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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';
    }
};

  • 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);
    }
}

############

class Solution {
    public char findKthBit(int n, int k) {
        if (k == 1 || n == 1) {
            return '0';
        }
        Set<Integer> set = new HashSet<>();
        int len = calcLength(n, set);
        if (set.contains(k)) {
            return '1';
        }
        // 中间，返回1
        if (k < len / 2) {
            return findKthBit(n - 1, k);
        } else {
            if (set.contains(len - k)) {
                return '1';
            }
            return r(findKthBit(n - 1, len - k + 1));
        }
    }

    private char r(char b) {
        if (b == '0') {
            return '1';
        }
        return '0';
    }

    private int calcLength(int n, Set<Integer> set) {
        if (n == 1) {
            return 1;
        }

        int ans = 2 * calcLength(n - 1, set) + 1;
        set.add(ans + 1);
        return ans;
    }
}

  • // 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';
    }
};

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