The challenge
Find the next higher number (int) with same ‘1’- Bits.
I.e. as much 1
bits as before and output next higher than input. Input is always an int in between 1 and 1«30 (inclusive). No bad cases or special tricks…
Examples:
Input: 129 => Output: 130 (10000001 => 10000010)
Input: 127 => Output: 191 (01111111 => 10111111)
Input: 1 => Output: 2 (01 => 10)
Input: 323423 => Output: 323439 (1001110111101011111 => 1001110111101101111)
The solution in Golang
Option 1:
package solution
func NextHigher(x int) int {
rightOne := x & -x
nextHigherOneBit := x + rightOne
rightOnesPattern := x ^ nextHigherOneBit
rightOnesPattern = (rightOnesPattern) / rightOne
rightOnesPattern >>= 2
next := nextHigherOneBit | rightOnesPattern
return next
}
Option 2:
package solution
import (
"strconv"
"strings"
)
func NextHigher(n int) int {
for i := n+1 ; i > 0 ; i ++ {
if strings.Count(strconv.FormatInt(int64(n), 2) , "1") == strings.Count(strconv.FormatInt(int64(i), 2) , "1") { return i } }
return 0
}
Option 3:
package solution
import (
"math/bits"
)
func NextHigher(n int) int {
bcount := bits.OnesCount32(uint32(n))
for x := uint32(n+1); ;x++ {
if bits.OnesCount32(x) == bcount {
return int(x)
}
}
return 0
}
Test cases to validate our solution
package solution_test
import (
. "github.com/onsi/ginkgo"
. "github.com/onsi/gomega"
)
var _ = Describe("Kata", func() {
It("Basic tests", func() {
Expect(NextHigher(128)).To(Equal(256))
Expect(NextHigher(1)).To(Equal(2))
Expect(NextHigher(1022)).To(Equal(1279))
Expect(NextHigher(127)).To(Equal(191))
Expect(NextHigher(1253343)).To(Equal(1253359))
})
})