I want to create a function that returns true if a number has consecutive digits or not,
example:
My solution right now is to transform the number into an array and loop through the number one by one, if the next number is equal to the current number then we just return true. But this has a complexity time of O(n) and I was wondering if anyone can come with a better solution.
Thank you
There is an arguably better solution where you don't need to convert the number into a string or array of numbers/character. It works as follows:
curr to -1.while num > 0 and do the following:next_curr = num % 10if next_curr == curr: return truecurr = next_currnum = num / 10 (integer division)This is a one pass O(log n) time complexity algorithm where n is the input number. The space complexity is O(1)
Note that while your algorithm was also O(log n) time complexity, it did 2 passes, and had a space complexity of O(log n) too.
I haven't written JS for some time now, but here's a possible implementation of the above algorithm in JS:
function sameAdjacentDigits(num) {
// to deal with negative numbers and
// avoid potential problems when using Math.floor later
num = Math.abs(num)
let curr = -1
while (num > 0) {
const nextCurr = num % 10
if (nextCurr == curr) return true
curr = nextCurr
num = Math.floor(num / 10)
}
return false
}
Use some regex, and then check what was found via the matcher
numbers_match = /(00|11|22|33|44|55|66|77|88|99)/;
numbers_match.match("11")
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/match
Easiest way to execute this is by using regex. Not sure what would be effectiveness of algorithm, but solution could be
/(\d)\1/