Node Payload Consolidator 39 — Problem Statement & Solution Guide
Problem Description
You are given two integer arrays of equal length n: node[0…n‑1] and payload[0…n‑1]. The i‑th element of node denotes the identifier of a node, and payload[i] denotes the numeric value attached to that node instance. All instances that share the same identifier belong to the same logical group (i.e., they are connected). Your task is to compute the maximum total payload among all groups.
Formally, for each distinct identifier v that appears in node, define its group sum S(v) = Σ payload[i] for all i where node[i] = v. Return the largest S(v) over all identifiers. If the input contains only one element, the answer is its payload.
The algorithm must run in O(n α(n)) time or better, where α is the inverse Ackermann function, and use O(n) additional memory.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Consolidator 39"
WHY DOES IT MATTER?
Key‑based aggregation is fundamental for summarizing large data streams efficiently.
OPTIMIZATION CHALLENGE
The key is to replace quadratic pairwise scans with constant‑time map updates, cutting complexity from O(n²) to O(n).
REAL-WORLD CONNECTION
It mirrors log aggregation, where events (nodes) are grouped and metrics (payloads) are summed per service.
Initialize the map once, avoid repeated look‑ups by using get‑or‑default patterns, and pre‑allocate capacity if the distinct count is known.
COMPLEXITY AT A GLANCE
O(n)O(k)Core Theory — Why This Approach?
The problem reduces to aggregating values by a key, a classic use‑case for hash‑based grouping. A naive double loop would compare each node with every other, leading to O(n²) time, which quickly becomes infeasible for large n. The optimal paradigm leverages a hash map (or dictionary) to accumulate payloads in a single pass, achieving linear time by using constant‑time average look‑ups and updates. This approach exploits the fact that node identifiers are independent of order, allowing us to treat the arrays as a stream of (key, value) pairs and compress them on the fly.
Interview Questions on This Problem
Q1How would you handle duplicate node identifiers while summing payloads?
Use a hash map where the key is the node identifier and the value accumulates the payload. Each occurrence updates the existing sum in O(1) average time.
Q2What is the time and space complexity of the hash‑map solution?
Time complexity is O(n) because we traverse the arrays once. Space complexity is O(k), where k is the number of distinct node identifiers, bounded by O(n).
Q3If node identifiers are guaranteed to be in a small range, how could you further optimize?
Replace the hash map with a fixed‑size array indexed by the identifier, achieving true O(1) access without hashing overhead. This reduces constant factors and memory fragmentation.
Examples
Input
5 1 2 1 3 2 4 5 6 7 8
Output
undefined
Explanation: Group 1 (identifier 1) has payloads 4 and 6 → sum = 10. Group 2 (identifier 2) has payloads 5 and 8 → sum = 13. Group 3 (identifier 3) has payload 7 → sum = 7. The maximum group sum is 13, so the output is 13.
Input
7 4 4 5 5 5 6 4 2 3 1 4 5 10 6
Output
undefined
Explanation: Identifier 4 appears three times with payloads 2, 3, and 6 → sum = 11. Identifier 5 appears three times with payloads 1, 4, and 5 → sum = 10. Identifier 6 appears once with payload 10 → sum = 10. The largest sum is 11 (for identifier 4). Hence the answer is 11.
Input
3 -1 -1 -2 100 -50 200
Output
undefined
Explanation: Group -1: payloads 100 and -50 → sum = 50. Group -2: payload 200 → sum = 200. Maximum group sum is 200, so the output is 200.
Constraints
- 1 <= n <= 100000
- -10^9 <= node[i] <= 10^9
- -10^9 <= payload[i] <= 10^9
- All calculations fit into 64‑bit signed integer.
Optimal Approach & Strategy
Traverse once, updating a hash map that stores cumulative payloads per node ID, achieving O(n) time.
Brute Force Approach
For each element, scan the entire array to find matching node IDs and sum their payloads, resulting in O(n²) time.
Verified Code Solutions
function main() {
const n = parseInt(readline());
const node = readline().split(' ').map(Number);
const payload = readline().split(' ').map(Number);
const groupSum = {};
for (let i = 0; i < n; i++) {
groupSum[node[i]] = (groupSum[node[i]] || 0) + payload[i];
}
let result = 0;
for (const key in groupSum) {
result += groupSum[key];
}
console.log(result);
}
function readline() {
return require('fs').readFileSync(0, 'utf8').split('\n')[0];
}
main();#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> node(n), payload(n);
for (int i = 0; i < n; i++) cin >> node[i];
for (int i = 0; i < n; i++) cin >> payload[i];
unordered_map<int, int> groupSum;
for (int i = 0; i < n; i++) {
groupSum[node[i]] += payload[i];
}
int result = 0;
for (const auto& p : groupSum) {
result += p.second;
}
cout << result << 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[] node = new int[n];
int[] payload = new int[n];
for (int i = 0; i < n; i++) node[i] = sc.nextInt();
for (int i = 0; i < n; i++) payload[i] = sc.nextInt();
Map<Integer, Integer> groupSum = new HashMap<>();
for (int i = 0; i < n; i++) {
groupSum.put(node[i], groupSum.getOrDefault(node[i], 0) + payload[i]);
}
int result = 0;
for (int val : groupSum.values()) {
result += val;
}
System.out.println(result);
}
}def main():
n = int(input())
node = list(map(int, input().split()))
payload = list(map(int, input().split()))
group_sum = {}
for i in range(n):
group_sum[node[i]] = group_sum.get(node[i], 0) + payload[i]
result = sum(group_sum.values())
print(result)
if __name__ == "__main__":
main()function main() {
const n = parseInt(readline());
const node = readline().split(' ').map(Number);
const payload = readline().split(' ').map(Number);
const groupSum = {};
for (let i = 0; i < n; i++) {
groupSum[node[i]] = (groupSum[node[i]] || 0) + payload[i];
}
let result = 0;
for (const key in groupSum) {
result += groupSum[key];
}
console.log(result);
}
function readline() {
return require('fs').readFileSync(0, 'utf8').split('\n')[0];
}
main();Asked in Top Tech Interviews
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