LeetCode - count-and-say challenge solution

Tue Jul 07 2020
Links:  question  |  code

This challenge may be "easy" but it was quite challenging, for me at least.

My solution was first creating a function that will create the next string from an existing string.

So when it gets "1" => "11", "11" => "21", "1211" and so on.

It took me some time to understand how this syntax work wo I'll try to explainit here. For n=1, the string is "1", that's our base case. For next n+1, is the "string representative of n.

So in case of n=2, that is the string represantive of n=1. n=1, is "1". How we describe "1", its "count"+"value" so "11".

Now let's show n=3, that is the string represantive of n=2, that is "11". so "11" is "count"+"value" so "2"+"1" => "21".

Lets do 4 and 5.

n=4, string represantive of n=3, that is "21". So "21" => "count"+"value" so "1211".

n=5, string represantive of n=4, that is "1211". So "1211" => "count"+"value" so "111221".

Let me explain n=5 in words, n=5 is the string represantive of n=4. n=4 is "1211", in words is:

  • one time the digit one
  • one time the digit two
  • two times the digit one
  • let's convert the above to numbers:
  • 11
  • 12
  • 21
  • lets merge them to a single string:
  • 111221
  • and that is our result for n=5

Now after we got the idea of how to calculate the string, lets get to the solution of our exercise.

def countAndSay(self, n: int) -> str:
    val = "1"
    i = 1
    while i < n:
        val = self.calcNext(val)
        i += 1
    return val

As you can Im starting from the base and get my way up until I get to the requested "n" result. One each iteration I'm calculating the next "representation" until I reach to the desired n, stop and return it. This is pretty easy, but let me share the code for generating the "string representation".

def calcNext(self, n: str) -> str:
    count = 0
    prev = -1
    res = []
    for i in range(len(n)):
        cur = n[i]
        if cur == prev:
            count += 1
            if count:
            count = 1
            prev = cur
    if count:
    return ''.join(res)

This is a bit more complex, and I saw other solution that were much shorter. But I always prefer to have code that is a bit more readable than having one liner tricks that make it harder to understand.