RogerBW's Blog

I’ve been doing the Weekly Challenges. The latest involved string and integer list manipulation. (Note that this ends today.)

Task 1: Divisible Average

You are given an array of positive integers.

Write a script to return the closest integer to the mean value of all even integers that are divisible by 3. If none is divisible, return 0.

Quite straightforward. In Perl:

sub divisibleaverage($a) {

First, "even integers that are divisible by 3" is "integers divisible by 6." We don't care about anything else.

  my @b = grep { $_ % 6 == 0 } @{$a};

If there are none left, bail out here.

  if (scalar @b == 0) {
    return 0;
  }

Otherwise, divide sum by list length and round to nearest integer.

  int(sum(@b) / (scalar @b) + 0.5);
}

Task 2: String Difference

You are given an array of equal length strings.

Write a script to find the difference integer array of each word in the given array. There will be one word with a different array; return that word.

Each words[i] can be converted into a difference integer array difference[i] of length n - 1 where difference[i][j] = words[i][j+1] - words[i][j]

In Rust.

fn stringdifference(a: Vec<&str>) -> String {

First I'll build the list of difference arrays.

    let mut da = Vec::new();

For each input string,

    for s in &a {
        let mut p = Vec::new();

walk across pairs of values and take their differences.

        for d in s.chars().collect::<Vec<_>>().windows(2) {
            p.push((d[1] as i32) - (d[0] as i32));
        }
        da.push(p);
    }

Now we'll look for the odd one out. Walk over all entries except the first.

    let mut dx = 0;
    for b in 1..a.len() {

If this one isn't the same as the first,

        if da[0] != da[b] {

If we've seen a different entry already,

            if dx > 0 {

Then it's the first that's the different one.

                dx = 0;
                break;
            } else {

Otherwise this is probably the different one (but keep looking).

                dx = b;
            }
        }
    }

Return the value of the different entry.

    a[dx].to_string()
}

Full code in all tagged languages is on codeberg.