Formatted question description: https://leetcode.ca/all/1545.html

# 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/

// 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';
}
};


Java

class Solution {
public char findKthBit(int n, int k) {
String str = "0";
for (int i = 2; i <= n; i++) {
String prevInvert = invert(str);
StringBuffer curr = new StringBuffer(str).append("1").append(new StringBuffer(prevInvert).reverse());
str = curr.toString();
}
return str.charAt(k - 1);
}

public String invert(String str) {
char[] array = str.toCharArray();
int length = array.length;
for (int i = 0; i < length; i++)
array[i] = (char) (1 - (array[i] - '0') + '0');
return new String(array);
}
}