Description
https://leetcode.com/problems/count-and-say/
The count-and-say sequence is a sequence of digit strings defined by the recursive formula:
countAndSay(1) = "1"
countAndSay(n)
is the way you would “say” the digit string fromcountAndSay(n-1)
, which is then converted into a different digit string.
To determine how you “say” a digit string, split it into the minimal number of groups so that each group is a contiguous section all of the same character. Then for each group, say the number of characters, then say the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.
For example, the saying and conversion for digit string "3322251"
:
Given a positive integer n
, return the nth
term of the count-and-say sequence.
Example 1:
Input: n = 1 Output: "1" Explanation: This is the base case.
Example 2:
Input: n = 4 Output: "1211" Explanation: countAndSay(1) = "1" countAndSay(2) = say "1" = one 1 = "11" countAndSay(3) = say "11" = two 1's = "21" countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"
Constraints:
1 <= n <= 30
Explanation
The base case is when n = 1. For other case, we call countAndSay(n – 1) to get the previous string and build the count and say string by grouping and counting the consecutive numbers.
Python Solution
class Solution:
def countAndSay(self, n: int) -> str:
if n == 1:
return "1"
else:
prev_sequence = self.countAndSay(n - 1)
result_sequence = ""
counter = {}
for i in range(0, len(prev_sequence)):
digit = prev_sequence[i]
if digit not in counter:
for key, value in counter.items():
result_sequence += str(value) + key
counter = {}
counter[digit] = 1
else:
counter[digit] += 1
for key, value in counter.items():
result_sequence += str(value) + key
return result_sequence
- Time complexity: O(N*M). M is the longest sequence length.
- Space complexity: O(N).
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