Most Optimistic Integer Rotations in Golang

The problem

Write perform MaxRot(n) which given a optimistic integer n returns the utmost quantity you bought doing rotations just like the above instance.

So MaxRot is similar to:

• MaxRot(56789) ought to return 68957
• MaxRot(38458215) ought to return 85821534

Examples:

Take a quantity: 56789. Rotate left, you get 67895.

Preserve the primary digit in place and rotate left the opposite digits: 68957.

Preserve the primary two digits in place and rotate the opposite ones: 68579.

Preserve the primary three digits and rotate left the remaining: 68597. Now it’s over since holding the primary 4 it stays just one digit which rotated is itself.

You’ve got the next sequence of numbers:

56789 -> 67895 -> 68957 -> 68579 -> 68597

..and you could return the best: 68957.

The answer in Golang

Possibility 1:

bundle answer
import "strconv"
func MaxRot(n int64) int64 {
  str := strconv.FormatInt(n, 10)
  max := n
  for i := 0; i<len(str)-1 ; i++ {
    str = str[:i]+str[i+1:]+string(str[i])
    num, _:= strconv.ParseInt(str, 10, 64)
    if max < num { max = num }
  }
  return max
}

Possibility 2:

bundle answer
import . "strconv"
func MaxRot(n int64) int64 {
    s, max := FormatInt(n,10), n
    for i := 0; i < len(s); i++ {
        s = s[:i] + s[i:][1:] + s[i:][:1]
        v,_ := ParseInt(s,10,64)
        if max < v {
            max = v
        }
    }
    return max
}

Possibility 3:

bundle answer
import "strconv"
func MaxRot(rot int64) int64 {
  rotRunes := []rune(strconv.FormatInt(rot, 10))  
  for index := 0; index < len(rotRunes) - 1; index++ {
    rightPart := append(rotRunes[index + 1:], rotRunes[index])
    rotRunes = append(rotRunes[:index], rightPart...)
    rotVariation, err := strconv.ParseInt(string(rotRunes), 10, 64)
    if err == nil && rotVariation > rot {
      rot = rotVariation
    }
  }
  return rot
}

Take a look at circumstances to validate our answer

bundle solution_test
import (
  . "github.com/onsi/ginkgo"
  . "github.com/onsi/gomega" 
)
func dotest(n int64, exp int64) {
    var ans = MaxRot(n)
    Anticipate(ans).To(Equal(exp))
}
var _ = Describe("Checks MaxRot", func() {
    It("ought to deal with primary circumstances", func() {
        dotest(38458215, 85821534)
        dotest(195881031, 988103115)
        dotest(896219342, 962193428)
    })
})