Return true if three or more unique vowels in word, false if less than three in word.

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1.00/5 (2 votes)

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The code I came up with works, but it's not pretty.
String->array->Dict

What would be a better option for less Time/Space Complexity?

I've got hurricane coming in, almost 9am here. Been coding since 4pm yesterday trying to cram before power loss(Hence why I posted 3 questions today.) I am going to sleep now, so I'll have to check after.

Me need sleepy.
Me likey and value your responsey.
Good night/morning/day/time/space.

function hasThreeVowels(str){
    let vowels = "aeiou";
    let newArray = []
    for( let i = 0; i < str.length; i++){
        if(vowels.includes(str[i])){
            newArray.push(str[i]);
        }
        
    }
    let uniqueArray = new Set(newArray);
    if(uniqueArray.size >= 3){
        return true;
    } else {
        return false;
    }
}

What I have tried:

Sleep more, code less, I will.

Posted 28-Sep-22 2:33am

Chris Aug2022

Updated 28-Sep-22 2:46am

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Chris Copeland · Accepted Answer · 2022-09-28T02:46:00

The shortest I was able to come up with was:

JavaScript

function hasThreeVowels(value) {
  const vowels = 'aeiou';
  return [...value].filter(e => vowels.indexOf(e.toLowerCase()) > -1).length >= 3;
}

To break it down:
[...value]
takes the string value and creates an array populated with all the characters.

.filter(e => vowels.indexOf(e.toLowerCase()) > -1)
Then takes the array of characters and only includes those which exist within the vowels string (case-insensitive).

.length >= 3
Then checks whether the resulting array is greater-than-or-equal to three, which is the point of the function.