Optimal Path Weight — Problem Statement & Solution Guide
Problem Description
You are given a linear sequence of integer weights representing nodes in a one-dimensional grid. The objective is to determine the total accumulated weight by traversing the entire sequence from the first element to the last. This traversal is mandatory; you must include every element in the summation regardless of its sign. The result represents the net weight of the complete path.
Implement a function that accepts an array of integers and returns the sum of all its elements. The solution must efficiently handle large datasets and correctly process negative values, zero, and positive integers. If the input array is empty, the function should return 0, as the sum of an empty set is defined as the additive identity.
Your implementation should be optimized for time complexity, ensuring that it processes the input in a single pass. The output is a single integer representing the final accumulated weight.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Path Weight"
WHY DOES IT MATTER?
The linear scan pattern is fundamental because it guarantees the lowest possible time and space complexity for problems that require examining every element. It also simplifies reasoning about correctness and makes the code easier to maintain and test.
OPTIMIZATION CHALLENGE
The key insight is that the sum operation is associative; therefore, you can accumulate the result incrementally without storing intermediate sums, eliminating the need for auxiliary data structures.
REAL-WORLD CONNECTION
Consider a sensor network that streams temperature readings. To compute the total energy consumption over a period, each sensor reports its reading, and a central server sums them in a single pass—mirroring the linear scan pattern used here.
When interviewing, emphasize that the algorithm is O(n) and O(1), and mention that built‑in functions like Python's sum() or Java's streams can be used for brevity, but a manual loop is often preferred for clarity and control over data types.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a classic linear scan over a one‑dimensional array of integers. The optimal algorithm simply iterates once, adding each element to a running total. This yields a time complexity of O(n) and constant auxiliary space, which is optimal because every element must be examined at least once to guarantee correctness.
A naive approach might attempt to use nested loops or dynamic programming to compute partial sums, which would unnecessarily increase time complexity to O(n^2) or add extra memory for prefix arrays. Such approaches are overkill for a simple summation and would fail on large inputs due to quadratic time or linear space overhead. The key insight is that the sum of a sequence is associative and commutative, allowing a single pass accumulation without any need for intermediate storage.
In practice, this pattern is a building block for many algorithms: computing prefix sums, sliding window aggregates, and cumulative metrics in streaming data. Recognizing when a problem can be solved with a single pass is essential for writing efficient code and for passing interview questions that test algorithmic thinking and optimization awareness.
Interview Questions on This Problem
Q1What is the most efficient way to compute the total weight of a sequence of integers, and why is it efficient?
Use a single linear scan that adds each element to a running total. This approach runs in O(n) time and O(1) space, which is optimal because every element must be processed at least once.
Q2How would you handle potential integer overflow when summing a large array of weights in a production system?
Use a 64‑bit integer type (e.g., long in Java, long long in C++) or a big integer library if the range exceeds 64 bits. Additionally, consider checking for overflow during accumulation or using built‑in functions that detect overflow.
Q3In a distributed system that aggregates logs from multiple nodes, how can you efficiently compute the global sum of event weights?
Each node can compute a local sum in O(k) time for its k events, then send the local sum to a central aggregator. The aggregator combines these local sums in O(m) time for m nodes, achieving overall O(total_events) time with minimal communication overhead.
Examples
Input
nums = [1, 2, 3, 4, 5]
Output
15
Explanation: Start with sum = 0. Add 1 -> sum = 1. Add 2 -> sum = 3. Add 3 -> sum = 6. Add 4 -> sum = 10. Add 5 -> sum = 15. The final accumulated weight is 15.
Input
nums = [-1, -2, -3]
Output
-6
Explanation: Start with sum = 0. Add -1 -> sum = -1. Add -2 -> sum = -3. Add -3 -> sum = -6. The negative weights accumulate to a total of -6.
Input
nums = [10, -5, 0, 5, -10]
Output
0
Explanation: Start with sum = 0. Add 10 -> sum = 10. Add -5 -> sum = 5. Add 0 -> sum = 5. Add 5 -> sum = 10. Add -10 -> sum = 0. The positive and negative weights cancel each other out, resulting in a net weight of 0.
Input
nums = []
Output
0
Explanation: The input array is empty. By definition, the sum of no elements is 0. The function returns 0 immediately.
Constraints
- 0 <= nums.length <= 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 over the array, adding each element to a single accumulator variable. This achieves O(n) time and O(1) space, the optimal solution.
Brute Force Approach
A naive method might use nested loops or build a prefix sum array, leading to O(n^2) time or O(n) extra space. This is unnecessary for a simple summation.
Verified Code Solutions
function sumArray(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}#include <bits/stdc++.h>
using namespace std;
int main(){
vector<int> nums = {1, 2, 3, 4, 5};
int sum = 0;
for(int x : nums) sum += x;
cout << sum << endl;
return 0;
}public class Solution {
public int sumArray(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def sum_array(nums):
total = 0
for num in nums:
total += num
return totalfunction sumArray(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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