README_EN.md

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481. Magical String

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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: Simulate the Construction Process

According to the problem, we know that each group of numbers in the string $s$ can be obtained from the digits of the string $s$ itself.

The first two groups of numbers in string $s$ are $1$ and $22$, which are obtained from the first and second digits of string $s$, respectively. Moreover, the first group of numbers contains only $1$, the second group contains only $2$, the third group contains only $1$, and so on.

Since the first two groups of numbers are known, we initialize string $s$ as $122$, and then start constructing from the third group. The third group of numbers is obtained from the third digit of string $s$ (index $i=2$), so at this point, we point the pointer $i$ to the third digit $2$ of string $s$.

1 2 2
    ^
    i

The digit pointed by pointer $i$ is $2$, indicating that the third group of numbers will appear twice. Since the previous group of numbers is $2$, and the numbers alternate between groups, the third group of numbers is two $1$s, i.e., $11$. After construction, the pointer $i$ moves to the next position, pointing to the fourth digit $1$ of string $s$.

1 2 2 1 1
      ^
      i

At this point, the digit pointed by pointer $i$ is $1$, indicating that the fourth group of numbers will appear once. Since the previous group of numbers is $1$, and the numbers alternate between groups, the fourth group of numbers is one $2$, i.e., $2$. After construction, the pointer $i$ moves to the next position, pointing to the fifth digit $1$ of string $s$.

1 2 2 1 1 2
        ^
        i

Following this rule, we simulate the construction process sequentially until the length of string $s$ is greater than or equal to $n$.

The time complexity is $O(n)$, and the space complexity is $O(n)$.

Python3

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)

Java

class Solution {
    public int magicalString(int n) {
        List<Integer> s = new ArrayList<>(List.of(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++

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

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

function magicalString(n: number): number {
    const s: number[] = [1, 2, 2];
    for (let i = 2; s.length < n; ++i) {
        let pre = s[s.length - 1];
        let cur = 3 - pre;
        for (let j = 0; j < s[i]; ++j) {
            s.push(cur);
        }
    }
    return s.slice(0, n).filter(x => x === 1).length;
}

Rust

impl Solution {
    pub fn magical_string(n: i32) -> i32 {
        let mut s = vec![1, 2, 2];
        let mut i = 2;

        while s.len() < n as usize {
            let pre = s[s.len() - 1];
            let cur = 3 - pre;
            for _ in 0..s[i] {
                s.push(cur);
            }
            i += 1;
        }

        s.iter().take(n as usize).filter(|&&x| x == 1).count() as i32
    }
}