Shifted Cycle Metric — Problem Statement & Solution Guide
Problem Description
You are given an array of integers nums. The *shifted cycle metric* is defined as the sum of all elements in the array. Your task is to compute this metric and output the resulting integer.
**Input**
The first line contains a single integer n (1 ≤ n ≤ 10^5), the number of elements in the array. The second line contains n space‑separated integers nums[i] (−10^9 ≤ nums[i] ≤ 10^9).
**Output**
Print a single integer: the sum of all elements in nums.
The sum is guaranteed to fit within a signed 64‑bit integer.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shifted Cycle Metric"
WHY DOES IT MATTER?
Understanding how to reduce a problem to a single linear pass is fundamental for any large‑scale data processing task; it prevents unnecessary quadratic work and keeps runtimes predictable.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the metric is a pure sum, so no intermediate data structures or nested loops are required—just a moving accumulator.
REAL-WORLD CONNECTION
Think of a distributed log aggregation service that needs to compute the total traffic volume each day; instead of re‑scanning logs for every metric, it maintains a running total as events stream in.
When you see a problem that asks for a global property (sum, max, min) of a collection, immediately consider a single‑pass solution before reaching for complex data structures.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The shifted cycle metric reduces to a simple aggregation: the sum of all elements in the array. While the definition sounds exotic, the underlying operation is a linear scan that accumulates a running total. A naive approach might attempt to recompute partial sums for every possible sub‑array or use nested loops, which would explode to O(n²) time and quickly exceed limits for n up to 10⁵. The optimal paradigm leverages the additive property of integers—addition is associative and commutative—allowing a single pass with a constant‑space accumulator. This is a classic example of the "two‑pointer" or "sliding window" mindset, where the window spans the entire array, and the pointers converge without any inner iteration, yielding O(n) time and O(1) extra space.
Interview Questions on This Problem
Q1How would you compute the shifted cycle metric for an array of up to 10⁵ integers while ensuring you stay within the time limits?
Iterate once over the array, maintaining a 64‑bit accumulator (e.g., long long) that adds each element; this runs in O(n) time and O(1) extra space.
Q2If the array contains values up to ±10⁹, why is it important to choose the right numeric type for the accumulator?
The sum can reach ±10⁹ × 10⁵ = ±10¹⁴, which exceeds 32‑bit integer range; using a 64‑bit signed type prevents overflow.
Q3Can you extend the shifted cycle metric to support dynamic updates (e.g., point updates) and still answer queries in O(1)?
Maintain a global sum variable; for each update subtract the old value and add the new one, updating the sum in O(1) time.
Examples
Input
5 1 2 3 4 5
Output
undefined
Explanation: The array contains the numbers 1, 2, 3, 4, and 5. Summing them: 1 + 2 + 3 + 4 + 5 = 15. Therefore the shifted cycle metric is 15.
Input
4 -1 0 2 -3
Output
undefined
Explanation: The elements are -1, 0, 2, and -3. Adding them together: (-1) + 0 + 2 + (-3) = -2. The metric equals -2.
Input
1 1000000000
Output
undefined
Explanation: With only one element, the sum is the element itself. Thus the metric is 1,000,000,000.
Constraints
- 1 <= n <= 10^5
- -10^9 <= nums[i] <= 10^9
- The absolute value of the sum will not exceed 9·10^18, fitting in a 64‑bit signed integer.
Optimal Approach & Strategy
Traverse the array once, adding each element to a 64‑bit accumulator; this yields O(n) time and O(1) extra space.
Brute Force Approach
Use two nested loops to sum every possible sub‑array and then add those sums together, resulting in 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 nums = lines[1].split(' ').map(Number);
let result = 0;
for (let i = 0; i < n; i++) {
result += nums[i];
}
console.log(result);
});
}
main();#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> nums(n);
for (int i = 0; i < n; i++) {
cin >> nums[i];
}
int result = 0;
for (int i = 0; i < n; i++) {
result += nums[i];
}
cout << result << 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().trim());
String[] parts = br.readLine().trim().split(" ");
int[] nums = new int[n];
for (int i = 0; i < n; i++) {
nums[i] = Integer.parseInt(parts[i]);
}
int result = 0;
for (int i = 0; i < n; i++) {
result += nums[i];
}
System.out.println(result);
}
}import sys
def main():
input = sys.stdin.read
data = input().split()
n = int(data[0])
nums = list(map(int, data[1:n+1]))
result = sum(nums)
print(result)
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 nums = lines[1].split(' ').map(Number);
let result = 0;
for (let i = 0; i < n; i++) {
result += nums[i];
}
console.log(result);
});
}
main();Asked in Top Tech Interviews
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