Accelerated Stack Horizon — Problem Statement & Solution Guide
Problem Description
Accelerated Stack Horizon
You are given a sequence of N integers that represent a set of numerical metrics collected over time. Your task is to compute the total sum of all values in the sequence. The input will consist of two lines: the first line contains the integer N (the length of the sequence), and the second line contains N space‑separated integers. The output should be a single integer – the sum of the N values.
The problem is intentionally straightforward to serve as a warm‑up for more complex graph traversal techniques that may be introduced later. It tests basic input handling, iteration, and integer arithmetic.
Input Format
- The first line contains a single integer N (1 ≤ N ≤ 10^5).
- The second line contains N integers a_1, a_2, …, a_N, each satisfying -10^9 ≤ a_i ≤ 10^9.
Output Format
- Output a single integer: the sum a_1 + a_2 + … + a_N.
The sum will always fit within a signed 64‑bit integer.
Core Theory — Why This Approach?
In the Accelerated Stack Horizon problem, the sequence of numerical metrics can be structurally modeled as a directed path graph, a specific type of Directed Acyclic Graph (DAG) where each element index i represents a vertex with a single directed edge pointing to index i+1. Under this formulation, computing the sum of the array elements is equivalent to accumulating vertex weights along a complete topological traversal of the path graph. While general graph traversal algorithms like BFS or DFS require O(V+E) space to store adjacency relations and traversal states, the linear chain structure of this problem allows us to perform an implicit traversal. By leveraging the sequential physical memory layout of the array, we bypass the need for an explicit stack or recursion. We can aggregate the total system metrics in a single pass, optimizing the auxiliary space complexity to O(1) while maintaining a strict linear time complexity of O(N). This approach bridges the gap between abstract graph aggregation and highly performant, cache-friendly linear scans.
Interview Questions on This Problem
Q1How does modeling this linear array sum as a path graph traversal help us understand more complex graph-accumulation problems, and why is an explicit graph representation unnecessary here?
A linear sequence of length N is isomorphic to a path graph where each element is a vertex with an out-degree of 1. Modeling it this way helps us see that we are traversing a DAG in topological order. An explicit graph representation (like an adjacency list) is unnecessary because the edge relation is implicitly defined by index adjacency (i to i+1), allowing us to traverse the topology using simple pointer arithmetic instead of heavy pointer-chasing structures.
Q2What are the precise time and space complexities of this traversal, and how would they change if we transitioned from this implicit path graph to an explicit adjacency list representation?
The implicit traversal takes O(N) time and O(1) auxiliary space because we only maintain a single accumulator variable. If we transitioned to an explicit adjacency list, the space complexity would degrade to O(V + E) = O(N) to store the vertices and directed edges in memory, and the time complexity would remain O(N) but incur significant allocation and cache-miss overhead.
Q3How should the traversal handle extreme edge cases such as an empty sequence or a single-node graph, and what do these cases represent in graph terminology?
An empty sequence represents a null graph (zero vertices), which should return a sum of 0. A single-node sequence represents a trivial graph with one vertex and zero edges, returning the value of that node. Both are handled correctly by initializing the accumulator to 0 and loop-guarding against the array length, preventing out-of-bounds errors on empty structures.
Q4If the problem is modified such that arbitrary jump edges are introduced (creating cycles) and we must sum all unique node weights reachable from the first node, how must our algorithm adapt?
With arbitrary edges and cycles, a simple linear scan can cause infinite loops or duplicate summation. We must adapt by implementing an explicit graph traversal (such as DFS or BFS) and maintaining a hash set or boolean array of visited nodes. We would only add a node's weight to our sum the first time it is visited, ensuring the traversal remains O(V + E) time.
Examples
Input
5 1 2 3 4 5
Output
15
Explanation: The sequence contains five numbers: 1, 2, 3, 4, and 5. Adding them together: 1+2+3+4+5 = 15. Therefore, the output is 15.
Input
3 -1 0 1
Output
0
Explanation: The sequence is -1, 0, and 1. Their sum is -1+0+1 = 0. The output is 0.
Input
4 1000000000 -1000000000 500000000 -500000000
Output
0
Explanation: The numbers are 1,000,000,000; -1,000,000,000; 500,000,000; and -500,000,000. Summing them: 1,000,000,000 + (-1,000,000,000) + 500,000,000 + (-500,000,000) = 0. Hence the output is 0.
Constraints
- 1 <= N <= 10^5
- -10^9 <= a_i <= 10^9
- The absolute value of the sum will not exceed 10^14
Optimal Approach & Strategy
Use Depth-First Search to maintain a running state in O(N) time and O(1) auxiliary space.
Brute Force Approach
Iterate over all pairs/subarrays using nested loops and calculate the metric in O(N^2) time.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}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):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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