count-and-say.py

"""

38. Count and Say

Medium

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 from countAndSay(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 substrings such that each substring contains exactly one unique digit. Then for each substring, say the number of digits, then say the digit. Finally, concatenate every said digit.

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

"""
"""
i.e. 

n = 1  ->  1        ( 1, when n = 1 )
n = 2  ->  11       ( there is 1 "1" when n = 2 )
n = 3  ->  21       ( there is 2 "1" when n = 3 )
n = 4  ->  1211     ( there is 1 "2" and 1 "1"  when n= 4 )
n = 5  ->  111221   ( there is 1 "1" and 1 "2" and  2 "1"  when n = 5)
n = 6  ->  312211   ( there is 3 "1" and 2 "2" and 1 "1" when n = 6 )

.
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.
"""

# V0
# IDEA : ITERATION
class Solution:
    def countAndSay(self, n: int) -> str:
        
        val = ""
        res = "1"
        
        for _ in range(n-1):
            cnt = 1
            for j in range(len(res)-1):
                if res[j]==res[j+1]:
                    cnt+=1
                else:
                    val += str(cnt) + res[j]
                    cnt = 1
            val += str(cnt)+res[-1]
            res = val
            val = ""
        return res

# V0 
# IDEA : RECURSION
# for example : 
# n = 4 --> s = self.countAndSay(3) 
#           while n = 3 --> s = self.countAndSay(2)
#                               while n = 2 --> s = self.countAndSay(1)
#                                       while n = 1 --> 1

# DEMO 
# s :  1
# res :  11
# s :  11
# res :  21
# s :  21
# res :  1211
# s :  1211
# res :  111221
# s :  111221
# res :  312211
# s :  312211
# res :  13112221
# s :  13112221
# res :  1113213211
# s :  1113213211
# res :  31131211131221
# s :  31131211131221
# res :  13211311123113112211
# 13211311123113112211
class Solution(object):
    def countAndSay(self, n):  # recursion 
        if n == 1: return '1'  # end loop of recursion 
        s = self.countAndSay(n-1)
        res, count = '', 0
        for i in range(len(s)):
            count += 1
            if i == len(s) - 1 or s[i] != s[i+1]:  # end before meet the last string element 
                res += str(count)
                res += s[i]
                count = 0
        return res

# V1
# IDEA : ITERATION
# https://leetcode.com/problems/count-and-say/discuss/962795/My-Python-Solution-40ms
class Solution:
    def countAndSay(self, n: int) -> str:
        
        val = ""
        res = "1"
        
        for _ in range(n-1):
            cnt = 1
            for j in range(len(res)-1):
                if res[j]==res[j+1]:
                    cnt+=1
                else:
                    val += str(cnt) + res[j]
                    cnt = 1
            val += str(cnt)+res[-1]
            res = val
            val = ""
        return res

# V1'
# https://blog.csdn.net/XX_123_1_RJ/article/details/80957046
# IDEA : RECURSION  
class Solution(object):
    def countAndSay(self, n):  # recursion 
        if n == 1: return '1'  # end loop of recursion 
        s = self.countAndSay(n-1)
        res, count = '', 0
        for i in range(len(s)):
            count += 1
            if i == len(s) - 1 or s[i] != s[i+1]:  # end before meet the last string element 
                res += str(count)
                res += s[i]
                count = 0
        return res

# V1' 
# https://blog.csdn.net/XX_123_1_RJ/article/details/80957046
# IDEA : ITERATION 
class Solution(object):
    def countAndSay1(self, n):  # iteration 
        if n == 1: return '1'
        res = '1'
        while n > 1:
            s, res, count = res, '', 0
            for i in range(len(s)):
                count += 1
                if i == len(s) - 1 or s[i] != s[i + 1]:  # end before meet the last string element 
                    res += str(count)
                    res += s[i]
                    count = 0
            n -= 1
        return res

# V1''
# https://www.jiuzhang.com/solution/count-and-say/#tag-highlight-lang-python
class Solution:
    # @return a string
    def count(self,s):
        t=''; count=0; curr='#'
        for i in s:
            if i!=curr:
                if curr!='#':
                    t+=str(count)+curr
                curr=i
                count=1
            else:
                count+=1
        t+=str(count)+curr
        return t
    def countAndSay(self, n):
        s='1'
        for i in range(2,n+1):
            s=self.count(s)
        return s
        
# V2 
class Solution:
    # @return a string
    def countAndSay(self, n):
        seq = "1"
        for i in range(n - 1):
            seq = self.getNext(seq)
        return seq

    def getNext(self, seq):
        i, next_seq = 0, ""
        while i < len(seq):
            cnt = 1
            while i < len(seq) - 1 and seq[i] == seq[i + 1]:
                cnt += 1
                i += 1
            next_seq += str(cnt) + seq[i]
            i += 1
        return next_seq