Task 1: Consecutive Odds
Task
You are given an array of integers.
Write a script to print 1 if there are THREE consecutive odds in the given array otherwise print 0.
My solution
This task is pretty straight forward, so doesn't need much explanation. I have a variable odds. As I iterate through the array, I add one to the odds variable if it is an odd number. If it is even, I reset odds value as zero.
If at any time the odds value is 3, I print 1 and exit. Otherwise I will print 0 if we have exited the loop without finding three consecutive odd values.
Examples
$ ./ch-1.py 1 5 3 6
1
$ ./ch-1.py 2 6 3 5
0
$ ./ch-1.py 1 2 3 4
0
$ ./ch-1.py 2 3 5 7
1
Task 2: Widest Valley
Task
Given a profile as a list of altitudes, return the leftmost widest valley. A valley is defined as a subarray of the profile consisting of two parts: the first part is non-increasing and the second part is non-decreasing. Either part can be empty.
My solution
This is a really interesting challenge, and I had a few attempts at trying to come up with the best solution. Often, the best solution is just try everything to get the correct solution. That's what I did here.
For this task, I have a variable valley that iterates from 0 to one less than the length of the array. This is the deepest point in the valley.
From this I calculate the start point of the valley. I do this by setting start at the current position, and working backwards until we find a decreasing number or hit the start of the array. Likewise, I calculate the end value by doing the same rightwards.
Once I have the start and end point, if the difference between these values is greater than or equal to the length of the current solution, I replace the solution variable.
Why equals? The difference between start and end is one less than the numbers of elements it contains. For example if start is 7 and end is 9, it contains three items (7, 8 and 9), even though the difference is two.
I'll be first to admit that this could actually be optimised a lot. For example if the previous or next value is lower than the current point, we know that we aren't at the bottom of the cliff. However, when dealing with such a short list, the benefits are negligible, and adds unnecessary complications. YMMV.
Examples
$ ./ch-1.py 1 5 3 6
1
$ ./ch-1.py 2 6 3 5
0
$ ./ch-1.py 1 2 3 4
0
$ ./ch-1.py 2 3 5 7
1
$ ./ch-2.py 1 5 5 2 8
1, 5, 5
$ ./ch-2.py 2 6 8 5
2, 6, 8
$ ./ch-2.py 9 8 13 13 2 2 15 17
13, 13, 2, 2, 15, 17
$ ./ch-2.py 2 1 2 1 3
2, 1, 2
$ ./ch-2.py 1 3 3 2 1 2 3 3 2
3, 3, 2, 1, 2, 3, 3
Top comments (1)
Okay. Looking at other peoples solutions (which I only do after submitting my pull request), it seems like I might be on the wrong track with the first task. I've taken "THREE consecutive odds" to mean three consecutive numbers that are odd. It seems everyone else has taken it to mean three consecutive odd numbers (like 5, 7, 9) to appear anywhere in the array. The example solution does not prove either way is wrong.
There are others who have taken the approach I took, so its definitely open to interpretation. Oh for the love of the English language.