We've got more trouble! Today's challenge comes from jon_pot on Codewars. You are asked to:
Write a function tripledouble(num1,num2) which takes numbers num1 and num2 and returns 1 if there is a straight triple of a number at any place in num1 and also a straight double of the same number in num2. If this isn't the case, return 0
Good luck, and happy coding!
Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!
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Top comments (12)
JavaScript
A quick solution using JavaScript, I need to improve it later. Live demo on CodePen.
Hmmm.... reading other answers I might have misinterpreted what was required in this challenge. What is it exactly? For example, if num1 = 121 and num2 = 141, would they be triple troubled numbers? (because 7*3 = 21, and 7*2 =14) Or would it be num1 = 8777and num2 = 877 be triple troubled numbers? (because 7 repeats 3 times in the first one and 2 in the second one)
I'm getting caught up a few days behind, but I understood this is three digits in a row. Which I think is confirmed by the linked challenge examples.
So
num1 = 8777 and num2 = 877
would return true for this!Starting on my version now
To achieve that, the only change would be in the
if
statement. Instead of doingx*3
andx*2
, it would be''+x+x+x
and''+x+x
respectively:A python solution. Assumes numbers are strings.
Here's my Perl solution. Works even if the first triple is not a double in the second number. Also uses regex backreferences, but only uses a regex for the first number. The much faster
index
is used to find the exact string in the second number.any
from List::Util lets us search all "triples" from the global regex, and as long as one of them matches as a double in$num2
, this will return a true value.Java:
Rust Solution!
Used a regex to find groups of 3 digits. Then checked if the digits were the same. If they were we check the second number for the digit twice, again using a regex!
A day late to the party, but here's my solution:
Gist: gist.github.com/kerrishotts/85214d...
Ruby solution:
Seems like a good problem for regex with backreferences. JavaScript:
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