LeetCode in Kotlin
762. Prime Number of Set Bits in Binary Representation
Easy
Given two integers left and right, return the count of numbers in the inclusive range [left, right] having a prime number of set bits in their binary representation.
Recall that the number of set bits an integer has is the number of 1’s present when written in binary.
- For example,
21written in binary is10101, which has3set bits.
Example 1:
Input: left = 6, right = 10
Output: 4
Explanation:
6 -> 110 (2 set bits, 2 is prime)
7 -> 111 (3 set bits, 3 is prime)
8 -> 1000 (1 set bit, 1 is not prime)
9 -> 1001 (2 set bits, 2 is prime)
10 -> 1010 (2 set bits, 2 is prime)
4 numbers have a prime number of set bits.
Example 2:
Input: left = 10, right = 15
Output: 5
Explanation:
10 -> 1010 (2 set bits, 2 is prime)
11 -> 1011 (3 set bits, 3 is prime)
12 -> 1100 (2 set bits, 2 is prime)
13 -> 1101 (3 set bits, 3 is prime)
14 -> 1110 (3 set bits, 3 is prime)
15 -> 1111 (4 set bits, 4 is not prime)
5 numbers have a prime number of set bits.
Constraints:
1 <= left <= right <= 1060 <= right - left <= 104
Solution
class Solution {
fun countPrimeSetBits(left: Int, right: Int): Int {
var count = 0
val notPrime = BooleanArray(33)
notPrime[0] = true
notPrime[1] = true
sieve(notPrime)
for (i in left..right) {
var num = i
var setBits = 0
while (num > 0) {
num = num and num - 1
setBits++
}
if (!notPrime[setBits]) {
count++
}
}
return count
}
private fun sieve(notPrime: BooleanArray) {
for (i in 2..32) {
if (!notPrime[i]) {
var j = 2 * i
while (j <= 32) {
notPrime[j] = true
j += i
}
}
}
}
}