Given an integer n
, you must transform it into 0
using the following operations any number of times:
- Change the rightmost (
0th
) bit in the binary representation ofn
. - Change the
ith
bit in the binary representation ofn
if the(i-1)th
bit is set to1
and the(i-2)th
through0th
bits are set to0
.
Return the minimum number of operations to transform n
into 0
.
Example 1:
Input: n = 3 Output: 2 Explanation: The binary representation of 3 is "11". "11" -> "01" with the 2nd operation since the 0th bit is 1. "01" -> "00" with the 1st operation.
Example 2:
Input: n = 6 Output: 4 Explanation: The binary representation of 6 is "110". "110" -> "010" with the 2nd operation since the 1st bit is 1 and 0th through 0th bits are 0. "010" -> "011" with the 1st operation. "011" -> "001" with the 2nd operation since the 0th bit is 1. "001" -> "000" with the 1st operation.
Constraints:
0 <= n <= 109
class Solution:
def minimumOneBitOperations(self, n: int) -> int:
if n <= 1:
return n
for i in range(64):
if (n >> i) == 1:
base = 1 << i
break
return 2 * base - 1 - self.minimumOneBitOperations(n - base)
func minimumOneBitOperations(n int) int {
if n <= 1 {
return n
}
base := 0
for i := 0; i < 64; i++ {
if (n >> i) == 1 {
base = 1 << i
break
}
}
return (base << 1) - 1 - minimumOneBitOperations(n-base)
}