One way is to keep the unique elements in a separate array or set and skip arr.push([text[i], count]); when text[i] is in this array or set. Regards, suic

@Spindle: It's javascript.

What programming language?
It would also help to format your code.

Thanks)

• Implement it yourself
• Use libraries
• Find a solution which doesn't require flattening

You choose.

Can you tell alternate option, please)

flat is available in Node 11+

If you post your with proper markdown (remember to use a spoiler tag), then someone can help you.

I can't decide if I love it or hate it. Is it a cool puzzle? Yes. Does it belong here? Sure, why not. Do you have to "prepare" before writing your code? Yes, and I suppose that is not a bad thing to learn. Because if you don't, you're gonna be mad after this kata.

Thank you! I thought, that I needed to plus numbers over a hundred. I make life difficult for myself:D

• Sum all the digits of n.
• Subtract the sum from n, and it is your new n.
• If the new n is in the list below return the associated fruit, otherwise return back to task 1.

10 -> 1 + 0 = 1
10 - 1 = 9
9 -> apple

Somebody can me explain? Why in the task wrote for check /(SubtractSum(10), "apple");/ BUT in list 'fruits', 10 poit takes "10-pineapple"...

oh damn, so it is spelling mistakes. let me try that and I will update here lol thank you