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3.2 An Example of Transformation within a Vector in Rust

3 Basic Vector Algorithm Problems in Rust

Full Course on Data Structures and Algorithms in Rust

By: Chrysanthus Date Published: 3 Feb 2026

The reader is advised to read all the lessons (tutorials) in this full course, in the order presented.

Question

Given an vector A, your task is to output an vector B of the same length, by applying the following transformation:

 - For each i from 0 to N - 1 inclusive, B[i] = A[i - 1] + A[i] + A[i + 1].
 - If an element in the sum A[i - 1] + A[i] + A[i + 1] does not exist, use 0 in its place.
 - For instance, B[0] = 0 + A[0] + A[1].

N is the number of elements in the vector.

Example:

For A = [4, 0, 1, -2, 3]:

 - B[0] = 0 + A[0] + A[1] = 0 + 4 + 0 = 4
 - B[1] = A[0] + A[1] + A[2] = 4 + 0 + 1 = 5
 - B[2] = A[1] + A[2] + A[3] = 0 + 1 + (-2) = -1
 - B[3] = A[2] + A[3] + A[4] = 1 + (-2) + 3 = 2
 - B[4] = A[3] + A[4] + 0 = (-2) + 3 + 0 = 1

So, the output should be solution(A) = [4, 5, -1, 2, 1].

Employ a time complexity of O(N) and a space complexity of O(N).

Note:

Consider the expression:

    A[i - 1] + A[i] + A[i + 1]

In the evaluation of, A[-1] + A[0] + A[1], the value A[-1] should not exist, because the index -1 is out of bounds.
In the evaluation of, A[N-2] + A[N-1] + A[N], the value A[N] should not exist, because the index N is out of bounds.

These are the only two edge cases.

Solution

Just apply the statement,

    B[i] = A[i - 1] + A[i] + A[i + 1]

in a for-loop, taking into consideration, the two edge cases above. The program is (read through the code and comments):

fn solution(a: &[i32]) {
    let n = a.len();    //length of A
    let mut b = vec![0; n];

    for i in 0..n {
        b[i] = a[i];    //for A[i] 
        
        if i > 0 {    //to avoid A[-1]
            b[i] += a[i - 1];
        }
        
        if i < n - 1 {    //to avoid A[N]
            b[i] += a[i + 1];
        }
    }

    // Output
    for i in 0..n {
        print!("{}, ", b[i]);
    }
    println!();
}

fn main() {
    let a = vec![4, 0, 1, -2, 3];
    solution(&a);
}

The output is:

    4, 5, -1, 2, 1, 

The time complexity is O(N) for the first for-loop. The space complexity is actually O(2N) for the two arrays. However, the coefficient (multiplier) is always omitted, to quote O(N).





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