Question

The Python lookNsayNth(), given below, is a generator function that generates the digits (as characters) of the “n-th” “look-and-say number”. E.g., 1113… generates the characters ‘1’, ‘1’, ‘1’, ‘3’, etc. Use this generator to print out the 1,000,000-th to 1,000,019-th digits of the 200th “look-and-say” number. Check: your number will start with 211121… (Keep in mind that the 1st digit has index 0 if the digits were put in an array). Note that the 200th “look-and-say” number has somewhere around 10**23 digits. Extra credit for printing them all!

The python "look and say number"

def lookNsay():
        '''Generate the "look and say" sequence:
        1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ...
        Note that strings are returned
        '''
        s = '1'                         # s is the previous result, start with 1
        while True:
                yield s
                c,n, result = s[0], 1, ''
                for t in s[1:]:
                        if t==c:
                                n += 1
                        else:
                                result += str(n) + c
                                c,n = t, 1
                result += str(n) + c
                s = result


def saythis(prev): # Supporting function
        '''Generate characters of a "look-and-say" number given a generator of
        characters for the previous number.
        '''
        c,n = next(prev), 1
        for t in prev:
                if t==c:
                        n += 1
                else:
                        yield str(n)
                        yield c
                        c,n = t, 1
        yield str(n)
        yield c
        
        
def lookNsayNth(n = 1):
        '''Generate characters of the n-th value of the "look-and-say" sequence
        recursively
        '''
        if n == 1:
                yield '1'
        else:
                yield from saythis(lookNsayNth(n-1)) 
                

Answer 1

# Python 3 program to find
# n'th term in look and
# say sequence

# Returns n'th term in
# look-and-say sequence
def countnndSay(n):
  
   # Base cases
   if (n == 1):
       return "1"
   if (n == 2):
       return "11"

   # Find n'th term by generating
   # all terms from 3 to n-1.
   # Every term is generated using
   # previous term
  
   # Initialize previous term
   s = "11"
   for i in range(3, n + 1):
      
       # In below for loop,
       # previous character is
       # processed in current
       # iteration. That is why
       # a dummy character is
       # added to make sure that
       # loop runs one extra iteration.
       s += '$'
       l = len(s)

       cnt = 1 # Initialize count
               # of matching chars
       tmp = "" # Initialize i'th
               # term in series

       # Process previous term to
       # find the next term
       for j in range(1 , l):
          
           # If current character
           # does't match
           if (s[j] != s[j - 1]):
              
               # Append count of
               # str[j-1] to temp
               tmp += str(cnt + 0)

               # Append str[j-1]
               tmp += s[j - 1]

               # Reset count
               cnt = 1
          
           # If matches, then increment
           # count of matching characters
           else:
               cnt += 1

       # Update str
       s = tmp
   return s;

# Driver Code
N = 3
print(countnndSay(N))

# This code is contributed
# by ChitraNayal