Formatted question description: https://leetcode.ca/all/481.html
481. Magical String
Level
Medium
Description
A magical string S consists of only ‘1’ and ‘2’ and obeys the following rules:
The string S is magical because concatenating the number of contiguous occurrences of characters ‘1’ and ‘2’ generates the string S itself.
The first few elements of string S is the following: S = “1221121221221121122……”
If we group the consecutive ‘1’s and ‘2’s in S, it will be:
1 22 11 2 1 22 1 22 11 2 11 22 ……
and the occurrences of ‘1’s or ‘2’s in each group are:
1 2 2 1 1 2 1 2 2 1 2 2 ……
You can see that the occurrence sequence above is the S itself.
Given an integer N as input, return the number of ‘1’s in the first N number in the magical string S.
Note: N will not exceed 100,000.
Example 1:
Input: 6
Output: 3
Explanation: The first 6 elements of magical string S is “12211” and it contains three 1’s, so return 3.
Solution
Generate the string directly. Obviously, the first three characters are “122”. Read each character and generate a group using a different character than the character before, until the string’s length is at least n
. Loop over the first n
characters and count the number of 1’s.

class Solution { public int magicalString(int n) { StringBuffer sb = new StringBuffer(); int index = 0; int digit = 1; while (sb.length() < n) { if (index >= sb.length()) { for (int i = 0; i < digit; i++) sb.append(digit); } else { int curDigit = sb.charAt(index)  '0'; for (int i = 0; i < curDigit; i++) sb.append(digit); } digit = 3  digit; index++; } String str = sb.substring(0, n).toString(); int ones = 0; for (int i = 0; i < n; i++) { if (str.charAt(i) == '1') ones++; } return ones; } }

Todo

class Solution(object): def magicalString(self, n): """ :type n: int :rtype: int """ s = "122" p = 2 while len(s) < n: s += str((3  int(s[1]))) * int(s[p]) p += 1 return s[:n].count("1")