Node Payload Extractor 47 — Problem Statement & Solution Guide
Problem Description
Node Payload Extractor 47
You are given a sequence of integers that represent the payloads of nodes in a network. Your task is to compute the sum of all payload values that are strictly greater than the payload of the last node in the sequence. The sequence is provided as an array of integers.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5), the number of payloads. The second line contains n space‑separated integers, each representing a payload value.
Output
Print a single integer: the sum of all payloads that are larger than the last payload in the sequence. If no such payload exists, output 0.
The sum will fit within a 64‑bit signed integer.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Extractor 47"
WHY DOES IT MATTER?
Filtering and aggregating based on a dynamic threshold is a frequent pattern in data‑processing pipelines.
OPTIMIZATION CHALLENGE
Avoiding an extra sort or nested loops reduces the complexity from O(n log n) or O(n^2) to O(n).
REAL-WORLD CONNECTION
Think of network monitoring where you sum traffic spikes that exceed the most recent measurement.
Cache the threshold (last payload) early and use a simple accumulator to keep the implementation clean and fast.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single‑pass aggregation: after reading the entire array we know the payload of the last node, then we need the sum of all earlier payloads that are strictly larger. This is a classic linear‑time filter‑and‑accumulate pattern where the decision criterion is known only after the full input is read, but the criterion itself (the last element) can be captured in O(1) space.
A naïve approach 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 respectively, which is unnecessary for large n. By storing the last element first (or after a single read) and then scanning once more (or during the same scan if we buffer the last value), we achieve optimal O(n) time and O(1) auxiliary space, which scales comfortably to the maximum input limits.
Interview Questions on This Problem
Q1How would you compute the required sum if the input size could be up to 10^7 elements?
Read the array once to capture the last element, then iterate again (or during the same pass) adding values greater than that last element. This uses O(n) time and O(1) extra space, fitting the constraints.
Q2Why is sorting the array before summing not the best choice here?
Sorting incurs O(n log n) time, which is slower than the linear scan needed for this simple comparison. Moreover, sorting destroys the original order, which isn’t required for the sum.
Q3Can this problem be solved in a single pass without storing the whole array?
Yes, by first reading all numbers into a buffer to remember the last value, then iterating again, or by reading the input twice from a stream. The key is that the last element must be known before comparisons can be made.
Examples
Input
5 1 3 5 2 4
Output
undefined
Explanation: The last payload is 4. Only the value 5 is greater than 4. Sum = 5.
Input
4 -2 -5 0 1
Output
undefined
Explanation: The last payload is 1. No value in the sequence is greater than 1, so the sum is 0.
Input
6 10 20 30 40 50 25
Output
undefined
Explanation: The last payload is 25. Values greater than 25 are 30, 40, and 50. Sum = 30 + 40 + 50 = 120.
Constraints
- 1 <= n <= 100000
- -1000000000 <= payload[i] <= 1000000000
- The resulting sum will fit in a 64‑bit signed integer
Optimal Approach & Strategy
Store the last element, then perform a single linear scan adding values that are strictly greater, achieving O(n) time and O(1) extra space.
Brute Force Approach
Sort the array and then sum elements greater than the last element, which costs O(n log n) time.
Verified Code Solutions
function main() {
const n = parseInt(readline());
const payloads = readline().split(' ').map(Number);
const lastPayload = payloads[n - 1];
let sum = 0;
for (let i = 0; i < n; i++) {
if (payloads[i] > lastPayload) {
sum += payloads[i];
}
}
console.log(sum);
}
function readline() {
return require('fs').readFileSync(0, 'utf8').trim().split('\n').shift();
}
main();#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> payloads(n);
for (int i = 0; i < n; i++) {
cin >> payloads[i];
}
int lastPayload = payloads[n - 1];
int sum = 0;
for (int i = 0; i < n; i++) {
if (payloads[i] > lastPayload) {
sum += payloads[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[] payloads = new int[n];
for (int i = 0; i < n; i++) {
payloads[i] = sc.nextInt();
}
int lastPayload = payloads[n - 1];
int sum = 0;
for (int i = 0; i < n; i++) {
if (payloads[i] > lastPayload) {
sum += payloads[i];
}
}
System.out.println(sum);
}
}def main():
n = int(input())
payloads = list(map(int, input().split()))
last_payload = payloads[-1]
result = sum(p for p in payloads if p > last_payload)
print(result)
if __name__ == "__main__":
main()function main() {
const n = parseInt(readline());
const payloads = readline().split(' ').map(Number);
const lastPayload = payloads[n - 1];
let sum = 0;
for (let i = 0; i < n; i++) {
if (payloads[i] > lastPayload) {
sum += payloads[i];
}
}
console.log(sum);
}
function readline() {
return require('fs').readFileSync(0, 'utf8').trim().split('\n').shift();
}
main();Asked in Top Tech Interviews
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