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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;
}
}

• class Solution:
def magicalString(self, n: int) -> int:
s = [1, 2, 2]
i = 2
while len(s) < n:
pre = s[-1]
cur = 3 - pre
s += [cur] * s[i]
i += 1
return s[:n].count(1)

############

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")


• class Solution {
public:
int magicalString(int n) {
vector<int> s = {1, 2, 2};
for (int i = 2; s.size() < n; ++i) {
int pre = s.back();
int cur = 3 - pre;
for (int j = 0; j < s[i]; ++j) {
s.emplace_back(cur);
}
}
return count(s.begin(), s.begin() + n, 1);
}
};

• func magicalString(n int) (ans int) {
s := []int{1, 2, 2}
for i := 2; len(s) < n; i++ {
pre := s[len(s)-1]
cur := 3 - pre
for j := 0; j < s[i]; j++ {
s = append(s, cur)
}
}
for _, c := range s[:n] {
if c == 1 {
ans++
}
}
return
}

• function magicalString(n: number): number {
const cs = [...'1221121'];
let i = 5;
while (cs.length < n) {
const c = cs[cs.length - 1];
cs.push(c === '1' ? '2' : '1');
if (cs[i] !== '1') {
cs.push(c === '1' ? '2' : '1');
}
i++;
}
return cs.slice(0, n).reduce((r, c) => r + (c === '1' ? 1 : 0), 0);
}


• impl Solution {
pub fn magical_string(n: i32) -> i32 {
let n = n as usize;
let mut s = String::from("1221121");
let mut i = 5;
while s.len() < n {
let c = s.as_bytes()[s.len() - 1];
s.push(if c == b'1' { '2' } else { '1' });
if s.as_bytes()[i] != b'1' {
s.push(if c == b'1' { '2' } else { '1' });
}
i += 1;
}
s.as_bytes()[0..n].iter().filter(|&v| v == &b'1').count() as i32
}
}