SpeedTyping

Tez yozishni mashq qilib ko'ring. Tepada Select bor Hard Medium va Easy birini tanlab qanchalik tez yozishingizni sinab bilib oling.

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Kanal @jstutorial

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Dasturchilar uchun #Coding_Challenge Algoritmni rivojlantirishga yordam beradigan top saytlar to'plami.

1. www.hackerrank.com
2. www.coderbyte.com
3. https://exercism.io
4. www.codeforces.com
5. www.topcoder.com
6. www.leetcode.com
7. www.codechef.com
8. www.geeksforgeeks.org
9. https://edabit.com
10. https://codingbat.com
11. www.codewars.com
12. https://www.topcoder.com
13. www.freecodecamp.org/learn
14. https://the-winter.github.io/codingjs/
15. https://www.codingame.com/
16. https://codesignal.com/
17.https://the-winter.github.io/codingjs
18. https://robocontest.uz/


Kanalimiz @JsTutorial

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Ushbu algoritmni yechishga harakat qilib ko'ring. Yechimlaringizni @jsfamily_bot ga yuborishingiz mumkin. Eng optimal yechim kanalda e'lon qilinadi.

Funksiya ixtiyoriy uzunligi 2 ga bo'linadigan so'z qabul qiladi. Ushbu so'z teng ikkiga bo'linadi. Masalan book bo'lsa bo bilan ok shaklida. Keyin ushbu bo'lingan so'zlardan unli harflar sanaladi. Agarda ikkalasida ham unli harflar soni teng bo'lsa funksiya true qaytarsin aks holda false qaytarsin.
Harflar katta kichik bo'lishi mumkin. Uni ahamiyati yo'q

function halvesAreAlike(str){
let volwes=['a','i','e','o','u','A','I','E','O','U']
let a=str.slice(0,str.length/2)
let b=str.slice(str.length/2)
let count1=0, count2=0
for(let aelement of a){
if(volwes.includes(aelement)){
count1++
}
}

for(let belement of b){
if(volwes.includes(belement)){
count2++
}
}

return count1===count2

}
console.log(halvesAreAlike('AbCdEfGh'))

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