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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;
    }
}

  • 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
    }
}

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