Optimal Capacity Window — Problem Statement & Solution Guide
Problem Description
You are provided with a sequence of integers representing discrete capacity units. Your objective is to determine the aggregate total of these units. This computation requires traversing the sequence exactly once to accumulate the sum of all elements. The result represents the cumulative capacity of the entire dataset.
Input Specification:
The first line contains a single integer n, denoting the length of the sequence. The second line contains n space-separated integers, where each integer represents a capacity unit value.
Output Specification:
Print a single integer representing the total sum of all elements in the sequence.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Capacity Window"
WHY DOES IT MATTER?
Linear‑time accumulation is a foundational pattern for any problem that requires a global aggregate (sum, max, min, xor, etc.) because it guarantees optimal performance regardless of input size.
OPTIMIZATION CHALLENGE
The key insight is recognizing that addition is associative and commutative, allowing each element to be processed exactly once without auxiliary structures.
REAL-WORLD CONNECTION
Think of a power meter that continuously adds the instantaneous power draw to a total energy counter; it never revisits past readings, mirroring the single‑pass sum.
In an interview, initialize your accumulator to the neutral element (0 for sum) and update it inside the loop; avoid using built‑in functions that may hide the linear scan if the problem explicitly asks for a single traversal.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to computing the sum of a list of integers, which is a classic example of a linear scan algorithm. By iterating once over the array and maintaining an accumulator, we can aggregate the total in O(n) time, where n is the number of elements. This approach leverages the associative property of addition, allowing each element to be combined independently without needing to revisit prior elements.
A naive approach might attempt to use nested loops or repeated summations of sub‑arrays, which would inflate the time complexity to O(n^2) and quickly become infeasible for large inputs. The optimal paradigm—single-pass accumulation—avoids redundant work and keeps auxiliary memory to a constant, making it both time‑ and space‑efficient. This pattern underpins many real‑world streaming and analytics tasks where data must be processed on the fly.
Interview Questions on This Problem
Q1How would you compute the sum of a massive array that cannot fit into memory?
Process the data in chunks (or use a streaming iterator), maintaining a running total for each chunk and aggregating the partial sums. This keeps memory usage O(1) while still achieving O(n) total time.
Q2What issues arise when summing integers in languages with fixed‑width integer types, and how can you mitigate them?
Overflow can occur if the cumulative sum exceeds the maximum representable value. Mitigate by using a larger data type (e.g., long long in C++, BigInteger in Java) or by checking for overflow during accumulation.
Q3Explain how you would adapt the sum algorithm to compute the sum of absolute values in a single pass.
During the linear scan, add Math.abs(currentElement) to the accumulator instead of the raw value. This still preserves O(n) time and O(1) space.
Examples
Input
5 1 2 3 4 5
Output
15
Explanation: The array contains [1, 2, 3, 4, 5]. Summing these values: 1 + 2 = 3; 3 + 3 = 6; 6 + 4 = 10; 10 + 5 = 15. The final output is 15.
Input
3 -10 20 -5
Output
5
Explanation: The array contains [-10, 20, -5]. Summing these values: -10 + 20 = 10; 10 + (-5) = 5. The final output is 5.
Input
1 100
Output
100
Explanation: The array contains a single element [100]. The sum is simply 100. The final output is 100.
Input
4 0 0 0 0
Output
0
Explanation: The array contains [0, 0, 0, 0]. Summing these values: 0 + 0 + 0 + 0 = 0. The final output is 0.
Constraints
- 1 <= n <= 10^5
- -10^9 <= nums[i] <= 10^9
- The sum of all elements will fit within a 64-bit signed integer.
Optimal Approach & Strategy
Iterate once, maintaining a running total; this yields O(n) time and O(1) extra space.
Brute Force Approach
Repeatedly sum sub‑arrays or use nested loops, leading to O(n^2) time.
Verified Code Solutions
/**
* @param {number[]} nums
* @return {number}
*/
var solve = function(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
sum += nums[i];
}
return sum;
};#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
int solve(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};
int main() {
int n;
cin >> n;
vector<int> nums(n);
for (int i = 0; i < n; i++) {
cin >> nums[i];
}
Solution sol;
cout << sol.solve(nums) << endl;
return 0;
}class Solution {
public int solve(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}class Solution:
def solve(self, nums: list[int]) -> int:
total = 0
for num in nums:
total += num
return total/**
* @param {number[]} nums
* @return {number}
*/
var solve = function(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
sum += nums[i];
}
return sum;
};Asked in Top Tech Interviews
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