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Cesar Del rio
Cesar Del rio

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#12 - Fibonacci digit sequence CodeWars Kata (6 kyu)


You are given three non negative integers a, b and n, and making an infinite sequence just like fibonacci sequence, use the following rules:

step 1: use ab as the initial sequence.
step 2: calculate the sum of the last two digits of the sequence, and append it to the end of sequence.
repeat step 2 until you have enough digits
Your task is to complete the function which returns the nth digit (0-based) of the sequence.

0 <= a, b <= 9, 0 <= n <= 10^10
16 fixed testcases
100 random testcases, testing for correctness of solution
100 random testcases, testing for performance of code
All inputs are valid.
Pay attention to code performance.


For a = 7, b = 8 and n = 9 the output should be 5, because the sequence is:
78 -> 7815 -> 78156 -> 7815611 -> 78156112 -> 781561123 -> 7815611235 -> ...
and the 9th digit of the sequence is 5.

For a = 0, b = 0 and n = 100000000 the output should be 0, because all the digits in this sequence are 0.

My solution:

function find(a,b,n){
  let r = a.toString() + b.toString()
  n = +n.toString().slice(-4);

  while (r.length <= n){
   let x = r.split('')
   r += (+x[x.length-1] + +x[x.length-2]).toString()

  return +r.charAt(n)
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First I started concatinating the 2 first numbers as a string, then I changed the n value, because if it was too high, the code performance would be awful and it tooka lot to returning the result, so I just used the last 4 numbers of the n variable because after some cycles, the result is the same.

let r = a.toString() + b.toString()
n = +n.toString().slice(-4);

After that I used a while loop that would keep iterating until the string "r" length is equal to "n", inside the loop I made a variable "x" that splitted the "r" string into an array and after that the "r" string will concatenate and be equal to the sum of the last 2 elements of the "x" array, for making this strings a number I just added the + operator before the values, and after summing them I just made them string again so for the next loop cycle it would be a string again.

r += (+x[x.length-1] + +x[x.length-2]).toString()

At the end I just returned the character at the "n" position of the "r" string, and converted it to a number because I used the + operator before returning the value.

return +r.charAt(n)

Comment how would you solve this kata and why? 👇🤔

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Discussion (2)

namhle profile image
Nam Hoang Le

It could work the string length is growing too much. Just keep the last two characters instead 👌🏻

cesar__dlr profile image
Cesar Del rio Author

Yeah, maybe I could just keep the last two characters and keep concatinating them to the main string, so I don't have to continue splitting the main string in every loop.

Good contribution 🙌