Leetcode: 1545. Find Kth Bit in Nth Binary String

14 Nov 2020

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”)


Solution:

class Solution {
public:
    char findKthBit(int n, int k) {
        return n == 1            ? '0' :
               k == 1 << (n - 1) ? '1' :
               k < 1 << (n - 1)  ? findKthBit(n - 1, k) :
                                   reverse(findKthBit(n - 1, (1 << n) - k));
    }
    
    static char reverse(char c) { return c == '0' ? '1' : '0'; }
};