# LeetCode Find Kth Bit in Nth Binary String

5484. 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

class Solution {
private:
string InvertStr(string &s) {
string ans = "";
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '0')ans.push_back('1');
else ans.push_back('0');
}
return ans;
}
public:
char findKthBit(int n, int k) {
string s = "0";
for (int i = 2; i <= n; ++i) {
string inv_str = InvertStr(s);
reverse(inv_str.begin(), inv_str.end());
s = s + "1" + inv_str;
}
return s[k - 1];
}
};