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Solutions to weekly challenge 73 by Bob Lied.
https://perlweeklychallenge.org/blog/perl-weekly-challenge-073/
* TASK #1 >
** Initial thoughts
A window of an array implies slices. Minimum of an array implies List::Util::min.
** Post Solution Thoughts
Repeatedly doing the scan for minimum is a lot of extra work. The cost of
doing an array of 50 with a window of 3 is minimal, but what about an array
of 100000 with a window of 200? So much extra work. I did a second solution
with indices that didn't actually take slices. It's also not necessary to
find the minimum each time: as we slide the window right, if it's less than
the minimum, it becomes the new minimum for as long as it's in the window.
The only time we need to scan for minimum is when the window slides the
minimum off the left edge.
I used the task as an excuse to build a slightly more structured program,
with the solution in a module and a separate test file.
** Problem Statement
# You are given an array of integers @A and sliding window size $S.
# Write a script to create an array of min from each sliding window.
* TASK #2 > Smallest Neighbor
** Initial Thoughts
There's probably a clever solution involving a reduce() operation, but the
direct solution of scanning and keeping track of the minimum is probably
going to be the way to go.
** Post Solution Thoughts
The minimum keeps getting smaller as we move from left to right. Test sets
with a zero in the data create ambiguous answers, but there's nothing in the
specification that says we have to do anything about that. Negative numbers
work just fine.
** Problem Statement
# You are given an array of integers @A.
# Write a script to create an array that represents the smallest element to the
# left of each corresponding index. If none found, then use 0.
# Example 1:
# Input: @A = (7,8,3,12,10)
# Output: (0,7,0, 3, 3)
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