Formatted question description: https://leetcode.ca/all/481.html

# 481. Magical String

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")