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1545. Find Kth Bit in Nth Binary String (Medium)

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

Related Topics:
String

Solution 1. Recursion

The length of the string len is 2^n - 1.

If k - 1 == len / 2, then this is the middle of the string, return 1 unless n == 1.

If k - 1 < len / 2, this is at the left part of Sn, which is the same as findKthBit(n - 1, k).

If k - 1 > len / 2, this is the i = k - 1 - len / 2-th bit in the right part, which is the invert of findKthBit(n - 1, len / 2 - i + 1).

// OJ: https://leetcode.com/problems/find-kth-bit-in-nth-binary-string/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    char findKthBit(int n, int k) {
        if (n == 1) return '0';
        int len = pow(2, n) - 1;
        if (k - 1 == len / 2) return '1';
        if (k - 1 < len / 2) return findKthBit(n - 1, k);
        int i = k - 1 - len / 2;
        return findKthBit(n - 1, len / 2 - i + 1) == '0' ? '1' : '0';
    }
};