The challenge
he count-and-say sequence is the sequence of integers with the first five terms as following:
1. 1 2. 11 3. 21 4. 1211 5. 111221
1
is read off as "one 1"
or 11
.11
is read off as "two 1s"
or 21
.21
is read off as "one 2
, then one 1"
or 1211
.
Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.
Note: Each term of the sequence of integers will be represented as a string.
Example 1:
Input: 1 Output: "1" Explanation: This is the base case.
Example 2:
Input: 4 Output: "1211" Explanation: For n = 3 the term was "21" in which we have two groups "2" and "1", "2" can be read as "12" which means frequency = 1 and value = 2, the same way "1" is read as "11", so the answer is the concatenation of "12" and "11" which is "1211".
The solution with Python
def countAndSay(self, n: int) -> str: # helper function def countAndSayHelper(s): # return value result = [] # loop counter i = 0 # loop while less than while i < len(s): c = 1 # increment counters while i+1 < len(s) and s[i] == s[i+1]: i += 1 c += 1 # add string to the return list result.append(str(c)+s[i]) # increment i += 1 # return a string from the created list return ''.join(result) # set default return string value s = "1" # loop through input size for i in range(n-1): # use helper function s = countAndSayHelper(s) # return answer return s