Maximized Cycle Metric — Problem Statement & Solution Guide
Problem Description
Given an array of integers, compute the sum of all its elements. The array may contain positive, negative, or zero values. The result should be returned as a single integer. Input is provided as a list of integers, and output is the total sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Maximized Cycle Metric"
WHY DOES IT MATTER?
Aggregating values in a single pass is a foundational pattern for many real‑world metrics—such as total transaction volume, cumulative latency, or overall resource consumption—where speed and memory efficiency are non‑negotiable.
OPTIMIZATION CHALLENGE
The key insight is recognizing that addition is associative and can be performed incrementally, eliminating the need for nested loops or auxiliary data structures, thereby collapsing both time and space complexity to their minimal bounds.
REAL-WORLD CONNECTION
Think of a distributed logging system that streams events to a central collector; the collector continuously adds each event's size to a running total to monitor bandwidth usage, mirroring the linear‑scan sum in a high‑throughput pipeline.
During an interview, write the loop first, then immediately declare the accumulator with the widest safe type; this pre‑emptively addresses overflow concerns and demonstrates attention to production‑grade robustness.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The sum‑of‑array problem is a classic example of a linear‑time aggregation operation. At its core, it relies on the associative property of addition, which guarantees that the order of summation does not affect the final result. This property enables a single pass over the input, accumulating a running total in a constant‑space accumulator. Naïve alternatives—such as nested loops that recompute partial sums or recursive divide‑and‑conquer without tail‑call optimization—introduce unnecessary overhead and can cause stack overflow on large inputs. The optimal paradigm leverages an iterative scan (also known as a prefix‑sum scan) that runs in O(n) time and O(1) auxiliary space, making it scalable to arrays with millions of elements.
When the input size grows to the order of 10^7 or more, cache locality and branch prediction become critical. A tight loop that adds each element to a 64‑bit accumulator minimizes cache misses and avoids the overhead of function calls. Moreover, handling mixed sign integers requires using a data type wide enough to prevent overflow; in most languages, a 64‑bit signed integer (long long, long, or BigInt in JavaScript) is sufficient for typical constraints. The combination of linear traversal, constant extra memory, and careful type selection constitutes the optimal solution for this problem.
Interview Questions on This Problem
Q1How would you compute the sum of an integer array in a language that only supports 32‑bit integers when the result may exceed 32‑bit range?
Promote the accumulator to a 64‑bit type (e.g., long long in C++ or long in Java) before the loop starts, ensuring each addition is performed in the wider type. If the language lacks native 64‑bit support, use a big‑integer library or manually handle overflow by splitting the sum into high and low 32‑bit parts.
Q2Can you modify the linear‑time sum algorithm to also return the maximum prefix sum in a single pass?
Yes. Maintain two variables: one for the running total and another for the maximum prefix seen so far. After each addition, update the max‑prefix variable if the current total exceeds it. This still runs in O(n) time and O(1) space.
Q3Why might a recursive divide‑and‑conquer sum (splitting the array in half) be less efficient than an iterative scan, even though both have O(n) time complexity?
The recursive approach incurs extra function‑call overhead and uses O(log n) stack space, which can degrade performance due to increased instruction count and potential cache misses. An iterative scan avoids these costs, leading to better constant factors and lower memory usage.
Examples
Input
[1,2,3,4]
Output
10
Explanation: 1+2+3+4 equals 10.
Input
[-5,0,5]
Output
0
Explanation: -5+0+5 equals 0.
Input
[1000000000,-1000000000,123456789]
Output
123456789
Explanation: 1000000000-1000000000+123456789 equals 123456789.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- The sum fits within a 64‑bit signed integer.
Optimal Approach & Strategy
The optimal solution iterates once over the array, maintaining a running total in a 64‑bit accumulator, achieving O(n) time and O(1) auxiliary space.
Brute Force Approach
A naive method would recompute partial sums for every possible sub‑array, leading to O(n²) time, or use recursion without tail‑call optimization, which adds overhead.
Verified Code Solutions
function solution(nums) { return nums.reduce((a, b) => a + b, 0); }class Solution { public: int solution(vector<int>& nums) { int sum = 0; for (int num : nums) { sum += num; } return sum; } };class Solution { public int solution(int[] nums) { int sum = 0; for (int num : nums) { sum += num; } return sum; } }def solution(nums): return sum(nums)function solution(nums) { return nums.reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.