TL;DR

On with TASK #2 from The Weekly Challenge #226. Enjoy!

# The challenge

You are given an array of non-negative integers, @ints.

Write a script to return the minimum number of operations to make every element equal zero.

In each operation, you are required to pick a positive number less than or equal to the smallest element in the array, then subtract that from each positive element in the array.

Example 1:

Input: @ints = (1, 5, 0, 3, 5)
Output: 3

operation 1: pick 1 => (0, 4, 0, 2, 4)
operation 2: pick 2 => (0, 2, 0, 0, 2)
operation 3: pick 2 => (0, 0, 0, 0, 0)


Example 2:

Input: @ints = (0)
Output: 0


Example 3:

Input: @ints = (2, 1, 4, 0, 3)
Output: 4

operation 1: pick 1 => (1, 0, 3, 0, 2)
operation 2: pick 1 => (0, 0, 2, 0, 1)
operation 3: pick 1 => (0, 0, 1, 0, 0)
operation 4: pick 1 => (0, 0, 0, 0, 0)


# The questions

No specific questions, I’d probably ask if it makes sense to check for very big numbers or big arrays.

Well, OK, I’ll take the bait.

Technically speaking, if the array contains a 0, it’s impossible to take a positive integer that is less than, or equal to the smallest element in the array (which would be 0). So I’ll assume that this reads like the smallest positive element in the array.

# The solution

As we’re asked the minimum number of bites that we can take, it makes no sense to take bites that are less than the smallest positive value still left in the array. Hence, at each step we would be removing exactly the smallest positive integer, which leaves us with all what remains from other values that are higher than that. As a consequence, we’re going to need exactly as many removals as there are non-zero distinct elements in the array.

For this reason, our algorithm will be: count how many unique, positive values are in the array and that will be our answer.

Perl goes first now:

#!/usr/bin/env perl
use v5.24;
use List::Util 'uniq';
say scalar grep { $_ } uniq @ARGV;  Raku: #!/usr/bin/env raku use v6; sub MAIN (*@ints) { put @ints.unique.grep({$_}).elems }


Stay safe folks!

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