Start a new Kumite
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Kumite (ko͞omiˌtā) is the practice of taking techniques learned from Kata and applying them through the act of freestyle sparring.

You can create a new kumite by providing some initial code and optionally some test cases. From there other warriors can spar with you, by enhancing, refactoring and translating your code. There is no limit to how many warriors you can spar with.

A great use for kumite is to begin an idea for a kata as one. You can collaborate with other code warriors until you have it right, then you can convert it to a kata.

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DTB4vs.jorelcb

Multiples of 3 or 5

Code
Diff
  • package kata

func Multiple3And5(number int) int {
	sum := 0
	for i := 3; i < number; i++ {
		if i%15 == 0 {
			sum += i
			continue
		}
		if i%3 == 0 {
			sum += i
		}
		if i%5 == 0 {
			sum += i
		}
	}
	return sum
}
    • package kata
    • func Multiple3And5(number int) int {
    • sum := 0
    • for i := 3; i < number; i++ {
    • if i%15 == 0 {
    • sum += i
    • continue
    • }
    • if i%3 == 0 {
    • sum += i
    • }
    • if i%5 == 0 {
    • sum += i
    • }
    • }
    • return sum
    • }
engelthehypvs.linzinha

Palindrome

No need to "branch to boolean". You can just return the condition.
You can also add " " to the list of characters to remove (which was just punctuation before) instaead of having a separate call to replace.

Code
Diff
  • import string

def is_palindrome(s: str) -> bool:
    forward = s.lower().translate(str.maketrans('', '', string.punctuation + ' '))
    return forward == forward[::-1]
    • import string
    • def is_palindrome(s: str) -> bool:
    • forward = s.lower().replace(" ", "").translate(str.maketrans('', '', string.punctuation))
    • reverse = forward[::-1]
    • if forward == reverse:
    • return True
    • else:
    • return False
    • forward = s.lower().translate(str.maketrans('', '', string.punctuation + ' '))
    • return forward == forward[::-1]
Vars0vievs.psychoandtech

double value

Code
Diff
  • int doubleValue(int x) {
  return x*2;
}
    • int doubleValue(int x) {
    • return x<<1;
    • }
    • return x*2;
    • }
YurichBROvs.0x0x0x0x

Pyramid of X's

Uglifyed it to it's limit. But I like it.

Code
Diff
  • pyramid_of_x=lambda n:['\n'.join(['*'*(i+1) for i in range(n)]),'Not enough building blocks!'][n<2]
    • def pyramid_of_x(n):
    • if n < 2:
    • return 'Not enough building blocks!'
    • rows = ['*' * i for i in range(1, n + 1)]
    • return '\n'.join(rows)
    • pyramid_of_x=lambda n:['\n'.join(['*'*(i+1) for i in range(n)]),'Not enough building blocks!'][n<2]
JonathanP4vs.poacosta87

Add numbers of an array

const addArr

Code
Diff
  • const addArr = (arr) => arr.reduce((acc, i) => acc + i, 0) || null
    • const addArr = (arr) => {
    • let sum = 0;
    • for (let i = 0; i < arr.length; i++) {
    • sum += arr[i];
    • }
    • return sum || null;
    • };
    • const addArr = (arr) => arr.reduce((acc, i) => acc + i, 0) || null
katonprayogivs.rishadharis

SQL Left Join 1 (BIE)

Code
Diff
  • select 
  city_name, 
  sum(confirmed_cases) confirmed_cases , 
  sum(recovered_cases) recovered_cases, 
  sum(death_cases) death_cases
from cases
  join dati
  on dati.code = cases.dati_code
group by city_name
order by confirmed_cases desc;
    • -- Code Here
    • select
    • city_name,
    • sum(confirmed_cases) confirmed_cases ,
    • sum(recovered_cases) recovered_cases,
    • sum(death_cases) death_cases
    • from cases
    • join dati
    • on dati.code = cases.dati_code
    • group by city_name
    • order by confirmed_cases desc;
user5006542vs.user5006542

Return 100 efixty

Code
Diff
  • def return_hundred():
    a = 99 + 1
    return a
    • def return_hundred():
    • kk = 100
    • return kk * 10000000000000000000000000000000000000000000000 // 10000000000000000000000000000000000000000000000
    • a = 99 + 1
    • return a
IamWongjosvs.rishadharis

Basic Use Case 2 LIMIT (BIE)

Code
Diff
  • -- Buatlah query sql untuk mendapatkan 10 pegawai dengan gaji terbesar.
select * from employees order by salary desc limit 10;
    • --- Code Here
    • -- Buatlah query sql untuk mendapatkan 10 pegawai dengan gaji terbesar.
    • select * from employees order by salary desc limit 10;
neetha88vs.rishadharis

Basic Ordering 3 Multiple Order (BIE)

Code
Diff
  • -- Code Here

select * 
from transactions
where customer is not null
order by store, total_price desc
    • --- Code Here
    • -- Code Here
    • select *
    • from transactions
    • where customer is not null
    • order by store, total_price desc
ahmdxrzkyvs.rishadharis

Basic Ordering 2 DESC (BIE)

Code
Diff
  • -- Code Here
SELECT *
FROM transactions
ORDER BY total_price DESC;
    • --- Code Here
    • -- Code Here
    • SELECT *
    • FROM transactions
    • ORDER BY total_price DESC;