Verified Cycle Metric — Problem Statement & Solution Guide
Problem Description
Given an array of integers, the verified cycle metric is defined as the sum of all elements in the array. Your task is to compute this metric and output it as a single integer. The array may contain negative numbers, zeros, and positive numbers. The sum can be large, so it should be stored in a 64‑bit signed integer type. The input format consists of two lines: the first line contains the integer n (the number of elements), and the second line contains n space‑separated integers that make up the array. The output is a single line containing the computed sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Verified Cycle Metric"
WHY DOES IT MATTER?
The reduce (or fold) pattern is foundational because many real‑world metrics—totals, averages, checksums—are computed by aggregating values. Mastery of this pattern shows a candidate can write concise, efficient code for a broad class of problems.
OPTIMIZATION CHALLENGE
The key insight is recognizing that each element contributes linearly and independently, allowing us to discard any auxiliary data structures and keep only a running total, thereby achieving O(1) extra space.
REAL-WORLD CONNECTION
Think of a distributed logging system that needs to compute the total number of bytes transferred across all servers each day; each server streams its counters to a central aggregator that simply adds them up, mirroring the single‑pass sum.
During an interview, read the input as you compute the sum—don’t store the array unless the problem explicitly requires it. This demonstrates both correctness and resource‑conscious thinking.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The Verified Cycle Metric reduces to the classic problem of computing the arithmetic sum of a sequence of integers. In algorithmic terms, this is a linear reduction (also known as a fold or accumulate) where each element contributes exactly once to the final result, making the operation O(n) in time and O(1) in auxiliary space. Naïve alternatives—such as nested loops that recompute partial sums or using high‑overhead data structures like linked lists—inflate the runtime to O(n²) or introduce unnecessary memory overhead, which becomes prohibitive for inputs that approach the limits of 10⁶ or 10⁷ elements.
The optimal paradigm leverages a single pass over the array while maintaining a 64‑bit accumulator (e.g., long long in C++ or int64 in Go). This approach guarantees that overflow is avoided for the problem’s constraints, as the accumulator can represent values up to ±9.22×10¹⁸. Moreover, streaming the input directly into the accumulator eliminates the need to store the entire array in memory, a technique especially valuable when the data originates from a file or network socket.
From a theoretical perspective, this problem exemplifies the "reduce" pattern in functional programming and the "prefix sum" concept in algorithm design. Understanding why a single linear scan suffices—and why more complex structures like segment trees or hash maps are overkill—helps candidates demonstrate mastery of time‑space trade‑offs, a core skill evaluated in high‑stakes technical interviews.
Interview Questions on This Problem
Q1How would you compute the Verified Cycle Metric for a stream of integers where the total count is unknown beforehand?
Initialize a 64‑bit accumulator to zero and read each integer from the stream, adding it to the accumulator on the fly. Since each element is processed exactly once, the algorithm remains O(n) time and O(1) extra space, regardless of the stream length.
Q2Why might using a 32‑bit integer type cause failures on this problem, and how do you prevent it?
If the sum of the array exceeds the 32‑bit signed range (±2,147,483,647), overflow will wrap around, yielding incorrect results. To prevent this, declare the accumulator as a 64‑bit signed integer (e.g., long long) and ensure all intermediate additions are performed in that type.
Q3Can you modify the algorithm to also return the maximum sub‑array sum in a single pass? Explain the trade‑offs.
Yes, by maintaining two accumulators: one for the total sum and another for Kadane's algorithm (current max ending here and global max). This adds O(1) extra space and still runs in O(n) time, but it introduces additional conditional logic that may slightly increase constant factors.
Examples
Input
4 1 2 3 4
Output
undefined
Explanation: The array contains 1, 2, 3, and 4. Adding them together gives 1+2+3+4 = 10, which is the verified cycle metric.
Input
3 -5 0 5
Output
undefined
Explanation: The elements are -5, 0, and 5. Their sum is -5+0+5 = 0, so the metric is 0.
Input
3 1000000000 1000000000 -2000000000
Output
undefined
Explanation: The sum of 1,000,000,000 + 1,000,000,000 + (-2,000,000,000) equals 0, which is the metric.
Constraints
- 1 <= n <= 100000
- -1000000000 <= nums[i] <= 1000000000
- The absolute value of the sum will not exceed 9,223,372,036,854,775,807 (fits in a 64‑bit signed integer)
Optimal Approach & Strategy
The optimal solution scans the array once, maintaining a 64‑bit accumulator, achieving O(n) time and O(1) extra space.
Brute Force Approach
A naive method would repeatedly recompute partial sums using nested loops, leading to O(n²) time.
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 = parseInt(lines[0]);
const arr = lines[1].split(' ').map(Number);
let sum = 0;
for (let i = 0; i < n; i++) {
sum += arr[i];
}
console.log(sum);
});
}
main();#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin >> n;
vector<long long> arr(n);
for (int i = 0; i < n; ++i) {
cin >> arr[i];
}
long long sum = 0;
for (int i = 0; i < n; ++i) {
sum += arr[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));
int n = Integer.parseInt(br.readLine());
String[] parts = br.readLine().split(" ");
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = Long.parseLong(parts[i]);
}
long sum = 0;
for (int i = 0; i < n; i++) {
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[1:1+n]))
total = sum(arr)
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 = parseInt(lines[0]);
const arr = lines[1].split(' ').map(Number);
let sum = 0;
for (let i = 0; i < n; i++) {
sum += arr[i];
}
console.log(sum);
});
}
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