Pipeline Vector Analyzer 1 — Problem Statement & Solution Guide
Problem Description
You are given an array of integers nums and an integer K. Your task is to compute the sum of all elements in nums that are strictly greater than K. The algorithm should run efficiently for large inputs.
Input: The first line contains two space‑separated integers N (the number of elements) and K. The second line contains N space‑separated integers representing the array nums.
Output: Output a single integer – the sum of all nums[i] such that nums[i] > K. If no element satisfies the condition, output 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Vector Analyzer 1"
WHY DOES IT MATTER?
The "single‑pass aggregation" pattern is fundamental because many real‑world metrics (e.g., total sales above a quota, count of high‑priority events) require scanning large datasets once and accumulating a result without extra storage. Mastering this pattern ensures you can handle massive inputs efficiently.
OPTIMIZATION CHALLENGE
The key insight is recognizing that ordering or additional data structures provide no benefit for a simple threshold comparison; thus, eliminating sorting or prefix‑sum tables reduces the time complexity from O(N log N) or O(N) with extra space to pure O(N) with O(1) auxiliary memory.
REAL-WORLD CONNECTION
Think of a network router that monitors packet sizes and needs to report the total volume of packets exceeding a certain size threshold. The router processes each packet once, adds its size to a counter if it meets the condition, and never stores the entire traffic log—mirroring the algorithm’s constant‑space, linear‑time nature.
During an interview, write the loop first, then immediately discuss edge cases (empty array, negative numbers, overflow) and I/O efficiency. Mention that the solution is optimal and that any more complex approach would be over‑engineering.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem asks for the sum of all array elements that are strictly greater than a threshold K. A naive solution would iterate over the array and, for each element, check the condition and add it to an accumulator – this is already linear, but the key is to understand why any more complex approach (like sorting or using prefix sums) would be unnecessary and even detrimental. Sorting the array would introduce O(N log N) time and extra space, which is wasteful because the condition "greater than K" does not require ordering; we only need a single pass to evaluate each element independently. The optimal paradigm here is the classic "single‑pass aggregation" pattern, where we maintain a running total while scanning the input once, achieving O(N) time and O(1) auxiliary space.
When N grows to the order of 10^7 or more, even constant‑factor overhead matters. Reading input efficiently (e.g., using buffered I/O) and using a 64‑bit accumulator to avoid overflow become crucial. The algorithm’s simplicity also makes it cache‑friendly: each element is accessed exactly once in sequential order, leading to optimal memory‑access patterns on modern CPUs. This combination of linear time, constant extra space, and low constant factors makes the solution scalable for large inputs.
Interview Questions on This Problem
Q1How would you adapt the solution if the array is provided as a real‑time data stream where you cannot store all elements?
Maintain a running sum and, for each incoming element, add it to the sum only if it exceeds K. Since you never need to retain past elements, the memory usage stays O(1) and the time per element is O(1), making the approach suitable for unbounded streams.
Q2If the numbers can be as large as 10^18, what changes would you make to avoid overflow?
Use a 64‑bit integer type (e.g., long long in C++ or long in Java) for both the individual elements and the accumulator. Additionally, consider using arbitrary‑precision libraries if the sum could exceed 64‑bit limits.
Q3Can you compute the sum of elements greater than K in parallel on a multi‑core system? Outline the steps.
Divide the array into roughly equal chunks, assign each chunk to a separate thread, and let each thread compute a local sum of elements > K. After all threads finish, combine the local sums in a final reduction step. This yields O(N/p) time per thread (where p is the number of cores) plus O(p) for the reduction, while keeping space O(1) per thread.
Examples
Input
5 10 12 7 15 3 20
Output
undefined
Explanation: Elements greater than 10 are 12, 15, and 20. Their sum is 12 + 15 + 20 = 47.
Input
8 -3 -5 -2 0 4 -1 7 -8 2
Output
undefined
Explanation: Elements greater than -3 are -2, 0, 4, -1, 7, 2. Their sum is -2 + 0 + 4 + (-1) + 7 + 2 = 10. Wait, recalculation: -2 + 0 = -2; -2 + 4 = 2; 2 + (-1) = 1; 1 + 7 = 8; 8 + 2 = 10. The correct sum is 10, not 14. Adjusted output: 10.
Input
4 100 10 20 30 40
Output
undefined
Explanation: No element exceeds 100, therefore the sum is 0.
Constraints
- 1 <= N <= 100000
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
- The sum fits in a 64‑bit signed integer.
Optimal Approach & Strategy
Use a single loop with a 64‑bit accumulator, adding only elements > K, and avoid any sorting or auxiliary containers. This yields O(N) time and O(1) extra space.
Brute Force Approach
Iterate over the array, and for each element, check if it is greater than K; if so, add it to the sum. This already runs in O(N) time but may use extra space if implemented with unnecessary data structures.
Verified Code Solutions
function 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 nums = lines[1].split(' ').map(Number);
let sum = 0;
for (let i = 0; i < N; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
console.log(sum);
});
}
main();#include <iostream>
#include <vector>
using namespace std;
int main() {
int N, K;
cin >> N >> K;
vector<int> nums(N);
for (int i = 0; i < N; i++) {
cin >> nums[i];
}
long long sum = 0;
for (int i = 0; i < N; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
cout << sum << endl;
return 0;
}import java.util.*;
import java.io.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int N = Integer.parseInt(st.nextToken());
int K = Integer.parseInt(st.nextToken());
st = new StringTokenizer(br.readLine());
int[] nums = new int[N];
for (int i = 0; i < N; i++) {
nums[i] = Integer.parseInt(st.nextToken());
}
long sum = 0;
for (int i = 0; i < N; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
System.out.println(sum);
}
}def main():
import sys
input = sys.stdin.read
data = input().split()
N = int(data[0])
K = int(data[1])
nums = list(map(int, data[2:2+N]))
total = 0
for num in nums:
if num > K:
total += num
print(total)
if __name__ == "__main__":
main()function 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 nums = lines[1].split(' ').map(Number);
let sum = 0;
for (let i = 0; i < N; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
console.log(sum);
});
}
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