Matrix Vessel Extractor 34 — Problem Statement & Solution Guide
Problem Description
You are given a rectangular grid of integers with dimensions n rows and m columns, followed by an integer K. Your task is to compute the sum of all grid entries that are strictly greater than K. The input begins with three space‑separated integers n, m, and K. The next n lines each contain m space‑separated integers describing the grid. Output a single integer – the required sum. If no element exceeds K, output 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Vessel Extractor 34"
WHY DOES IT MATTER?
Binary search on a sorted structure turns linear scans into logarithmic lookups, dramatically cutting runtime for threshold‑based queries.
OPTIMIZATION CHALLENGE
The key is reducing repeated O(N) traversals to a single O(N log N) sort plus cheap O(log N) lookups.
REAL-WORLD CONNECTION
Databases use indexed columns to locate rows above a value in O(log N) instead of scanning every record.
Cache the sorted array and suffix sums in contiguous memory to exploit CPU cache lines and avoid repeated allocations.
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
The optimal solution treats the matrix as a flat list, sorts it, and builds a suffix‑sum array. Binary search then locates the first element > K in O(log(N)) time, allowing the sum of all larger values to be retrieved in O(1) from the pre‑computed suffix sums. A naïve scan of every cell costs O(N) time and O(1) space, which is acceptable for a single query but becomes prohibitive when N (n·m) reaches 10^7 or when multiple thresholds are asked, because each scan repeats the full traversal. By sorting once (O(N log N)) and using binary search, we amortize the cost and answer each threshold instantly, embodying the “pre‑process‑then‑query” paradigm common in binary‑search‑based optimization problems.
Interview Questions on This Problem
Q1Why does sorting the matrix values enable O(log N) query time for the sum of elements greater than K?
Sorting arranges values in monotonic order, so the boundary where values exceed K can be found with binary search. Once the index is known, a suffix‑sum array gives the total in constant time.
Q2What is the trade‑off between using a counting sort versus a comparison sort for this problem?
Counting sort runs in O(N + R) where R is the value range, offering linear time when R is small, but it uses O(R) extra space. Comparison sort is O(N log N) with O(N) space, independent of value magnitude.
Q3How would you adapt the solution if you needed to answer Q independent queries for different K values?
Pre‑process once: sort the values and build a suffix‑sum array. Each query then runs binary search in O(log N) and retrieves the sum in O(1), giving total O(N log N + Q log N).
Examples
Input
2 3 5 1 6 3 7 2 9
Output
22
Explanation: The grid contains six numbers: 1,6,3,7,2,9. Elements greater than K=5 are 6,7,9. Their sum is 6+7+9 = 22.
Input
3 2 0 -1 2 3 -4 5 6
Output
16
Explanation: Elements larger than 0 are 2,3,5,6. Adding them yields 2+3+5+6 = 16.
Input
1 5 10 11 9 12 10 15
Output
38
Explanation: Numbers exceeding K=10 are 11,12,15. Their sum is 11+12+15 = 38.
Constraints
- 1 <= n, m <= 10^3
- 1 <= n * m <= 10^5
- -10^9 <= grid[i][j] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Flatten, sort, build suffix sums, binary‑search the first > K, then read the pre‑computed sum.
Brute Force Approach
Iterate over every cell, add the value to the answer if it is > K.
Verified Code Solutions
function solution(metrics, K) { return metrics.filter(m => m > K).reduce((a, b) => a + b, 0); }class Solution { public: int solution(vector<int>& metrics, int K) { int sum = 0; for (int metric : metrics) { if (metric > K) sum += metric; } return sum; } };class Solution { public int solution(int[] metrics, int K) { int sum = 0; for (int metric : metrics) { if (metric > K) sum += metric; } return sum; } }def solution(metrics, K): return sum(m for m in metrics if m > K)function solution(metrics, K) { return metrics.filter(m => m > K).reduce((a, b) => a + b, 0); }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.