kangaroo | OpenObject client/wrapper for Ruby , based on ActiveModel | REST library

You need a list of objects of class data. You can also use list comprehension when generating the objects

Generally, yes, you never want a non-void function to end without a return value. It's bad form and invokes undefined behavior. It might "happen" to work but you should not rely on that.

One alternative is to break out of the loop when you find the result you want, and then return based on whether you exhausted the loop. Some people prefer that approach. Others feel you should return as soon as you know you have the result you want, at least as long as you don't have to do any cleanup. So you could either do:

Assume you only need Yes/No answer, here is how to improve your original O(N^2) code to O(NlogN). You can modify to show which 2 are watched at same time

Initial remark: You want to check if there exists an integer n with

x1 + n * v1 = x2 + n * v2

In other words:

x1 - x2 = n(v2 - v1)

Now if v1 != v2 and (x1 - x2) / (v2 - v1) is an integer (or (x1 - x2) % (v2 - v1) == 0), the answer is YES. If v1 == v2 and x1 == x2, the answer is also YES In all other cases, the answer is NO.

Trying to calculate this using a loop isn't exactly an optimal approach.

Now to your code. And I must admit I didn't spot the mistake yet, but I have a few comments on it.

You terminate your loop as soon as either x1 == x2 or min <= max. We'll have to think about the meaning of the variables here.

x1 is the position of kangaroo1

x2 is the position of kangaroo2

Obviously, if both are equal, you have a "hit" (answer is YES).

min is the position of the slower kangaroo

max is the position of the faster kangaroo

As a hint, it would be a good idea to choose names that reflect this meaning. min and max aren't that well chosen. The double bookkeeping doesn't make it easier to understand what's going on (you keep the same position in two variables), and the comparison to check which velocity is which doesn't add to clarity.

Let me rewrite your algorithm a little

There are a few things that need to be changed. Changed the import line for each svg to import BellIcon from './icons/bell.svg';

Then, in the webpack, change to:

Specify key attribute for the enclosing div. Key in li element is redundant:

A canvas element can be sized arbitrarily by a style sheet, its bitmap is then subject to the 'object-fit' CSS property.

Source

The width/height of the canvas element are different from the width/height of the canvas element's bitmap. This means that you can use only CSS styles to fix this problem.

Here's a regex-based approach to the problem. We form a regex from lookup_word by adding .* between each letter in the word. Then we attempt to match the regex against word. Since .* is inherently greedy, you will get the longest possible match inside word. You can then compare the length of the matched string to the length of lookup_word, and if the matched string is longer, then lookup_word is a kangaroo word:

The issue is you're calling the shuffle function in updateQuestion. So it updates the questionArray everytime updateQuestion method is called.

Solution

Remove shuffleQuestions(); from updateQuestion method and add it before updateQuestion(); in onCreate method such that shuffling happens once in the beginning when the class is loaded.

Updated source code