LeetCode in Kotlin
2595. Number of Even and Odd Bits
Easy
You are given a positive integer n.
Let even denote the number of even indices in the binary representation of n (0-indexed) with value 1.
Let odd denote the number of odd indices in the binary representation of n (0-indexed) with value 1.
Return an integer array answer where answer = [even, odd].
Example 1:
Input: n = 17
Output: [2,0]
Explanation:
The binary representation of 17 is 10001.
It contains 1 on the 0th and 4th indices.
There are 2 even and 0 odd indices.
Example 2:
Input: n = 2
Output: [0,1]
Explanation:
The binary representation of 2 is 10.
It contains 1 on the 1st index.
There are 0 even and 1 odd indices.
Constraints:
1 <= n <= 1000
Solution
@Suppress("NAME_SHADOWING")
class Solution {
fun evenOddBit(n: Int): IntArray {
var n = n
var flag = 1
var even = 0
var odd = 0
while (n != 0) {
val bit = n and 1
if (bit == 1 && flag == 1) {
even++
}
if (bit == 1 && flag != 1) {
odd++
}
flag = Math.abs(flag - 1)
n = n ushr 1
}
return intArrayOf(even, odd)
}
}