Dynamic Stack Horizon — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums of length N. The dynamic stack horizon is defined as the sum of all elements in nums. Your task is to compute this sum and output it as a single integer.
Input format: The first line contains a single integer N, the number of elements in the array. The second line contains N integers separated by spaces, representing the elements of nums.
Output format: Print one integer – the dynamic stack horizon, i.e., the sum of all elements in the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Dynamic Stack Horizon"
WHY DOES IT MATTER?
Summation is a fundamental reduction pattern that appears in virtually every quantitative domain—from financial transaction totals to sensor data aggregation—making it a core building block for more complex algorithms.
OPTIMIZATION CHALLENGE
The key insight is recognizing that each element's contribution is independent and additive, allowing a single pass without any need for nested loops, sorting, or auxiliary data structures.
REAL-WORLD CONNECTION
Think of a distributed logging system where each server reports the number of processed requests; the central monitor adds these counts to obtain the total system throughput, mirroring the linear accumulation of array elements.
During an interview, start by stating the O(N) accumulator approach, then immediately discuss edge cases like empty arrays and integer overflow to demonstrate thoroughness.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The dynamic stack horizon problem reduces to computing the aggregate sum of a sequence of integers. In algorithmic terms, this is a classic example of a linear reduction where each element contributes exactly once to the final result. A naive approach might attempt to recompute partial sums repeatedly or use nested loops, leading to O(N^2) time, which quickly becomes infeasible for large N. The optimal paradigm leverages a single pass accumulation: initialize an accumulator to zero and iteratively add each array element, achieving O(N) time with O(1) auxiliary space. This pattern exemplifies the broader concept of "scan" or "prefix sum" operations, foundational in parallel computing and data analytics, where the goal is to transform a list into its cumulative aggregates efficiently.
Interview Questions on This Problem
Q1How would you modify the solution if the array size could be up to 10^7 and the sum might overflow a 32‑bit integer?
Use a 64‑bit integer type (e.g., long long in C++ or long in Java) for the accumulator to prevent overflow, and read input using fast I/O methods to handle the large volume efficiently.
Q2Can you compute the sum of a subarray [L, R] repeatedly after a single preprocessing step?
Yes, by building a prefix‑sum array where prefix[i] stores the sum of the first i elements; the sum of any subarray is prefix[R+1] - prefix[L], allowing O(1) query time after O(N) preprocessing.
Q3If the array is streamed and you cannot store all elements, how would you maintain the dynamic stack horizon?
Maintain a running total variable; for each incoming element, add it to the total. This uses O(1) memory and processes each element in O(1) time, perfectly suited for streaming scenarios.
Examples
Input
5 1 2 3 4 5
Output
15
Explanation: The array contains five elements: 1, 2, 3, 4, and 5. Adding them together gives 1+2+3+4+5 = 15, which is the required sum.
Input
3 -1 0 1
Output
0
Explanation: The elements are -1, 0, and 1. Their sum is -1+0+1 = 0, so the output is 0.
Input
4 1000000000 -1000000000 500000000 -500000000
Output
0
Explanation: Adding the four numbers: 1000000000 + (-1000000000) + 500000000 + (-500000000) = 0. The result is 0.
Input
1 42
Output
42
Explanation: With only one element, the sum equals that element itself, 42.
Constraints
- 1 <= N <= 100000
- -1000000000 <= nums[i] <= 1000000000
- The sum of all elements fits within a 64‑bit signed integer
- Input numbers are separated by single spaces
Optimal Approach & Strategy
Iterate once, accumulating each element into a single total variable, achieving O(N) time and O(1) extra space.
Brute Force Approach
A naive method would recompute sums for every possible sub‑range or use nested loops, resulting in O(N^2) time.
Verified Code Solutions
const fs=require('fs');
const data=fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx=0;
const N=data[idx++];
let sum=0;
for(let i=0;i<N;i++){
sum+=data[idx++];
}
console.log(sum);#include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int N; if(!(cin>>N)) return 0;
long long sum=0;
for(int i=0;i<N;i++){
long long x; cin>>x; sum+=x;
}
cout<<sum<<"\n";
return 0;
}import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int N=sc.nextInt();
long sum=0;
for(int i=0;i<N;i++){
sum+=sc.nextLong();
}
System.out.println(sum);
}
}import sys
data=sys.stdin.read().strip().split()
if not data:
sys.exit()
N=int(data[0])
nums=list(map(int,data[1:1+N]))
print(sum(nums))const fs=require('fs');
const data=fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx=0;
const N=data[idx++];
let sum=0;
for(let i=0;i<N;i++){
sum+=data[idx++];
}
console.log(sum);Asked in Top Tech Interviews
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