481. Magical String

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

Python Code
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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)

Java Code
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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;
    }
}

C++ Code
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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);
    }
};

Go Code
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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
}

TypeScript Code
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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);
}

Rust Code
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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
    }
}