, inflating time or space complexity, which becomes prohibitive for large m and n (e.g., 10^5 elements). The optimal paradigm leverages the fact that the condition (value > K) is independent per cell, allowing a single traversal that accumulates the result in O(m·n) time and O(1) auxiliary space, adhering to the principle of stream processing.
By treating the grid as a flat stream, we avoid unnecessary storage or recomputation. This aligns with the broader algorithmic strategy of “scan and aggregate,” common in prefix sums, histogram building, and map‑reduce patterns, where each element contributes independently to a global metric. The simplicity of the approach also ensures cache‑friendly access patterns, which is crucial for performance on large datasets.
Interview Questions on This Problem
Q1What is the time and space complexity of summing all elements greater than K in an mĂ—n matrix?
Time complexity is O(m·n) and space complexity is O(1) beyond the input.
Q2How would you modify the solution to also count the number of elements exceeding K?
Maintain a second counter variable alongside the sum and increment it whenever a cell value > K.
Q3If the matrix is stored in a sparse format, how does the approach change?
Iterate only over stored non‑zero entries, checking each against K, which can reduce time to O(number of stored entries).
Examples
Input
grid = [[1, 2, 3], [4, 5, 6], [7, 8, 9]], K = 5
Output
30
Explanation: We iterate through each element in the 3x3 grid. The elements strictly greater than 5 are 6, 7, 8, and 9. Summing these values: 6 + 7 + 8 + 9 = 30. Wait, let's re-calculate. Elements > 5: 6, 7, 8, 9. Sum = 6+7+8+9 = 30. Let me re-read the prompt's example logic. The prompt says 'sum of all metrics greater than K'. Let's pick a different K to avoid confusion or just calculate correctly. Let's use K=4. Elements > 4: 5, 6, 7, 8, 9. Sum = 5+6+7+8+9 = 35. Let's stick to the first calculation but correct the sum. 6+7+8+9 = 30. Let's try another example to be safe. Let's use grid = [[1, 2], [3, 4]], K = 2. Elements > 2: 3, 4. Sum = 7. Let's create 3 distinct examples. Example 1: grid = [[1, 2, 3], [4, 5, 6]], K = 3. Elements > 3: 4, 5, 6. Sum = 15. Example 2: grid = [[10, 20], [30, 40]], K = 25. Elements > 25: 30, 40. Sum = 70. Example 3: grid = [[1, 1], [1, 1]], K = 0. Elements > 0: 1, 1, 1, 1. Sum = 4. Let's refine the first example in the JSON to be accurate. Input: grid = [[1, 2, 3], [4, 5, 6]], K = 3. Output: 15. Explanation: Elements greater than 3 are 4, 5, and 6. Their sum is 4 + 5 + 6 = 15.
Input
grid = [[10, 20], [30, 40]], K = 25
Output
70
Explanation: The grid contains four elements: 10, 20, 30, and 40. The threshold K is 25. We check each element: 10 is not greater than 25. 20 is not greater than 25. 30 is greater than 25. 40 is greater than 25. The sum of the qualifying elements is 30 + 40 = 70.
Input
grid = [[5, 5], [5, 5]], K = 5
Output
0
Explanation: All elements in the grid are equal to 5. The condition requires elements to be strictly greater than K (which is 5). Since 5 is not greater than 5, no elements qualify. Therefore, the sum is 0.
Constraints
- 1 <= m, n <= 100
- -10^9 <= grid[i][j] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
The same single-pass nested loop is already optimal; just ensure you use a primitive accumulator and avoid extra data structures.
Brute Force Approach
Use two nested loops to visit each cell, and for each cell perform a conditional check and possibly add to the sum.
Verified Code Solutions
function sumGreaterThanK(grid, K) {
let sum = 0;
for (const row of grid) {
for (const val of row) {
if (val > K) sum += val;
}
}
return sum;
}
const grid = [[1,2,3],[4,5,6],[7,8,9]];
const K = 5;
const result = sumGreaterThanK(grid, K);
console.log(result);#include <vector>
using namespace std;
int sumGreaterThanK(const vector<vector<int>>& grid, int K) {
int sum = 0;
for (const auto& row : grid) {
for (int val : row) {
if (val > K) sum += val;
}
}
return sum;
}
int main() {
vector<vector<int>> grid = {{1,2,3},{4,5,6},{7,8,9}};
int K = 5;
int result = sumGreaterThanK(grid, K);
return 0;
}
public class Solution {
public static int sumGreaterThanK(int[][] grid, int K) {
int sum = 0;
for (int[] row : grid) {
for (int val : row) {
if (val > K) sum += val;
}
}
return sum;
}
public static void main(String[] args) {
int[][] grid = {{1,2,3},{4,5,6},{7,8,9}};
int K = 5;
int result = sumGreaterThanK(grid, K);
System.out.println(result);
}
}
def sum_greater_than_k(grid, K):
total = 0
for row in grid:
for val in row:
if val > K:
total += val
return total
if __name__ == "__main__":
grid = [[1,2,3],[4,5,6],[7,8,9]]
K = 5
result = sum_greater_than_k(grid, K)
print(result)
function sumGreaterThanK(grid, K) {
let sum = 0;
for (const row of grid) {
for (const val of row) {
if (val > K) sum += val;
}
}
return sum;
}
const grid = [[1,2,3],[4,5,6],[7,8,9]];
const K = 5;
const result = sumGreaterThanK(grid, K);
console.log(result);
Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.