The 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".
class Solution:
def countAndSay(self, n: int) -> str:
if n==1:
return "1"
prev = self.countAndSay(n-1)
res = ""
count = 1
for i in range(len(prev)):
if i == len(prev)-1 or prev[i] != prev[i+1]:
res+= str(count)+ prev[i]
count = 1
else:
count+=1
return res
TC: O(n) SC:O(n)
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