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1545. Find Kth Bit in Nth Binary String

Given two positive integers n and k, the binary string Sn is formed as follows:

• S1 = "0"
• Si = Si-1 + "1" + reverse(invert(Si-1)) for i > 1

Where + denotes the concatenation operation, reverse(x) returns the reversed string x, and invert(x) inverts all the bits in x (0 changes to 1 and 1 changes to 0).

For example, the first 4 strings in the above sequence are:

• S1 = "0"
• S2 = "011"
• S3 = "0111001"
• S4 = "011100110110001"

Return the kth bit in Sn. It is guaranteed that k is valid for the given n.

Example 1:

Input: n = 3, k = 1
Output: "0"
Explanation: S3 is "0111001". The first bit is "0".

Example 2:

Input: n = 4, k = 11
Output: "1"
Explanation: S4 is "011100110110001". The 11th bit is "1".

Example 3:

Input: n = 1, k = 1
Output: "0"

Example 4:

Input: n = 2, k = 3
Output: "1"

Constraints:

• 1 <= n <= 20
• 1 <= k <= 2n - 1

Solutions (Rust)

1. Solution

impl Solution {
    pub fn find_kth_bit(n: i32, k: i32) -> char {
        let mut bits = vec![false];

        loop {
            if let Some(&b) = bits.get(k as usize - 1) {
                return char::from(b as u8 + b'0');
            }

            let mut x = bits.clone().iter().map(|&b| !b).rev().collect::<Vec<_>>();
            bits.push(true);
            bits.append(&mut x);
        }
    }
}