Question

481. Magical String

中文文档

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 s is 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 is s itself.

Given an integer n, return the number of 1's in the first n number in the magical string s.

 

Example 1:

Input: n = 6
Output: 3
Explanation: The first 6 elements of magical string s is "122112" and it contains three 1's, so return 3.

Example 2:

Input: n = 1
Output: 1

 

Constraints:

• 1 <= n <= 105

Solutions

Solution 1

class Solution:
  def magicalString(self, n: int) -> int:
    s = [1, 2, 2]
    i = 2
    while len(s) < n:
      pre = s[-1]
      cur = 3 - pre
      # cur 表示这一组的数字，s[i] 表示这一组数字出现的次数
      s += [cur] * s[i]
      i += 1
    return s[:n].count(1)

class Solution {
  public int magicalString(int n) {
    List<Integer> s = new ArrayList<>(Arrays.asList(1, 2, 2));
    for (int i = 2; s.size() < n; ++i) {
      int pre = s.get(s.size() - 1);
      int cur = 3 - pre;
      for (int j = 0; j < s.get(i); ++j) {
        s.add(cur);
      }
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      if (s.get(i) == 1) {
        ++ans;
      }
    }
    return ans;
  }
}

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
  }
}