LeetCode-in-Java

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 four strings in the above sequence are:

• S1 = "0"
• S2 = "0**1**1"
• S3 = "011**1**001"
• S4 = "0111001**1**0110001"

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 1^st bit is “0”.

Example 2:

Input: n = 4, k = 11

Output: “1”

Explanation: S₄ is “011100110110001”. The 11^th bit is “1”.

Constraints:

• 1 <= n <= 20
• 1 <= k <= 2n - 1

Solution

@SuppressWarnings("java:S1172")
public class Solution {
    public char findKthBit(int n, int k) {
        boolean flip = false;
        while (k != 1) {
            int base = floorTwo(k);
            if (base == k) {
                return flip ? '0' : '1';
            }
            flip = !flip;
            k = base - (k - base);
        }
        return flip ? '1' : '0';
    }

    private int floorTwo(int k) {
        while ((k & (k - 1)) > 0) {
            k &= k - 1;
        }
        return k;
    }
}

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