1667. Find Kth Bit In Nth Binary String¶

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1667. 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 = "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 1st bit is "0".

Example 2:

Input: n = 4, k = 11
Output: "1"
Explanation: S4 is "011100110110001".
The 11th bit is "1".

Constraints:

• 1 <= n <= 20
• 1 <= k <= 2n - 1

Solution¶

class Solution {
    public char findKthBit(int n, int k) {
        String dp[] = new String[n + 1];
        dp[1] = "0";
        if (n == k) if (k == 1) return '0';
        dp[2] = "011";
        for(int i = 3; i <= n; i++) {
            dp[i] = dp[i - 1] + "1" + reverse(invert(dp[i - 1]));
        }
        String ans = dp[n];
        return ans.charAt(k - 1);
    }

   private String invert(String s){
        int n = s.length();
        StringBuilder inv = new StringBuilder();
        for(int i = 0; i<n; i++){
            if(s.charAt(i) == '1'){
                inv.append('0');
            }else{
                inv.append('1');
            }
        }
        return inv.toString();
    }

    private String reverse(String s){
        StringBuilder rev = new StringBuilder(s);
        return rev.reverse().toString();
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here