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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))
fori > 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'; } };