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You might also get a task like this right to function, which counts vowels in the string, and it is

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not always vowels, but something that you must count in the string.

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And actually, it is not that difficult, because the main idea is that first, we must define something

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like a range of possible loopholes or possible symbols, and then just try and check every single letter

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inside our strength.

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Let's do this now.

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So first of all, here we want to create a function, find the walls, and we're getting history and

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as an argument.

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Now, second live, we must define what our walls.

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So here we can create walls.

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And let's define here a e o i and you, and we can solve it with two different ways, either with follow

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up or with reduce.

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And actually, both are fine.

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So here, first of all, I want to create a counter because we want to increase it every single time

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when we find our wall.

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Now we need a for loop, so here we can write lead character off.

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And here will restrain to lower our case.

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In this case, we're getting access to every single character inside our street.

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And we can check, OK, if the wealth includes our character, then we want to increase our counter.

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So here will be count plus plus.

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And after this for loop, we just must return our account that we created.

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And here, let's try to find the balls.

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And we have some strain.

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As you can see here, we're getting three vowels because we have here e three times.

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And actually, I would say that this code is OK, but we can write it better with reduced.

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This is why I want to comment.

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Told this let code and follow up and directly here.

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Return as a deuce.

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So here, first of all, we want to convert our string to lowercase.

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So here we can just write string to lowercase.

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And after this, we want to split it in characters.

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This is where here will be split empty string.

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Now we want to call here, reduce and get an excess, first of all, to our accumulator.

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And secondly, to every single character.

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And after this function, we want to define our count and it will be zero.

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Now, inside our reduce, we want to do exactly the same logic, but can return here that if our vowels

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array includes character that we're checking, then we want to increase our count.

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So here we're increasing our accumulator with +1.

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In that case, we don't increase it and just return accumulator.

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Let's check the small time relied in the page.

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Here is the same string and we're getting three.

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And actually, this code is better because first of all, we can read it line by line.

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And secondly, it is fully pure and functional.