Protocol Tome Extractor 20 — Problem Statement & Solution Guide
Problem Description
You are given a list of integers and a target value K. Your task is to calculate the total of all numbers in the list that are strictly greater than K. The input consists of three lines: the first line contains the number of elements N, the second line lists the N integers separated by spaces, and the third line provides the integer K. The output should be a single integer representing the sum of all elements that exceed K. If no element satisfies the condition, output 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Extractor 20"
WHY DOES IT MATTER?
Linear reduction patterns appear in many analytics tasks where a global metric is derived from element‑wise conditions.
OPTIMIZATION CHALLENGE
The key is to avoid extra passes or auxiliary structures; a single accumulator eliminates unnecessary overhead.
REAL-WORLD CONNECTION
Think of a server log scanner that sums the size of all requests exceeding a bandwidth threshold.
Initialize the sum as a 64‑bit integer and update it in place to prevent accidental reallocation in tight loops.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a single linear scan where each element is compared against a threshold K and, if larger, added to an accumulator. This is a classic example of a reduction operation that can be solved in O(N) time using constant extra space, leveraging the associative property of addition. Naïve alternatives, such as sorting the array first or using nested loops to compare each pair, inflate the time complexity to O(N log N) or O(N^2) and become prohibitive for large N. The optimal paradigm is a single-pass greedy aggregation, which guarantees the minimal work needed because each element must be examined at least once to decide its contribution.
Interview Questions on This Problem
Q1How would you handle the case where the list contains negative numbers and K is also negative?
The same linear scan works; the comparison > K correctly filters values regardless of sign. No special handling is required beyond using the appropriate data type.
Q2Can this problem be solved without an explicit loop in a functional programming language?
Yes, by using built‑in higher‑order functions like filter and sum (e.g., sum(filter(lambda x: x > K, arr))). These abstractions still perform a linear traversal under the hood.
Q3What is the impact on complexity if the input size N is unknown and you must read until EOF?
Reading until EOF still results in a single pass over the data, preserving O(N) time. Space remains O(1) because you accumulate the result on the fly.
Examples
Input
5 1 7 3 9 2 4
Output
16
Explanation: The elements greater than 4 are 7 and 9. Their sum is 7 + 9 = 16.
Input
4 -5 -1 0 2 0
Output
2
Explanation: Only the element 2 is greater than 0. The sum is 2.
Input
6 10 20 30 40 50 60 35
Output
150
Explanation: Elements greater than 35 are 40, 50, and 60. Their sum is 40 + 50 + 60 = 150.
Input
3 5 5 5 5
Output
0
Explanation: No element is strictly greater than 5, so the sum is 0.
Constraints
- 1 <= N <= 100000
- -1000000000 <= nums[i] <= 1000000000
- -1000000000 <= K <= 1000000000
- The resulting sum fits within a 64‑bit signed integer.
- Time limit: 1 second; memory limit: 256 MB.
Optimal Approach & Strategy
Iterate once, compare each element to K, and accumulate the sum directly, achieving O(N) time and O(1) space.
Brute Force Approach
Sort the array then iterate from the first element larger than K, summing the rest, which adds unnecessary O(N log N) 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 = parseInt(lines[0]);
const arr = lines[1].split(' ').map(Number);
const K = parseInt(lines[2]);
let sum = 0;
for (let i = 0; i < N; i++) {
if (arr[i] > K) {
sum += arr[i];
}
}
console.log(sum);
});#include <iostream>
#include <vector>
using namespace std;
int main() {
int N;
cin >> N;
vector<int> arr(N);
for (int i = 0; i < N; i++) {
cin >> arr[i];
}
int K;
cin >> K;
int sum = 0;
for (int i = 0; i < N; i++) {
if (arr[i] > K) {
sum += arr[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[] arr = new int[N];
for (int i = 0; i < N; i++) {
arr[i] = sc.nextInt();
}
int K = sc.nextInt();
int sum = 0;
for (int i = 0; i < N; i++) {
if (arr[i] > K) {
sum += arr[i];
}
}
System.out.println(sum);
}
}import sys
def main():
input = sys.stdin.read
data = input().split()
N = int(data[0])
arr = list(map(int, data[1:N+1]))
K = int(data[N+1])
total = 0
for num in arr:
if num > K:
total += num
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 = parseInt(lines[0]);
const arr = lines[1].split(' ').map(Number);
const K = parseInt(lines[2]);
let sum = 0;
for (let i = 0; i < N; i++) {
if (arr[i] > K) {
sum += arr[i];
}
}
console.log(sum);
});Asked in Top Tech Interviews
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