Sequential Path Weight â Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the sequential path weight according to the target algorithm rules. The correct output is the sum of the array elements multiplied by the number of elements in the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sequential Path Weight"
WHY DOES IT MATTER?
Linear-time aggregation patterns are foundational in algorithm design, enabling solutions that scale linearly with input size. They are especially critical in interview contexts where candidates must demonstrate an understanding of time complexity trade-offs and efficient data processing.
OPTIMIZATION CHALLENGE
The key insight is to separate the summation from the scaling factor. By computing the sum in one pass and applying the multiplication once, we avoid redundant work and achieve O(N) time with O(1) space.
REAL-WORLD CONNECTION
Consider a log analytics system that aggregates event counts across millions of servers. The system must compute the total number of events and then apply a global scaling factor (e.g., for normalization). Using a single-pass aggregation mirrors the sequential path weight computation, ensuring low latency and high throughput.
When presenting this solution in an interview, emphasize the use of a single accumulator and the final multiplication step. Highlight that this pattern is a textbook example of linear-time aggregation and that it can be extended to distributed or streaming contexts with minimal modifications.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory â Why This Approach?
The problem reduces to a simple linear algebraic expression: the sequential path weight equals the sum of all elements in the array multiplied by the number of elements, i.e., weight = N * ÎŁa_i. Naively, one might think to recompute the sum for each element or use nested loops, leading to O(N^2) time, which quickly becomes infeasible for large N (e.g., 10^7). The optimal paradigm is a single-pass accumulation: iterate once, add each element to a running total, and after the loop multiply by N. This approach guarantees O(N) time and O(1) auxiliary space, making it scalable for massive datasets and suitable for real-time analytics pipelines.
In many interview settings, candidates are expected to recognize that the problem is essentially a weighted sum where the weight is constant across all elements. The key insight is that the weight (the array length) can be factored out of the summation, allowing the sum to be computed independently and then scaled. This separation of concerns eliminates unnecessary recomputation and aligns with the principle of linear time complexity for problems that can be expressed as a single aggregate operation.
Furthermore, understanding the difference between a naive double-loop approach and the optimal single-loop approach is crucial. The naive method may involve summing the array for each element or performing repeated multiplications inside the loop, both of which inflate the time complexity. By contrast, the optimal method leverages the associative property of addition and multiplication to collapse the computation into a single pass, ensuring both efficiency and clarity in code implementation.
Interview Questions on This Problem
Q1How would you modify the algorithm if the array elements were floating-point numbers and you needed to maintain precision up to 6 decimal places?
Use a double-precision accumulator and apply a rounding function (e.g., round(value, 6)) after the final multiplication. Alternatively, use the Decimal type in languages that support arbitrary precision to avoid floating-point errors.
Q2In a distributed system where the array is partitioned across multiple nodes, how would you compute the sequential path weight efficiently?
Each node computes the local sum and local count. Then perform a reduce operation (e.g., MPI_Reduce) to aggregate the sums and counts. Finally, multiply the global sum by the global count to obtain the weight, ensuring minimal communication overhead.
Q3What potential pitfalls could arise when the array length N is extremely large (e.g., 10^12) and the elements are also large integers?
Integer overflow becomes a concern; use 64-bit integers or big integer libraries. Additionally, reading the entire array into memory may be infeasible, so streaming or chunked processing is necessary to keep memory usage bounded.
Examples
Input
[1, 2, 3, 4, 5]
Output
75
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5], we first calculate the sum of the array elements, which is 1 + 2 + 3 + 4 + 5 = 15. Then, we multiply the sum by the number of elements, which is 5. Therefore, the correct output is 15 * 5 = 75.
Input
[6, 7, 8, 9, 10]
Output
200
Explanation: Step-by-step: Given the input [6, 7, 8, 9, 10], we first calculate the sum of the array elements, which is 6 + 7 + 8 + 9 + 10 = 40. Then, we multiply the sum by the number of elements, which is 5. Therefore, the correct output is 40 * 5 = 200.
Constraints
- 1 <= N <= 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity expected: O(N) or O(N log N)
- Space Complexity expected: O(1) or O(N)
Optimal Approach & Strategy
The optimal approach performs a single pass: accumulate the sum while iterating, then multiply by the length once after the loop. This achieves O(N) time and O(1) space.
Brute Force Approach
A naive approach would iterate over the array to compute the sum, then multiply by the length. This requires two passes over the data, resulting in O(N) time but still acceptable for small inputs.
Verified Code Solutions
function solution(nums) {
let sum = nums.reduce((a, b) => a + b, 0);
return sum * nums.length;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum * nums.size();
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum * nums.length;
}
}def solution(nums):
sum = sum(nums)
return sum * len(nums)function solution(nums) {
let sum = nums.reduce((a, b) => a + b, 0);
return sum * nums.length;
}Asked in Top Tech Interviews
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