# Unique digits sequence in Java

## The challenge

Consider the following series:

`0,1,2,3,4,5,6,7,8,9,10,22,11,20,13,24...`

There is nothing special between numbers “ and `10`

.

Let’s start with the number `10`

and derive the sequence. `10`

has digits `1`

and `. The next possible number that does not have a `1` or a `

is `22`

. All other numbers between `10`

and `22`

have a `1`

or a “.

From `22`

, the next number that does not have a `2`

is `11`

. Note that `30`

is also a possibility because it is the next *higher* number that does not have a `2`

, but we must select the *lowest* number that fits and **is not already in the sequence**.

From `11`

, the next lowest number that does not have a `1`

is `20`

.

From `20`

, the next lowest number that does not have a `2`

or a “ is `13`

, then `24`

, then `15`

and so on.

Once a number appers in the series, it cannot appear again.

You will be given an index number and your task will return the element at that position. See test cases for more examples.

Note that the test range is `n <= 500`

.

## The solution in Java code

Option 1:

```
import java.util.ArrayList;
class Solution{
public static boolean uniqDigits(int a, int b){
String c = a + "";
String d = b + "";
for (int i = 0; i < d.length(); ++i)
if (c.indexOf(d.charAt(i)) != -1) return false;
return true;
}
public static int findNum(int n){
ArrayList<Integer> v = new ArrayList<>(); v.add(0);
int i = 1, cur = 0;
while (++i <= n+1){
int next = 0;
while (!uniqDigits(next, cur) || v.contains(next)) {
next++;
}
v.add(next);
cur = next;
}
return v.get(v.size()-1);
}
}
```

Option 2:

```
import java.util.*;
class Solution{
public static int findNum(int n) {
System.out.println(n);
ArrayList<Integer> arrayList = new ArrayList<>();
for (int i = 1; i < 1000; i++) {
arrayList.add(i);
}
ArrayList<Integer> finalArray = new ArrayList<>();
finalArray.add(0);
M1:
for (int i = 0; finalArray.size() < n+1;) {
String[] strings = Integer.toString(arrayList.get(i)).split("", 0);
for (String s : strings) {
if (Integer.toString(finalArray.get(finalArray.size()-1)).contains(s)) {
i++;
continue M1;
}
}
if (!finalArray.contains(arrayList.get(i))) {
finalArray.add(arrayList.get(i));
arrayList.remove(i);
i = 0;
}
}
Iterator<Integer> iterator = finalArray.iterator();
for (int i = 0; i < finalArray.size() - 1; i++) {
iterator.next();
}
return iterator.next();
}
}
```

Option 3:

```
import java.util.Comparator;
import java.util.Set;
import java.util.TreeSet;
import java.util.function.Predicate;
import java.util.regex.Pattern;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
class Solution{
public static int findNum(int n){
Set<String> set =
IntStream.rangeClosed(0, 1000)
.mapToObj(Integer::toString)
.collect(Collectors.toCollection(() -> new TreeSet<>(Comparator.comparing(Integer::parseInt))));
String res = "1";
for (int i = 0; i <=n; i++) {
Pattern pattern = Pattern.compile(String.format("[%s]", res));
res = set.stream().filter(Predicate.not(j -> pattern.matcher(j).find())).findFirst().get();
set.remove(res);
}
return Integer.parseInt(res);
}
}
```

## Test cases to validate our solution

```
import org.junit.Test;
import static org.junit.Assert.assertEquals;
import org.junit.runners.JUnit4;
public class SolutionTest{
@Test
public void basicTests(){
assertEquals(1,Solution.findNum(1));
assertEquals(5,Solution.findNum(5));
assertEquals(22,Solution.findNum(11));
assertEquals(103,Solution.findNum(100));
assertEquals(476,Solution.findNum(500));
}
}
```