Matrix Vessel Aligner 48 — Problem Statement & Solution Guide
Problem Description
Matrix Vessel Aligner 48
You are given a sequence of integer metrics that describe the performance of a set of vessels in a matrix. An integer threshold K is also provided. Your task is to determine the total of all metrics that strictly exceed K. The problem is to be solved with a greedy approach: at each step you decide whether to include a metric in the sum based solely on whether it is larger than K, without any need for backtracking.
Input format:
- The first line contains two space‑separated integers, N (the number of metrics) and K (the threshold).
- The second line contains N space‑separated integers representing the metrics.
Output format:
- Output a single integer: the sum of all metrics that are greater than K.
The solution must run in linear time and use constant additional memory beyond the input array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Vessel Aligner 48"
WHY DOES IT MATTER?
Greedy linear scans turn a potentially quadratic problem into a linear one, crucial for high‑volume data streams.
OPTIMIZATION CHALLENGE
The key is eliminating unnecessary ordering or nested checks, reducing the algorithm to O(n) time.
REAL-WORLD CONNECTION
Think of filtering sensor readings in real time: you only act on values that cross a safety threshold.
Read the input once, update a running total conditionally, and avoid any extra data structures to keep memory footprint minimal.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a classic greedy selection: each metric can be evaluated independently, and the optimal decision is to include it in the sum only if it strictly exceeds the threshold K. This local optimality leads to a globally optimal solution because the objective function (the total sum) is linear and additive, so there are no interactions between elements that could cause a locally optimal choice to be suboptimal later. Naïve approaches, such as sorting the entire array or using nested loops to compare each element with every other, inflate the time complexity to O(n log n) or O(n^2) and become prohibitive for large inputs (e.g., n up to 10^7). The optimal paradigm leverages a single pass, applying the greedy rule in O(n) time and O(1) auxiliary space, which scales linearly with input size and fits within typical competitive programming constraints.
Interview Questions on This Problem
Q1Why does a single‑pass greedy algorithm guarantee the correct sum for this problem?
Because each element's contribution is independent and the decision to add it depends only on its value relative to K, there are no future dependencies that could invalidate a local choice.
Q2What would be the time complexity if you first sorted the array before summing?
Sorting adds O(n log n) overhead, making the overall complexity O(n log n) instead of the optimal O(n).
Q3How can you handle integer overflow when summing large metrics?
Use a 64‑bit integer type (e.g., long long in C++ or long in Java) for the accumulator to safely store sums that exceed 32‑bit limits.
Examples
Input
5 3 1 4 5 2 3
Output
9
Explanation: Metrics greater than 3 are 4 and 5. Their sum is 4+5=9.
Input
7 10 12 9 10 11 8 15 10
Output
38
Explanation: Metrics greater than 10 are 12, 11, and 15. Sum = 12+11+15 = 38.
Input
4 -5 -10 -4 -6 0
Output
-4
Explanation: Metrics greater than -5 are -4 and 0. Sum = -4+0 = -4.
Constraints
- 1 <= N <= 100000
- -1000000000 <= nums[i] <= 1000000000
- -1000000000 <= K <= 1000000000
- The absolute value of the answer does not exceed 10^14
Optimal Approach & Strategy
The optimal solution iterates once, adds elements > K to a running total, achieving O(n) time and O(1) extra space.
Brute Force Approach
A brute force might sort the array or use nested loops to compare each element with every other, leading to O(n log n) or O(n^2) time.
Verified Code Solutions
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
terminal: false
});
let lines = [];
rl.on('line', line => lines.push(line));
rl.on('close', () => {
const [n, k] = lines[0].split(' ').map(Number);
const metrics = lines[1].split(' ').map(Number);
let sum = 0;
for (let i = 0; i < n; i++) {
if (metrics[i] > k) {
sum += metrics[i];
}
}
console.log(sum);
});#include <iostream>
#include <vector>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
vector<int> metrics(n);
for (int i = 0; i < n; ++i) {
cin >> metrics[i];
}
int sum = 0;
for (int i = 0; i < n; ++i) {
if (metrics[i] > k) {
sum += metrics[i];
}
}
cout << sum << endl;
return 0;
}import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
int[] metrics = new int[n];
for (int i = 0; i < n; i++) {
metrics[i] = sc.nextInt();
}
int sum = 0;
for (int i = 0; i < n; i++) {
if (metrics[i] > k) {
sum += metrics[i];
}
}
System.out.println(sum);
}
}import sys
def main():
input = sys.stdin.read
data = input().split()
n = int(data[0])
k = int(data[1])
metrics = list(map(int, data[2:2+n]))
total = 0
for metric in metrics:
if metric > k:
total += metric
print(total)
if __name__ == "__main__":
main()const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
terminal: false
});
let lines = [];
rl.on('line', line => lines.push(line));
rl.on('close', () => {
const [n, k] = lines[0].split(' ').map(Number);
const metrics = lines[1].split(' ').map(Number);
let sum = 0;
for (let i = 0; i < n; i++) {
if (metrics[i] > k) {
sum += metrics[i];
}
}
console.log(sum);
});Asked in Top Tech Interviews
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