Description
You are given a positive integer n.
Let even denote the number of even indices in the binary representation of n with value 1.
Let odd denote the number of odd indices in the binary representation of n with value 1.
Note that bits are indexed from right to left in the binary representation of a number.
Return the array [even, odd].
Example 1:
Input: n = 50
Output: [1,2]
Explanation:
The binary representation of 50 is 110010.
It contains 1 on indices 1, 4, and 5.
Example 2:
Input: n = 2
Output: [0,1]
Explanation:
The binary representation of 2 is 10.
It contains 1 only on index 1.
Constraints:
1 <= n <= 1000
Solution
Python3
class Solution:
def evenOddBit(self, n: int) -> List[int]:
res = [0, 0]
b = bin(n)[2:]
for i, x in enumerate(b[::-1]):
if x == "1":
if i % 2 == 0:
res[0] += 1
else:
res[1] += 1
return res