fixed in latest fork

Approved

S = \frac{a_1}{1 - r}

python new test framework is required. updated in this fork

sen yarra yemişsin

Is not clear what's the return value if the discriminant (b^2 - 4ac) is negative.
Should return false,-1, 0?

Since every character in the string has to be touched, assuming 'n' refers to the size of 'aeiouAEIOU' -> 10

String comparison method:

vowel, equal frequencies: 5.5 checks

not a vowel: 10 checks, constant

Set, assuming underlying b-tree has minimal depth:

vowel, equal frequency: 1 * .1 + 2 * .2 + 3 + .4 + 4 * .3 = 2.9 checks

not a vowel: upper bound of 4 checks, lower bound 3 checks

Comparing: vowel 5.5 vs 2.9 -> ~1.90 faster for set

not a vowel: 10 vs 3 to 4 -> 3.33 to 2.5 times faster, in theory

Note: this is in theory, the python interpreter and execution environment could make this worse or better

Obviously as 'n' gets (not much) larger, the set is going to be faster. For n = 10, I would say tradeoff is speed versus speed of coding

Can someone please explain what is happening here?
And why the time complexity of this solution is O(n)?

What does a negative ratio and a negative target value means in the context of this kata? Is it even specified? Does it even make any sense to expect any values?

when given a sequence and Sn (sum of the sequence) to solve for n, where n is the minimum number of terms needed to exceed (or equal) the sum of the sequence

Reference sometimes (albeit very rarely) fails at some boundary cases in random tests:

terms([2, 12, 72], 86)
3 should equal 4

Such edge cases should be added to sample tests as well.

Any difference with looping through smileys first then check membership in arr?

• The a parameter is pointless if 1 is always passed in as the argument
• The output should really be a tuple with exact values
• Tests are using the old test framework
• Finding roots of a quadratic equation is a duplicate to many katas

Unpublishing due to low satisfaction rating and many not fixed issues.

@eijni Thanks about the tip. I was not sure if you had to tag as a spoiler for comments also :) I'm new to this discuss board.

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